### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# An Introduction To Proof Through Real Analysis MA 30100

Purdue

GPA 3.97

### View Full Document

## 38

## 0

## Popular in Course

## Popular in Mathematics (M)

This 106 page Class Notes was uploaded by Dorothea Bode on Saturday September 19, 2015. The Class Notes belongs to MA 30100 at Purdue University taught by Staff in Fall. Since its upload, it has received 38 views. For similar materials see /class/208120/ma-30100-purdue-university in Mathematics (M) at Purdue University.

## Reviews for An Introduction To Proof Through Real Analysis

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/19/15

NOTES ON DMODULES AND CONNECTIONS WITH HODGE THEORY DONU ARAPURA These notes are almost entirely expository and result from my attempt to learn this material The rst part summarizes Dmodule theory up to RiemannHilbert The second part discusses vanishing cycles in its various forms This is needed in the next part Which summarizes the basics of Morihiko Saito s theory of Hodge modulesi My main motivation for going through all this was to convince myself that Saito s methods and the more naive construction in A yield the same mixed Hodge structure on the cohomology of a geometric variation of Hodge structure The proof of this is given in part 4 Readers interested in just this part7 may nd the note A2 on the ArXiv77 more convenie ti 1 gave some informal talks on this material at KlAS in Seoul in 2005 and TIFR Mumbai in 2008 1 would them for giving me the opportunity to do so CONTENTS 1 D modules 2 111 Weyl Algebra 2 116 Dnmodules 3 1122 lnverse and direct image 5 1129 Differential operators on af ne varieties 7 1135 Nonaf ne varieties 8 1139 Connections 10 1143 RiemannHilbert correspondence 11 1147 Perverse Sheaves 12 1156 GaussManin connections 13 2 Vanishing cycles 15 211 Vanishing cycles 15 213 Perverse Sheaves on a disk 16 216 KashiwaraMalgrange ltration 17 3 Hodge modules 19 311 Hodge theory background 19 316 Filtered D modules 20 317 Hodge modules on a curve 21 3111 VHS on a punctured disk 22 3118 Hodge modules introduction 23 3124 Hodge modules conclusion 24 3126 Mixed Hodge modules 25 3130 Explicit construction 27 3132 Real Hodge modules 28 4 Comparison of Hodge structures 29 2 DONU ARAPURA References 30 1 D MODULES 11 Weyl Algebra Fix a positive integer n The nth Weyl algebra D7 over C is the ring of differential operators with complex polynomial coefficients in n variables More formally Dn can be de ned as the noncommutative Calgebra generated by symbols 11 zn 81 m H n Em subject to relations zirj zjzi 66 810i 9in 1119i ifi i Bin 1101 l The last two relations stem from the Leibnitz rule 1 8izjfzj8if These relations can be expressed more succinctly using commutators as zizj 8i8j 0 11397le 5o There is a sense in which Dn is almost commutative that I want to explain From the de ning relations it follows that any P 6 Dn can be expanded uniquely as P Z 111181 where I J E N z Iii etc The maximum value of J1 Jn occurring in this sum is the order of P We write Fan for the space of operators of order at most Is It is easy to see that Fka E F1617 Thus the associated graded GTD FkFk1 k inherits a graded algebra structure Lemma 12 Given P Q 6 D we have 0rderP lt 0rderP 0rderQ Sketch lt7s enough to check this when P Q are monomials ie expressions of the form I n this case it is a straight forward consequence of induction and the de ning relations Corollary 13 GrDn is commutative Slightly more work yields Theorem 14 GrDn is isomorphic to the polynomial ring 111 In 51 R27 I want to sketch a slightly nonstandard proof of this First quantize Dn to obtain a ring Hn which has an additional variable 4 subject to the relations that q commutes with z and Bj an Beryl 4 The remaining relations are the same as for Dn the 17s and 87s commute among themselves I will call Hn the Heisenberg algebra since it nothing but the universal enveloping algebra of the Heisenberg Lie algebra We see from the relations that NOTES ON DeMODULES AND CONNECTIONS WITH HODGE THEORY 3 Dn nqil and the classical limit77 is the polynomial ring Rgn where q 7 A is the two sided ideal generated by this element Now form the Rees algebra Rees thk C CM Dn with t a central element The theorem will follow from the next result which is straightforward Lemma 15 1 GrDn Reest 2 The map Rees A Hn sending z gt gt Ii taj gt gt Bj and t gt gt q is an isomor phism In more geometric terms we have an identi cation between Gr D7 and the ring of polynomial functions on the cotangent bundle T C 16 Dnmodules The notion of a Dnmodule gives an abstract way to think about systems of linear partial differential equations in nvariables Since the ring D7 is noncommutative we have to be careful about distinguishing between left modules and right Dnmodules I will often be lazy and refer to a left Dnmodule simply as a Dnmodule Here are some examp es Example 17 D7 is automatically both a left and right Dnmodule Example 18 Let Rn Czlzn be the polynomial ring This is a left Dn module where 1 acts by multiplication and E by 312 39 Example 19 Given operators P1 PN 6 D7 the left resp right ideal EDnPi resp EPiDn are left resp right Dnmodules Likewise for the quotients Dn EDnPi resp Dn E PiDn Note that Rn Dn EDnBi Example 110 Given a nonzero polynomial Rnf 1 Czlznf 1 is a Dnmodule where the derivatives act by di erentiation of rational functions Example 111 Let F be any space of complex valued functions on C which is an algebra over the polynomial ring and and closed under di erentiation then it becomes a left Dnmodule as above In particular this applies to holomorphic G00 and G00 functions with compact support Example 112 The space of distributions is the topological dual of the space of G00 functions with compact support or test functions This has a right module structure de ned a follows Given a distribution 6 a test function 45 and P 6 D7 let 6Pq5 6Pq5 The rst four examples above are nitely generated The last example requires some thought In the special case f 11 we see immediately that sz can be obtained by repeated differentiation Fix a space of functions F as in example 111 Given a left Dnmodule M de ne the space of solutions by SolM HarnDnMF To justify this terminology consider the example 19 above We see immediately that there is an exact sequence OHSoMM HFLP FN 4 DONU ARAPURA Therefore Sol is the space of solutions of the system 0 ere is a symmetry between right and left modules that I will refer to as ipping Lemma 113 There is an involution P gt gt P of D7 determined by 11 and a 781 Given a right Dnmodule M the operation P m mPquot makes M into a left module which I denote by FlipRHL This gives an equivalence between the categories of nitely generated left and right modules The inverse operation will denote by FlipLHR Suppose that M is a nitely generated Dnmodule We de ne good ltration on M to be a ltration FpM such that 1 The ltration FpM 0 for p lt 0 and UFpM M 2 Each FpM is a nitely generated Rnsubmodule 3 FpDn FqM Q FpqM Lemma 114 Every nitely generated module possess a good ltration Proof Write it as a quotient of some free module D5 and take the image of FpDnN The ltration is not unique however it does lead to some well de ned invariants Given a module with good ltration the associate grade GrM FpMFPAM is a nitely generated GrDnmodule The annihilator of GrM gives an ideal in GrDn E R2 The zero set of this ideal de nes a subvariety ChM F C C2 called the characteristic variety or singular support Since GrM is graded with respect to the natural grading on Gr D7 we see that the annihilator is homogeneous Therefore Lemma 115 ChM F is invariant under the action oft E C by 1135 gt gt 51 We can view this another way Consider the Rees module ReesM F EBFPM This is a nitely generated module over H7 such that GrM Hnq ReesM F So in some sense ChM F is the classical limit of M as q A 0 Theorem 116 ChM F is independent of the ltration Thus we can and will drop F from the notation Example 117 In the previous examples we see that 1 The annihilator of GrDn is 0 so that ChDn C2 2 Taking Rn DnEDnBi yields GrRn 111 ts annihilator is the ideal 51 Therefore ChRn C X 0 3 Let M R z l where we r 11 Then 11 1 generate M Let FkM Fk 1 Fk 14 emf This gives a good ltration A simple computation shows that GNM g Clam ElE EB ClryilI So ChM is a union of the axes NOTES ON DeMODULES AND CONNECTIONS WITH HODGE THEORY 5 Theorem 118 Bernsteinls inequality For any nonzero nitely generated Dn module we have dirn Ch 2 n There are a number of ways to prove this Perhaps the most conceptual though not the easiest way is to deduce it from the involutivity of the annihilator Ga This means that this ideal is closed under the Poisson bracket induced from the symplectic structure of C2 C This implies that the tangent space of any smooth point p E ChM satis es Tpi Q T1 and the inequality follows Note that the KYaction of lemma 115 is precisely the natural action on the bers of the cotangent bundle We say that nitely generated Dnrnodule M is holonomic if dirn ChM n or if M 0 Thanks to Bernstein s inequality this is equivalent to dim ChM S n For example Rn and R z l are holonomic but D7 isnlt Proposition 119 The class of holonomic modules is closed under submodules quotients and extensions Therefore the full subcategory of holonomic modules is Abelian Proof One checks that ChM2 ChM1 U ChM3 for any exact sequence 0 A M1 A 2 A 3 A 0 D From the symplectic viewpoint holonomic modules are precisely the ones with Lagrangian characteristic varieties There is also a homological characterization of such modules Theorem 120 A nitely generated Dn module is holonomic if and only if Exti M Dn 0 fori n fM is holonomic then the module Ext M Dn is a nitely generated holonomic right Dnmodule The contravariant functor M gt gt FlipRHLEzt M Dn is an involution on the category of holonomic modules Corollary 121 Holonomic modules are artinian which means the descending chain condition holds Sketch Any descending chain in M gets ipped around to an ascending chain in N Ezt MDn Dn is known to be right and left noetherian so the same goes for N D It will follow that holonomic modules can built up from simple holonomic mod ules 122 Inverse and direct image Suppose that X C with coordinates I and Y Cm with coordinates yj Consider a map F X A Y given by Fzl In F1zl In where the E are polynornials Let OX 111 In and Oy Cy1 yn and let DX and Dy denote the corresponding Weyl algebras To avoid confusion I will label the derivatives 8 etc Then F determines an algebra homomorphism Oy A OX f gt gt f o F Given a left Dyrnodule M we de ne a left DX module F M called the inverse image as follows First de ne FMOx 0YM 6 DONU ARAPURA as an OXmodule We now de ne an action of the derivatives by the chain rule 8f 6F Ii mfaxi 19me 8xf m 6 Lemma 123 This formula determines a DXmodule structure on F M Example 124 Let X C with coordinates 11 In Y C 1 with coordi nates 1117111 Let pzl In 11 1711 Then p M zn c M Given m In and 6 acts in the usual way through the rst factor and remaining generators of DX act through the second There is a second description that is useful De ne DXHY FlDY OX OY DY This has the structure of a left DXmodule as above as well as a right Dymodule structure where Dy acts by right multiplication on itself in the above formula These two actions commute so they determine a so called bimodule structure If we ip both of these actions we get left Dy right DX bimodule DYHX FliPhXRFliP LDXHY Lemma 125 fF M DXny DY M Proof F M OX OyM OX OY Dy DY M DXny DY M Given a left DXmodule N the direct image FiN DYFX DX N is a left Dymodule This operation is sometimes denoted with an integral sign to suggest the analogy with integration along the bers Example 126 Let X C with coordinates 11 In Y Cn1 with coordi nates 11zn11 and suppose izl 11zn0 We have iiM MN a countable direct sum Here 11 8112 6 acts as componentwise using the given DXmodule structure 1711 acts by 0 8711 acts as the shift operator m1m2 gt gt 0m1m2 Thus it is more suggestive to write 24M 69 aiz1M 1 These operations are compatible with composition as one would hope Theorem 127 If F X A Y and G Y A Z are polynomial maps of a ne spaces then for any Dzmodule M and DXmodule M we have 1 G o FM E F G M 2 G o FN E GiFiN Theorem 128 These operations preserve nite generation and holonomicity NOTES ON DeMODULES AND CONNECTIONS WITH HODGE THEORY 7 Since any map can be factored as an embedding followed by a projection7 it suffices by the theorem 127 to check these two cases This theorem provides many additional examples of holonomic modules 129 Differential operators on af ne varieties Let X be a nonsingular af ne variety over C This is a complex manifold which can be described as the set of solutions to a system of polynomial equations in some C We write RX for ring regular polynomial functions on X This is a finitely generated commutative algebra A differential operator of order S h on X is a Clinear endomorphism T of RX such that Tf0f1 0 for all E RX Let DiffkX denote the space of these operators We de ne DX UDil fk X Lemma 130 DX becomes a ring under composition such that Di kDi m C Dz l qn39 We note the following characterization cf K27 lemma 17 which sometimes use u Proposition 131 DX is a quotient of the univeral enveloping algebra of the Lie algebra of vector elds DerCRX by the relations 5 for all 5 E DercRX and f E RX When X C DX7Diff DmFJ Everything that we have done so far generalizes to the setting of affine varieties For example Theorem 132 The associated graded with respect to Di is isomorphic to the ring of regular functions on the cotangent bundle T X We can de ne leftright Dmodules as before All of the previous examples generalize We give a new example Example 133 Let f E Rn and let X be complement of the zero set of f in C This is an a ne variety with coordinate ring R Let A EAldzi be an r X r matrix of lforms with coe cients in R satisfying the integrability condition AlAj 0 Then M RT carries a left DXmodule structure with 7 8v 7 811 Note that this construction is equivalent to de ning an integrable connection on M There are nontrivial examples only when X is nonsimply connected and in particular none unless f f l 1 v Aiv vEM The flipping operation for affine varieties is more subtle than before Let Lox denote the canonical module or equivalently the module of algebraic nforms7 where n dim X This has right DX module structure dual to left module structure on RX Heuristically7 this can be undertood by the equation XaPf X aPf 8 DONU ARAPURA where a is an nform P a differential operator and f a function and X is replaced by a compact manifold A rigorous de nition can be given via the Lie derivative L Given a vector eld 5 and an element of a E wX LSWEM Min 5 517m5n Zw51wl575ilv 5n Then w E 7L a can be shown to extend to a right action of the whole ring DX with the help of proposition 131 Under this action the difference aPf 7 aPf can be shown to be exact and so above integral formula would follow Lemma 134 fM is a left DXmodule then FlipLHR AUX RXM carries a natural right D X module structure This operation induces an equivalence between the categories of left and right DX modules39 its inverse is FlipRHL N wgl N Note that Luca Rn which was why we could ignore it The notions of characteristic variety and holonomocity can be de ned as before The characteristic variety of example 133 is X embedded in T X as the zero section Therefore it is holonomic Given a morphism of af ne varieties F X A Y we can de ne bimodules DXny DYFX and inverse and direct images as before 135 Nonaf ne varieties Now we want to generalize to the case where X is a nonsingular nonaf ne variety for example projective space P First recall that in its modern formulation a variety consists of a space X with a Zariski topology and a sheaf of commutative rings OX such that for any open set OX U is the space of regular functions Ha By de nition X possesses an open covering by af ne varieties Our rst task is to extend DX to this world Lemma 136 There exists a unique sheaf of noncommutative rings DX on X such that for any a ne open U DX U is the ring of di ferential operators on U We can de ne a ltration by subsheaves FpDX C DX as above The previous result globalizes easily to Theorem 137 The associated graded is isomorphic to quuX where 7r T X A X is the cotangent bundle A left or right DXmodule is sheaf of left or right modules over DX For example OX resp Lox has a natural left resp right DXmodule structure DX has both We have an analogue of lemma 134 in this setting so we can always switch from right to left We will be primarily interested in the modules which are coherent ie locally nitely generated over DX The notion of good ltration for a DXmodule M can be extended to this setting The associated graded GrM de nes a sheaf over the cotangent bundle and the characteristic variety ChM is its support This depends only on M as before and is Cinvariant We have Bernstein s inequality dim ChM 2 dimX and M is holonomic if equality holds We again have Proposition 138 The full subcategory of holomonic modules is an artinian Abelian category Given a morphism of varieties F X A Y we de ne DXHY OX F710Y FilDY NOTES ON DeMODULES AND CONNECTIONS WITH HODGE THEORY 9 where F 1 is the inverse image in the category of sheaves Ha This is a left DX right F lDy bimodule We de ne a right DX left F lDy bimodule by ipping both actions DYHX FlipgRFzz39p iingxay As an OXmodule it this is isomorphic to AUX DXHY F wgl Given a left Dymodule M we can de ne the naive1 inverse image as the DX module FM DXHY F71DY FilM For a left DXmodule N the naive direct image as the Dymodule FfN FDYEX DX N where F on the right is the sheaf theoretic direct image The above de nitions proceed in complete analogy with the af ne case The bad news is that the naive direct image is somewhat pathological For example the composition rule theorem 127 may fail The solution is to work in the setting of derived categories Let DbDX denote the bounded derived category of quasi coherent left DXmodules The objects of this category are bounded ie nite complexes of left DXmodules Two objects are isomorphic in DbDX if and only if they are quasiisomorphic in the usual sense ie possess isomorphic cohomology sheaves More formally DbDX is constructed in two steps rst one passes to the homotopy category KbDX where morphisms are homotopy classes of maps between complexes then the quasiisomorphisms are inverted by a procedure analo gous to localization in commutative algebra The details can be found for example in B0 GM The categories KbDX and DbDX are not abelian so that exact sequences are not meaningful inside them However these categories are trian gulated which means that they are equipped with a collection of diagrams called distinguished triangles and this provides a reasonable substitute The functors Fl 8 extend to derived functors RF between these derived categories In practice this involves replacing a given complex by an appropriate injective at resolution before applying the functor Taking cohomology sheaves yields functors Hi DbDX A MDX to the category of DXmodules We can de ne the inverse image F DbDy A DbDX by FM DXxy lt3ka F 1M and the direct image F DbDX A DbDy by FN RFDYEX 9ng N39 These behave well under composition At the end of the day we can compose these operations with Hi to get actual Dmodules 1This is nonstandard terminology 10 DONU ARAPURA 139 Connections Let E be a vector bundle on a nonsingular variety X ie a locally free OXmodule A DXmodule structure on E is the same thing as an integrable connection on E which is given locally as in example 133 Globally this is a Clinear map from the tangent sheaf V Z TX A EndE such that Vv is a derivation and such that V preserves Lie bracket Vltl 17 2l lV 17V52l Equivalently it is given by a Clinear map v z E a 9 E satisfying the Leibnitz rule and the having curvature V2 0 From the local description it is easy to see that the characteristic variety of an integrable connection is the zero section of T X Thus it is holonomic Conversely Proposition 140 M is a vector bundle with integrable connection and only its characteristic variety is the zero section of T X Corollary 141 fM is a holonomic module there exists an open dense set U Q X such that MlU is an integrable connection Proof We can assume that the support of M is X otherwise the statement is trivially true Then the map ChM A X is generically nite and therefore nite over some open U C X Since ChM T U is C invariant it must be the zero section B Given a morphism F X A Y and an integrable connection EV on Y The pullback of the associated Dymodule coincides with the pullback F E in the category of Omodules with its induced connection If E V is an integrable connection on X then the pushforward of the associated DXmodule does not come from a connection in general However there is one important case where it does see section 156 We nally discuss the notion of regular singularities which is a growth condition at in nity The classical condition is the following Example 142 Let A be an r X r matrix of rational lforms on lP l Let U be the complement of the poles in P1 of the entries of A and letj U A P1 the inclusion Then we can de ne a DUmodule structure on M 0 by dv dz M is holomonic The DXmodule jiM has regular singularities if and the dif ferential equation 81 0 has regular singularities in the classical sense this is the case the poles of A are simple 8v Av In general we have the following extension due to Deligne A vector bundle EV with a connection on a smooth variety X has regular singularities if there exists a nonsingular compacti cation X with D X 7 X a divisor with normal crossings such that EV extends to a vector bundle with a map V E A Qk ogD E NOTES ON DeMODULES AND CONNECTIONS WITH HODGE THEORY 11 The notion of regular singularities can be extended to arbitrary holonomic DX modulesi If M is a simple holonomic module with support Z then Mlz is gener ically given by an integrable connection as above Say that M has regular singu larities if this connection is regular In general M has regular singularities if each of its simple subquotients have regular singularities This notion behaves well with respect to the operations de ned earlieri See Be Bo K2 for details 143 RiemannHilbert correspondence In the 19th century Riemann com pletely analyzed the hypergeometric equation in terms of its monodromyi Hilbert in his 21st problem proposed that a similar analysis should be carried out for more general differential equations Here I want to explain a very nice interpretation and solution in Dmodule language due to Kashiwara Kawai and Me k out Fix a smooth variety X over C We can treat X as a complex manifold and we denote this by Xani Most algebraic objects give rise to corresponding analytic ones usually marked by an In particular DXanmodule is the sheaf of holomorphic differential operators Any DXmodule gives rise to a DXanmodulei Let 9 denote the sheaf of holomorphic p forms on Xani Recall that we have a complex 9 called the de Rham complex which is quasiisomorphic to the constant sheaf Cxani We can modify this to allow coef cients in any DXanmodule DRM 9 oxan Mdim X The symbol mean shift the complex n places to the left The differential is given in local coordinates by M1 A AdmP m Zdz Adzil and eajm 139 We can de ne a complex Hli 0X AQTX A DX 0X TX 4 DX with differentials dual to DRDX under the identi cation HOmrighterimodHOmQ 0X DXyDX E DX OX APTX The complex DXan 0Xan A39Txan gives a locally free resolution of OXani This comes down to the fact that it becomes a Koszul complex after taking the associated graded with respect to Fr Therefore DRM HOmDXan 0Xan AoTXan M RHOVnOXan M We can extend the de nition of DR to the derived category Db DX by using the last formulai We can now give classical version of the RiemannHilbert correspondence Proposition 144 fE is a holomorphic vector bundle with an integrable connec tion V DRE7 dim X is a locally constant sheaf of nite dimensional Cvector spaces The functor E gt gt DRE7 dimX induces an equivalence of categories between these categories Sketch DRE7 dimX E 2 mg E X i H gives a resolution of herV which is locally constanti Conversely given a locally constant sheaf L Oxan L can be equipped with an integrable connection such L is the kernel 12 DONU ARAPURA y imposing regularity assumptions7 Deligne was able to make this correspon dence algebraic DeQ The point is that regularity ensures that the holomorphic data extends to a compacti cation7 where GAGA applies For general Dmodules7 we impose holonomicity as well DRM will no longer be a locally constant sheaf in general7 but rather a complex with constructible cohomology Recall that a CXanmodule L is constructible if there exists an algebraic strati cation of X such that the restrictions L to the strata are locally constant with nite dimensional stalks Theorem 145 The de Rham functor DR induces an equivalence of categories between the subcategory DrhDX C DbDX of complexes with regular holonomic cohomology and the subcategory Dams CXW C DbCXan of complexes with con structible cohomolgy Moreover the inverse and direct images constructions are compatible under this correspondence here is one more aspect of this7 which is worth noting The duality M gt gt FlipRHLEzt M7 Dn constructed earlier can be generalized naturally to this set ting to M H FlipRHLRHomM DX This corresponds to the Verdier dual F gt gt DF RHamFCXan 2 dimX on the constructible derived category This operation arises in the statement Poincar Verdier duality Theorem 146 IfL C Di omm cxm then HiX7 L HiXDL When X is nonsingular and L C we have DL 2 CB dim X7 so this reduces to ordinary Poincare duality 147 Perverse Sheaves Let X be nonsingular The category of of regular holo nomic modules sits in the triangulated category Db DX as an Abelian subcategory lts image under DR is the Abelian category of complex perverse sheaves BED In spite of the name7 these objects are neither perverse nor sheaves7 but rather a class of well behaved elements of Db Cxan Example 148 DROX CXdimX is perverse Example 149 Suppose that X is complete eg projective Suppose that V is a vector bundle with an integrable connection V V A 9 logD V with logarithmic singularities along a normal crossing divisor Let U X 7 an j U A X be the inclusion Then L keerrV is a local system on U Then we have warm Randim X is perverse Perverse sheaves can be characterized by purely sheaf theoretic methods Theorem 150 F E Dionmacxa is perverse and only 1 For allj dimsupijF S ij 2 These inequalities also hold for the Verdier dual DF Note that these conditions work perfectly well with other coefficients7 such as Q to de ne full subcategories PervQXan C DwmtrQXan NOTES ON DeMODULES AND CONNECTIONS WITH HODGE THEORY 13 Example 151 Ifj U A X is as in the previous example then for any local system of Q vector spaces L the sheaves RjLdimX and jLdimX are per verse The rst condition can be checked directly For the second observse that DRjLdimX ijVdimX and likewise of the dual ofjLdimX Perverse sheaves have another source7 independent of Dmodules In the late 70s Goresky and Macpherson introduced intersection homology by a geometric construction by placing restrictions how chains met the singular set in terms of a function refered to as the perversity Their motivation was to nd a theory which behaved like ordinary homology for nonsingular spaces in general for example7 by satisfying Poincare duality When their constructions were recast in sheaf theoretic language Br7 GoM7 they provided basic examples of perverse sheaves Example 152 Suppose that Z C X is a possible singular subvariety Then the complex CZ computing the rational intersection cohomology of Z is after a suitable shift and extension to X aperverse sheaf on X his is more generally true for the complex CZ L computing interesection cohomology of Z with coe cients in a locally constant sheaf de ned on a Zariski open U C Z In the notation of BED this would be denoted by iijLdim Z wherej U A Z andi Z A X are the inclusions It turns out that CZ is self dual under Verdier duality7 and in general that DUCZ 2 OZ This implies Poincare duality for intersection cohomology In typical cases2 of example 1497 if L keerrV then RjiL is quasiisomorphic to the log complex 9X log D E V V Then ICX L can be realized by an explicit subcomplex of the log complex It can also be realized7 in many cases7 by a complex of C00 forms on X 7 D with L2 growth conditions Theorem 153 The category ofperverse sheaves over QC is Artinian and the simple objects are as in example 152 with Z and L irreducible Corollary 154 Simple perverse sheaves are generically local systems on their support The last results shouldn7t come as a surprise7 since as we have seen previously7 regular holomonomic Dmodules are generically given by integrable connections Example 155 When X is a smooth curve this can be made very explicit Simple perverse sheaves are either sky scraper sheaves Qgc with x E X or sheaves of the form jL1 where L is an irreducible local system on a Zariski open setj U A X There is a functor pHi DEDMWGQX A PervQXan7 which corresponds to the operation Hi D hDX A MrhDX under RiemannHilbert 156 GaussManin connections Suppose that f z X A Y is a smooth and proper morphism of relative dimension n Let SigY be the sheaf of relative differ entials Then for any DXmodule we have a relative de Rham complex DRXyM 93 M DRXyDX gives a resolution of Dng7 with augmentation SEAEDX gwx filWY DX A DYHX h recise meaning of typical here is that the residues of the connection should have no positive integer eigenvalues 14 DONU ARAPURA Then the direct image 1 RfDYltX le M RfDRXYDX le M RfDRXYM When M OX the ith cohomology sheaf of the direct image Ri DRXHOX Pilingy is locally free With an integrable connection called the Gauss Manin connectioni Under RiemannHilbert GaussManin corresponds to the locally constant sheaf Ri c This connection was constructed in preDmodule language as follows Assume dimY l for simplicity then the connection is the connecting map Rif SrY A Qi RifQSY associated to the sequence 0 A 9 QgYkl A Q A QgY A 0 When f is not smooth then by resolution of singularities the singular bers E of f can be assumed to have normal crossingsi Then the above discussion can be extended to the log complexes resulting in a map Rif SHElog E A Q logfE Rif S y ogE This gives a proof due to Katz that the GaussManin connection has regular singularitiesi NOTES ON D7MODULES AND CONNECTIONS WITH HODGE THEORY 15 2 VANISHING CYCLES 21 Vanishing cycles Vanishing cycle sheaves and their corresponding Dmodules form the basis for Saito7s constructions described later I will start with the classical picture Suppose that f z X 7gt C is a morphism from a nonsingu lar variety The ber X0 f 10 may be singular but the nearby bers X1 0 lt M lt 6 ltlt 1 are not The premiage of the edisk f lDe retracts onto X0 and f 1D5 7 7 D5 7 0 is a ber bundle Thus we have a monodromy action by the counterclockwise generator T E 7r1Ct on HiXtl From now on I will tend to treat algebraic varieties as an analytic spaces and will no longer be conscientious about making a distinction The image of the restriction map HiX0 Hif 1D 7gt HiXt lies in the kernel of T 7 1 The restriction is dual to the map in homology which is induced by the nonholomorphic collapsing map of X onto X0 the cycles which die in the process are the vanishing cyc es Let me reformulate things in a more abstract way following SGA7 The nearby cycle functor applied to F E DbX is R IJF i RppF where 3 is the universal cover of C C 7 0 and p z 3 gtltC X 7gt X i X0 f 10 7gt X are the natural maps The vanishing cycle functor R i F is the mapping cone of the adjunction morphism i F 7gt R IJF and hence it ts into a distinguished triangle can i F a IMF A RltIgtF a rim Both R IJF and R i F are somewhat loosely refered to as sheaves of vanishing cycles These objects possess natural monodromy actions by T If we give i F the trivial T action then the diagram with solid arrows commutesl i F 7gt EMF gt R F gt iF1 l T71 W i 7 v 0 R F gtR F gtO Thus we can deduce a morphism var which completes this to a morphism of tri anglesl In particular T 71 var a can One can also show that can 0 var T 71 Given p 6 X0 let B5 be an e ball in X centered at p Then f 1t Be is the so called Milnor ber The stalks HiR 1 Qp Hif 1t 73542 HiltRltIgtQgtp 7 Hif 1t n BBQ give the reduced cohomology of the Milnor berl And HiX07R 1 Q Hif 1t7Q is as the terminology suggests the cohomology of the nearby berl We have a long exact sequence HlHiX0Q 7gt HiXtQ HiX0lR Q 7gt The following is a key ingredient in the whole story BED 16 DONU ARAPURA Theorem 22 Gabber IfL is perverse then so are R IJLEl and R i LEl We set 1 1va piJL R IJLFH and 1 de quL R39i39Lklll 23 Perverse Sheaves on a disk Let D be a disk with the standard coordinate function t7 and inclusion j z D 7 0 Dquot A Dr For simplicity assume 1 E D Consider a perverse sheaf F on D Which is locally constant on D Then we can form the diagram can 171th W p LF Note that the objects in the diagram are perverse sheaves on 0 ie vector spaces This leads to the following elementary description of the category due to Deligne and Verdier cifl V7 sect Proposition 24 The category of perverse sheaves on the disk D which are locally constant on Dquot is equivalent to the category of quivers of the form c 7 lt7 lt15 ie nite dimensional vector spaces 4571 with maps as indicated To get a sense of Why this is true7 let us explain how to construct perverse sheaves associated to certain quivers We see immediately that 0 lt V corresponds to the sky scraper sheaf V0 Let L be a local system L on D With monodromy given by T L1 A Lll Then the perverse sheaf jLl corresponds to c gt L L1 lt71 main Where c is the projection7 v is induced by T 7 L Thus a quiver c 7 lt7 lt15 With c surjective arises from ijH7 Where L1 1 With T 1 v 0 c it is easy to classify the simple quivers and see that they are covered by these cases There are three types P1 0 lt C Which corresponds to Col P2 C lt7 0 Which corresponds to CD 1333 C lt 1 C With A f 0 This corresponds to jiLH7 Where L is a rank one local system With monodromy A f l The examples considered above are not just perverse but in fact intersection cohomology complexes in general7 they can be characterized by Lemma 25 A quiver c 7 lt7 lt15 corresponds to a direct sum of intersection cohomology complexes and only if 45 imagec EB hervi NOTES ON DeMODULES AND CONNECTIONS WITH HODGE THEORY 17 26 KashiwaraMalgrange ltration Some of the original motivation for the ideas in this section came from the study of BernsteinSato polynomials or 12 functions However this connection would take us took far a eld lnstead we start with the question Under RiemannHilbert what is the Dmodule analogue of vanishing cycles I want to start with a prototype which should make the remainder easier to swallow Example 27 Let Y C with coordinate t Fix a rational number r 6 710 and let M Odt lltT with an acting on the left in the usual way For each a E Q de ne VDLM C M to be the Cspan of tnT l n E Zn r 2 7a The following properties are easy to check 1 The ltration is exhaustive and left continuous UVD M M and Va M VaMfor0lt6ltltl 2 Each VDLM is stable under tiag 2 j 3 all04M Q Va1M and tVaM Q VDklM 4 The associated graded era a6rZ 0 otherwise G ManHM is an eigenspace of tan with eigenvalue 7a 4 implies that the set of indices where VDLM jumps is r Z and hence discrete Such a ltration is called discrete Let f z X A C be a holomorphic function and let i z X A X X C Y be the inclusion of the graph Let t be the coordinate on C and let VaDy DXX0module generated by l iij 2 7a for a E In particular t E VL1Dy and E E VlDy Note that VoDy C Dy is the subring preserving the ideal Let be a regular holonomic DXmodule It is called called quasiunipotent along X0 f 10 if pwf DRM is quasiunipotent with respect to the action of T E 7r1 Set M iiM Note that in the previous example instead of working in C X C we were projecting onto the second C since no information is lost in this case Theorem 28 Kashiwara Malgrange There exists at most one ltration VM on M indexed by Q such that l The ltration is exhaustive discrete and left continuous 2 Each yaM is a coherent Vopysubmodule 3 all04M Q Va1M and tVaM Q Va1M with equality for a lt 0 4 GrXM is a generalized eigenspace of tan with eigenvalue 7a IfM is quasiunipotent along X0 then VM exists Given a perverse sheaf L and A E C let MAL and qufL be the generalized A eigensheaves of 1 wa and 1 de under the Taction Note that N TiA gives a nilpotent endomorphism of these sheaves For various reasons it is better to work the logarithm N logI N N 7 N 2 H which is again nilpotent Saito S1 has de ned a modi cation Var of var which plays an analogous role for N 18 DONU ARAPURA Example 29 Continuing with example 27 note that M is a simple Dcmodule so it should correspond to a simple perverse sheaf L I claim that after restricting to a disk L PgH for A exp27rir in the above classi cation To see this set t exp27ri739 E Cquot Then the monondromy 7 gt gt 739 1 is given by tr gt gt Atr as required In this case p zL 1715211 p tL 19 sz Theorem 210 Kashivvam7 Malgrange Suppose that L DRM Leta E Q and A eh Then p1JfL ifaE 7170 191me 27a 6 7170 The endomorphisms ma a at t on the left corresponds to N7 can7 Var respectively on the right DRGrX17 NOTES ON DAMODULES AND CONNECTIONS WITH HODGE THEORY 19 3 HODGE MODULES 31 Hodge theory background A rational pure Hodge structure of weight m E Z consists of a finite dimensional vector space HQ with a bigrading H HQ C 69 H W pqm satisfying Hm qu Such structures arise naturally from the cohomology of com pact Kahler manifolds For smooth projective varieties we have further constraints namely the existence of a polarization on its cohomology If m is even odd a po larization on a weight m Hodge structure H is a antisymmetric quadratic form Q on HQ satisfying the HodgeRiemann relations Given a Hodge structure of weight m its Hodge filtration is the decreasing filtration FPH 69 HpCmip 17217 The decomposition can be recovered from the filtration by H W F17 Fq Deligne extended Hodge theory to singular varieties The key definition is that of a mixed Hodge structure This consists of a bi ltered vector space F with H W defined over Q such that F induces a pure Hodge structure of weight k on GrLVH This refines the notion of a pure Hodge structure A pure Hodge structure of weight k can be regarded as a mixed Hodge structure such that GrLVH H Mixed Hodge structures form a category in the obvious way Morphisms are rational linear maps preserving ltrations Theorem 32 Deligne The singular rational cohomology of an algebraic variety carries a canonical mixed Hodge structure Griffiths introduced the notion a variation of Hodge structure VHS to describe the cohomology of family of varieties y gt gt HmXy where z A Y is a smoot projective map A variation of Hodge structure of weight m on a complex manifold Y consists of the following data 1 A locally constant sheaf L of Q vector spaces with finite dimensional stalks 2 A vector bundle with an integrable connection E V plus an isomorphism DRE E L Cdim Y 3 A filtration F39 ofE by subbundles satisfying Griffiths7 transversality VFp Q 1771 4 The data induces a pure Hodge structure of weight m on each of the stalks L A polarization is a at pairing Q L X L A Q inducing polarizations in the stalks The main example is Example 33 If f z X A Y is smooth and projective L RmfiQ underlies a polarizable VHS of weight m E is the associated vector bundle with its Gauss Manin connection The key analytic fact which makes the rest of the story possible is the following lt was originally proved by Zucker for curves 20 DONU ARAPURA Theorem 34 Cattani Kaplan Schmid Kashiwara Kawai Let X the complement of a divisor with normal crossings in a compact Kahler manifold Then intersec tion cohomology with coe cients in a polarized VHS on X is isomorphic to L2 cohomology for a suitable complete Kahler metric on the X It follows that L2 cohomology is nite dimensional in this case there is no a priori reason why it should be When combined with the Kahler identities we get Corollary 35 Intersection cohomology with coe cients in a polarized VHS car ries a natural pure Hodge structure 36 Filtered Dmodules The notion of a VHS is no longer adequate to describe what s going on for nonsmooth mapsl Saito de ned the notion Hodge modules which gives a good theory of VHS with singularitiesl I start by describing the basic setting nextl Fix a smooth variety Xl We de ne the category MFThX to consist of regular coherent DXmodules with good ltrations and ltration preserving morphisms Although not Abelian it is an exact category so we can form its derived category BED Recall that among the requirements of goodness is that FqD FpM Q Fpqu lt suf ces to check this for q 1 After reindexing FPM FpM we see that this condition is just Grif ths transveralityl Although I will say nothing about proofs it is worth remarking that many technical issues are handled by passing to the Rees module GBFPM over the Rees algebral The previously de ned operations can be extended to the ltered settingl Given M F E MFThX and a proper morphism f z X A Yl We de ne direct image explicitlyl Let n dimX 7 dim Yl We break the de nition into cases 1 If f is an embedding FAN M F M FkDYeX FpeknMlnl k where FkDyAX is induced by the order ltrationl 2 If f is smooth then using the formulas of section 156 FpRfDRXYM RM H 3W Fle l l 3 In general factor f into a composition of the inclusion of the graph followed y a projection and apply the previous cases We can lter the cohomology modules by FpHiRfM7F imlHiFpRfM7F a HiRfM7Fl The direct image is called strict if the above maps are injectionsl Now add a rational structure by de ning the category MFThDXQl The ob jects consist of 1 A perverse sheaf L over 2 A regular holonomic DXmodule M with an isomorphism DRM g L C 3 A good ltration F on Ml Variations of Hodge structure give examples of such objects We can de ne direct images for these things by combining the previous constructionsl Given a morphism f z X A Y and an object M F L E MFThXQ the direct images are de ned by mm M F L HiRfM7F7pHiRfL NOTES ON DeMODULES AND CONNECTIONS WITH HODGE THEORY 21 The notation is chosen to avoid confusion with the usual direct image for con structible sheaves Sometimes Hi f or Ri r are used in the literature When QX OX with trivial ltration then for any smooth projective map f z X A Y pRifiM gives the standard example of a VHS up to shift 37 Hodge modules on a curve The category MFThX Q is really too big to do Hodge theory and Saito de nes the subcategory of polarizable Hodge modules which provides the right setting The de nition of this subcategory is extremely delicate and will be explained later To ease our journey we will give a direct describe Hodge modules on a smooth projective curve X xed for this section Given an inclusion of a point i z A X and a polarizable pure Hodge struc ture H F HQ the Dmodule pushforward iiH with the ltration induced by F together with the skyscraper sheaf HQ de nes an object of MFThX Letls call these polarizable Hodge modules of type 0 Given a polarizable variation of Hodge structure EF L over a Zariski open subset j U A X we de ne an object of MFThX as follows The underlyung perverse sheaf is jLl which is the intersection cohomology complex for L e can extend EV to a vector bundle with logarithmic connection on X in many ways The ambiguity depends on the eigenvalues of the residues of the extension which are determined mod Z Example 38 Suppose that EV ODd r locally Then E can be extended to the whole disk by the obvious way as OD but also as Opt for any n E Z In general writing the connection with respect to the new trivialization has the e ect of translating the residue r by n Then the multivalued function t pr gives a solution to Vf 0 ts monodromy is given by multiplication by ezp727rir n and this is independent of n For every half open interval I of length 1 there is a unique extension E with eigenvalues in I Let M E C jiE This is a DXmodule which corresponds to the perverse sheaf RjLl E C Let M C M be the sub DXmodule generated by E 1gt0l This corresponds to what we want namely jiLll We lter this by FPE 1gt l jFpE m Eltilgt 1 FpM ZEDXFpiE 1gt l Then M FMjLl de nes an object of MFThX that we call a polarizable Hodge module of type 1 A polarizable Hodge module is a nite direct sum of objects of these two types Let MHX1quot l denote the full subcategory of these Theorem 39 MHX1 l is abelian and semisimple ln outline the proof can be reduced to the following observations We claim that there are no nonzero morphisms between objects of type 0 and type 1 To see this we can replace X by a disk and assume that z and U above correspond to 0 and Dquot respectively Then by lemma 24 the perverse sheaves of type 0 and 1 correspond to the quivers 0lt onto lt7 171 22 DONU ARAPURA and the claim follows We are thus reduced to dealing with the types seperatedly For type 0 respectively 17 we immediately reduce it the corresponding statements for the categories of Hodge structures respectively VHS7 where it is standard In essence the polarizations allow one to take orthogonal complements7 and hence conclude semisimplicity Theorem 310 fM E MHXp0l then its cohomology carries a Hodge structure Here is the proof Since MHXnpquotl is semisimple7 we can assume that M is simple Then either M is supported at point or it is of type 1 1n the rst case7 H0M M is already a Hodge structure by de nition7 and the higher cohomologies vanish 1n the second case7 we appeal to theorem 34 or just the special case due to Zucker 311 VHS on a punctured disk We want to analyse Hodge modules locally This will provide important clues for the higher dimensional case We may as well concentrate on the only interesting case of modules of type 17 that is variations of Hodge structure The key statements are due to Schmid Sc Proposition 312 Jacobson7 Morosov Fix an integer m Let N be a nilpotent endomorphism of a nite dimensional vector space E over a eld of characteristic 0 or more generally an object in artinian abelian category linear over such a eld Then the there is a unique ltration 0C Wml C HWmC meH E called the monodromy ltration on E centered at m characterized by following properties 1 C Wkig 2 Nk induces an isomorphism Gr7VXkE E Griz7k Note that this result applies to the categories of perverse sheaves and holonomic modules Example 313 If N2 0 the ltration is simply C kerN C E The last part of the proposition is reminiscent of the hard Lefschetz theorem There is an analogous decomposition into primitive parts Corollary 314 GridE 69 NiPGrLZrm i where 130 E WW GTZME a STELMEH Things can be re ned a bit in the presence of a nondegenerate form Lemma 315 If S is a nondegenerate skew symmetric form on E which N preserves in nitesimally 5Nu7 v Su7 Nv 0 then PGr7VXkE carries a non degenerate skew form given by u7 v gt gt Su7 Nkv This induces a form Gr7VXkE for which the primitive decomposition is orthogona NOTES ON DeMODULES AND CONNECTIONS WITH HODGE THEORY 23 Let L5F be a polarized variation of Hodge structure of weight m over a punctured disk D arising for example from the cohomology of family of varieties over D After choosing a trivialization of 5 we can identify the bers over t E Dquot with a xed vector space E possessing a rational lattice EQ and a family of ags F Since L is locally constant it corresponds to a representation of 7r1D A AutEQ It is known that the counterclockwise generator T E 7r1D acts quasi unipotently Thus after passing to a branched cover D A D we can assume tha T acts unipotently Then N logT will be nilpotent So we get an associated mondromy ltration W as above centered at m Theorem 316 Schmid For an appropriate trivialization F converges in a flag variety as t A 0 The limit ltration lim F together with W yields a mixed Hodge structure called the limit mixed structure on E Moreover the polarization of the VHS induces one on the associated graded as in the above lemma ln geometric situations this yields a natural mixed Hodge structure on the co homology of the nearby ber A alternative construction of this was given by Steenbrink The key consequence of importance here is Corollary 317 GridE with the ltration induced from limF is a pure polar izable Hodge structure of weight h For Hodge modules in general a re ned version of the above statement is taken as an axiom 318 Hodge modules introduction In the next section we will de ne the full subcategories MHX n C MFThX of Hodge modules of weight n E Z in general Since this is rather technical we start by explaining the main results Theorem 319 Saito MHX n is abelian and its objects possess strict support decompositions ie that the maximal subquotient module with support in a given Z C X can be split o as a direct summand There is an abelian subcategory MHX nquotl of polarizable objects which is semisimple We essenitally checked these properties for polarizable Hodge modules on curves in section 37 They have strict support decompositions by the way we de ned them Let MHZ X n C MHX n denote the subcategory of Hodge modules with strict support in Z ie that all subquotient modules have support exactly T e main examples are provided by the following Theorem 320 Saito Any weight n polarizable variation of Hodge structure L over an open subset of a closed subset ULZLX can be extended to a polarizable Hodge module in MHZ X nquotl C MHXnp0l The underlying perverse sheaf of the extension is the associated intersection coho mology complex iijLdimU All simple objects of MHXnpquotl are of this form Finally there is stability under direct images Theorem 321 Saito Let f z X A Y be a projective morphism with relatively ample line bundle Z IfVl M F L E MHX n is polarizable then PRifm e MHYnz 24 DONU ARAPURA is strict Moreover a hard Lefschetz theorem holds W pR jfiM 2 PRJ39fiMU Corollary 322 Given a polarizable variation of Hodge structure de ned on an open subset of X its intersection cohomology carries a pure Hodge structure This cohomology satis es the Hard Lefschetz theorem The last statement was originally obtained in the geometric case in BED The above results yield a Hodge theoretic proof of the decomposition theorem of loc cit Corollary 323 With assumptions of the theorem RfiQ decomposes into a direct sum of shifts of intersection cohomology complexes 324 Hodge modules conclusion We now give the precise de nition of Hodge modules This is given by induction on dimension of the support This inductive process is handled via vanishing cycles We start by explaining how to extend the construction to MFThX7 Given a morphism f z X A C and a DXmodule7 we introduced the Kashiwara Malgrange ltration V on M earlier in section 26 Now suppose that we have a good ltration F on M The pair M7 F is said to be be quasiunipotent and regular along f 10 if the following conditions hold 1 tFpVa17 FPVHM for a lt 1 2 61FpGrXM Fp1GTX1M n BEGTXM for a 2 0 It is worth noting that the corresponding statements in the non ltered case come for free7 once we know that V exists of course Also the basic example of a variation of Hodge structure on the disk satis es these conditions with respect to the identity function Lemma 325 fa 0 then t GrXM E erilM and at erilM E GrXM We extend the functors 45 and 1 to MFThX7 Let fM7F7L GB GTXUWhFllleWfL 71lt04lt0 fgt1M7F7L GTXM7F7p f1L Here F actually denotes the ltration induced by it on the associated graded The elementary de nition for curves given earlier will turn to be equivalent De ne X gt gt MHX7 n to be the smallest collection of full subcategories of MFThX7 satisfying MHl lf M7 F7 L E MFThX7 has zero dimensional support7 then it lies in MHX7 n iff its stalks are Hodge structures of weight n MH2 lf M7 F7 L E MHSXn and f U A C is a general morphism from a Zariski open U Q X7 then a MFlU is quasiunipotent and regular with respect to f b MF7 L U decomposes into a direct sum of a module supported in f 10 and a module for which no sub or quotient module is supported in f 10 c If W is the monodromy ltration of ibf MFLlU with respect to the log of the unipotent part of monodromy centered at n 7 17 then GerVibfMFLlU E MHUi Likewise for GrEVof1M F7 L with W centered at n NOTES ON DeMODULES AND CONNECTIONS WITH HODGE THEORY 25 This is a lot to absorb so let me make few remarks about the de nition o If f 10 is in general position with respect to suppM the dimension of the support drops after applying the functors 45 and Thus this is an inductive de nition The somewhat technical condition b ensures that Hodge modules admit strict support decompositions The condition can be rephrased as saying that M F L splits as a sum of the image of can and the kernel of vari A re nement of lemma 25 shows that L will then decompose into a direct sum of intersection cohomology complexesi This is ultimately needed to be able to invoke theorem 34 when the time comes to construct a Hodge structure on cohomolo Although MFThX is not an abelian category the category of compati ble pairs consisting of a Dmodule and perverse sheaf isi Thus we do get a W ltration for ibf M F LlU in c by proposition 312 by rst suppressing F and then using the induced ltrationi There is a notion of polarization in this setting Given M F L E MHZ X n a polarization is a pairing 5 L L A pr dim X7n satisfying certain axioms The key conditions are again inductivei When Z is a point 5 should correspond to a polarization on the Hodge structure at the stalk in the usual sense In general given a germ of a function f Z A C which is not identically zero 5 should induce a polarization on the nearby cycles GTEVibfldill using the same recipe as lemma 315 Once all the de nitions are in place the proofs of the theorems involve a rather elaborate induction on dimension of supportsi 326i Mixed Hodge modules Saito has given an extension of the previous theory by de ning the notion of mixed Hodge module 1 will start with recalling the older de nition of a variation of mixed Hodge structure 1 A locally constant sheaf L of Q vector spaces with nite dimensional stalks 2 An ascending ltration W C L by locally constant subsheaves 3 A vector bundle with an integrable connection E V plus an isomorphism DRE L Cdim Y 4 A ltration F39 ofE by subbundles satisfying Griffiths7 transversality VFp Q 1771 5 GrizL OX GrizL F39 OX Gr7lXL is a variation of pure Hodge structure of weight mi The data induces a mixed Hodge structure on each of the stalks Ly and hence the name Steenbrink and Zucker SZ showed that additional conditions are required to get a good theoryi While these conditions are rather technical they do hold in most natural examples A variation of mixed Hodge structure over a punctured disk D is admissible if 1 The pure variations Gr L are polarizablei 2 There exists a limit Hodge ltration lim Ftp compatible with the one on Cry L constructed by Schmidi 3 There exists a so called relative monodromy ltration U on E Lt W with respect to the logarithm N of the unipotent part of monodromyi This means that NUk Q 1162 and U induces the monodromy ltration on GridE constructed earlier up to suitable a shift Note that relative monodromy ltrations need not exist a priori 26 DONU ARAPURA For a general base X the above conditions are required to hold for every restric tion to a punctured disk Note that pure variations of Hodge structure are automatically admissible Let me now turn to the general case I will de ne a premixed Hodge module on X to consist of 1 A perverse sheaf L de ned over Q together ltration W of L by perverse subsheavesi 2 A regular holonomic DXmodule M with a ltration WM which corre sponds to L C W C under RiemannHilberti 3 A good ltration F on Mr These objects form a category and Saito de nes the subcategory of mixed Hodge modules MHMX by a rather delicate induction The key points are that for MFL W to be in MHMX we require 0 the associated graded objects GrizV M F L yield polarizable Hodge mod ules of weight k for any germ of a function f on X the relative monodromy ltration U respi U for ibf M F L respi f1M F L with respect to W exists The premixed Hodge modules ibf M F L U and 45f1M F L U are in fact mixed Hodge modules on f 10 The main properties are summarized below Theorem 327 Saito 1 MHMX is abelian and it contains each MHXnpquotl as a full abelian subcategory 2 MHMpoint is the category of polarizable mixed Hodge structures 3 If U Q X then any admissible variation of mixed Hodge structure extends to an object in Finally we have Theorem 328 Saito There is a realization functor real DbMHMX A Di omwX and re ned direct image and inverse image operations f DbMHMX a DbMHMY f DbMHMY a DbMHMX for each morphism f z X A Y such that realfVl RfirealUVl real lLfi real Similar statements hold for various other standard operations such as tensor prod ucts Putting this together with previous statements yields Corollary 329 The cohomology of a smooth variety U with coe cients in an admissible variation of mixed Hodge structure L carries a canonical mixed Hodge structure The cup product HiU L H1 U L 4 Hi U L o L is compatible with these structures NOTES ON DrMODULES AND CONNECTIONS WITH HODGE THEORY 27 When the base is a curve this was rst proved by Steenbrink and Zucker SZ 330i Explicit construction I want to give a bit more detail on the construction of the mixed Hodge structure in corollary 329 Let U be a smooth it dimensional varietyi We can choose a smooth compacti cation j U A X such that D X 7 U is a divisor with normal crossings Fix an admissible variation of mixed Hodge structure L W EFV on Ur Extend EV to a vector bundle E with logarithmic connection with the eigenvalues of its residues of the extension in I as in section 37 Let M UEI C jiEi This is a DXmodule which corresponds to the perverse sheaf BALM C Filter this by FpM ZEDXFpiE 1gt The rest of the story is somewhat more complicated so I will rst state the outcome Theorem 331 There exists cdmpatible ltratizms W on Rji n and M extending W over U such that RjLn W M F becomes a mixed Hodge module This second ltration W is not easy to describe So llll give an indication in the special case where the original variation of mixed Hodge is pure and D f 10 for some morphism f z X A lP li In this case RjLn ts into an exact sequence of perverse sheaves 0 ijLUL A BALM A K A 0 where K coheTN 17de71 A pdehil To construct W we can take jLn as the rst step The rest is obtained by pulling back the image of the monodromy ltration from Ki In general D is always given locally as f 10 for some Thus this description can be extended to the general case once it is shown that the locally de ned ltrations must patchi Note that it is always true that the associated graded GTZYV decomposes into a direct sum of intersection cohomology complexes of pure polarized VHslsi This provides the crucial link to the earlier work of Cattani Kaplan Schmid and Kashiwara Kawai theorem 34 Finally let me sketch where the mixed Hodge structure comes fromi We have an isomorphism HiU L C 2 Hi93log D EM and the ltrations de ning the Hodge structure can be displayed rather explicitly from this The ltration Fmgaogp E gt1gt EB 9 Fp iE gt1 induces the Hodge ltration on cohomologyi We de ne a ltration W on this complex by intersecting WnM under the inclusion 93logD E gt1gt c n Mn Then this ltration induces WikHiU L C HiX Wkngaogp E gt1gti 28 DONU ARAPURA Since we have a corresponding ltration on legL7 it follows that W is de ned over In more technical terms7 this data constitutes a cohomological mixed Hodge complexi 332 Real Hodge modules Saito has also developed a theory of mixed Hodge modules over the reals7 where the underlying perverse sheaf is de ned over R Some modi cations in the basic set up are necessary eigi Kashiwara Malgrange ltrations are now indexed by R but the basic theory goes through pretty much as before In particular7 the cohomology of real admissible variations of mixed Hodge structures carry real mixed Hodge structuresi NOTES ON DeMODULES AND CONNECTIONS WITH HODGE THEORY 29 4 COMPARISON OF HODGE STRUCTURES The cohomological mixed Hodge complex described in the previous section re duces to the one given by Deligne DeS for constant coef cients Consequently Proposition 41 Let X be smooth variety then Saito s mixed Hodge structure cor 329 on HquotX7 agrees with Deligne s From now on let f z X A Y be a smooth projective morphism of smooth quasiprojective varieties We recall two results Theorem 42 Deligne Del The Leray spectral sequence Egg HWY Run HWX Q degenerates In particular Egg 2 GrinJr KX for the associated Leray ltrationquot L Theorem 43 There exists varieties Y17 and morphisms Y A Y such that LpHiXQ herHiX7 Q A HiXpQ where Xp f lYp It follows that each L17 is a ltration by sub mixed Hodge structures When com bined with the earlier isomorphism7 we get a mixed Hodge structure on H1 Y7 Rq which I will call the naive mixed Hodge structure On the other hand7 quiQ car ries a pure hence admissible variation of Hodge structure7 so we can apply Saito7s result 329 to get another Hodge structure Proposition 44 The naive mixed Hodge structure coincides with Saito s Proof Deligne actually proved a stronger version of the above theorem which im plies that 2gt Rm 2 Rimlez non canonically in DbX7 The theorem is quite general and thanks to 3217 it even applies if regard these as objects in DbMHMY Note that the Leray ltration is induced by the truncation ltration LpHiX7 imageHiY7 75i7priQ A Under 27 TSPRW 2 EBRiwei 13931 Therefore we have a non canonicall isomorphism of mixed Hodge structures HiltXQgt EB HWY Rpm pqi where the right side is equipped with Saito s Hodge structure Under this isomor phism7 L17 maps to Hi pRpfQ ea Hi p1Rp 1fQEBm The proposition now follows E This result goes back to Zucker Z when Y is a curve 30 DONU ARAPURA REFERENCES A D Arapura The Leray spectral sequence is motivic Invent Math 2005 A2 D Arapura Mixed Hodge structures associated to geometric variations mathAG0611837 Be J Bernstein Algebraic theory of Dimodules Unpub Notes BBD J Bernstein A Beilinson P Deligne Faisceaum perverse Asterisque 1982 Bo A BoreI et al Algebraic Dimodules Acad Press 1987 Br JrL Brylinski Co7homologie intersection et faisceaum pervers SeIn Bourbaki 1982 BZ JrL Brylinski S Zucker An overview of iecent advances in Hodge theory Sev Complex Var VI SpringeriVerIag 1990 CKS E Cattani A Kaplan W Schmid L2 and intersection cohomologies for a polarizable variation of Hodge structure Invent Math 1987 Del P Deligne Theoreme de Lefschetz et criteres de degenerescence de suites spectrales PubI IHES 35 1968 De2 P Deligne Equation di erentielles a point singular r gulier SpringeriVerIag 1969 De3 P Deligne Theorie de Hodge II III IHES 1972 1974 SGA7 P Deligne et al SGA 7 II SpringeriVerIag E F El Zein Theorie de Hodge a coe icients CR Acad Sci Paris 307 1988 Ga 0 Gabber The integrability of the characteristic variety Amer J 1981 GM S Gelfand Y Manin Methods of homological algebra SpringeriVerIag 1996 GoM M Goresky R Macpherson Intersection homology II Invent Math 1983 Ha R Haltshorne Algebraic geometry SpringeriVerIag 1977 K1 M Kashiwara A study of a variation of mized Hodge structure RIMS 1986 K2 M Kashiwara Dimodules and microlocal calculus AMS 2000 KK M Kashiwara T Kawai Poincare lemma for variations of Hodge structure RIMS 1987 Od T Oda Introduction to algebraic analysis on complex manifolds Alg and Anal Varieties Tokyo 1981 PS C Peters J Steenbrink Mixed Hodge structures Preliminary Manuscript Sb C Sabbah Hodge theory singularities and Dimodules Lect Notes 2007 S1 M Saito Modules de Hodge polarizables RIMS 1988 S2 M Saito Mized Hodge modules and admissible variations CR Acad Sci Paris 309 1989 S3 M Saito Mized Hodge modules RIMS 1990 S4 M Saito Mized Hodge modules and applications Proc ICM Kyoto 1990 Sc W Schmid Variation of Hodge structure Invent Math 1973 V IL Verdier Extension of a perverse sheaf over a closed subspace Asterisque 130 1985 SZ J Steenbrink S Zucker Variations of mized Hodge structure Invent Math 1983 Z S Zucker Hodge theory with degenerating coe icients Annals 1979 DEPARTMENT OF MATHEMATICS PURDUE UNIVERSITY WEST LAFAYETTE IN 47907 USA Chapter 1 Systems 11 On Line In this introductory section we will pose no exercises but instead will detail how to use Maple to solve problems in linear algebra For the novice lVIaple user this section is essential reading and reference For the experienced Maple user this section can be examined for evidence of any new tips and ideas for working with Maple In addition this section will reveal the working style guiding the solutions in the rest of this manual lVIaple is a symbolic as well as a numeric language For users with ex perience in numerical computation Matlab BASIC FORTRAN 7 etc there are differences in thinking that accompany calculating in Maple The presence of symbolic variables adds a dimension missing in strictly numeric languages For one thing Maple is capable of computing answers in exact symbolic form so round off error need not be an issue in didactic ex periments For another it really helps to have a vision as to how to mix symbolic and numeric calculations lest operations performed numerically impinge 011 the operations which will be later performed symbolically We illustrate some sample Maple calculations beginning with simple arithmetic As we present the Maple syntax for various calculations we will explain the Maple Release 4 interface as it appears under Windows on a PC Release 4 has been engine red to have the same functionality 011 all major platforms Macintosh P NIX so only local differences in file structures and networks might possuly differ Enter Maple syntax at the prompt the gt at the left edge of the screen gt 23 57 2 11 llaple On Line Lines entered into Maple must have terminal punctuation typically the semicolon Simple arithmetic is done automatically A new prompt is generated upon execution of the command and that follows upon pressing the ENTER key Spaces are generally ignored by Maple and are used here to improve readability gt x 2x lVIaple applies the rules of algebra to symbolic expressions and some of the simplifications are immediate thers must be requested by a variety of special commands Assignments are made using the two characters colon 7 and equal two characters that must not be separated by a space gt f x 2 We have just entered an 11quot sign or formula and assigned it to the name f To evaluate this CXpI39LbblOIl at x 3 use t 1e subs command to substitute x 3 into f gt snbsx3f We next articulate Lopez s Large Law see the article Tips for Instructor aple IL 01 3 NO 2 1996 published by DirkLa which state never use on the left of an assignment a variable in use some where 011 the right Thus it is not a good idea to make the assignment x z x 1 Although this is an extremely common construction in numeric languages it is most unwise in a symbolic language Moreo gtr Lopez s Large Law precludes assigning values to the 7 working 39 z etc A strict distinction between the letters being used s 011 the left and 7 variables on the right makes Maple more ince we have demonstrated how to make an assignment to the name f 7 we next show how to erase that assignment gtf gtf First erasing consists of assigning f to its 7 letter value which is what the single quotes accomplish Second entering interrogates Maple for what it knows about that name Maple echoes the contents of that name It is possible to create unfathomable Maple Worksheets Violations of Lopez s Large Law can lead to such dif culties and so can misunderstand 11 llaple On Line 3 ing Maple s file management For eXample suppose Lopez s Large Law has been violated y gt x3 lVIuch later in the Worksheet the variable X is used symbolically as gt f xsinPix What happened Why didn t the formula 1 sin 7r 1 appear Since X has been assigned the value 3 Maple computed 5in3 7r and got 0 Violating Lopez s Large Law indeed has consequences It is not enough to er X Once the assignment to l39 is the number 0 it remains the number 0 Having a clear strategy for working with Maple prevents needless l39rustration The neXt timebomb is more subtle Suppose that fhas the value 0 from the above calculations And suppose that all record of the eX1stence of l39 in the Worksheet is removed by deleting the appearance of 139 from the Worksheet The letter is still assigned the value 0 In fact it s worse than that If another blank Worksheet is opened via the menu options FileNew this new Worksheet will still have 139 assigned the value 0 This is because both Worksheets share the same memory and what is known to Maple in one W39orksheet is known to the other On some platforms with the right initialization of Maple each Worksheet can have its own attached memory Unless you know for sure how your copy of Maple is installed it is best to assume that all Worksheets share a single memory and always under all circumstances remember that merely removing the appearance of an assignment from a Worksheet does not remove that assignment from Maple s memory Suppose at this point in your eXperiments with a Worksheet you have eXcised the last inputoutput pair and the bottom of your Worksheet no longer contains the neXt prompt How do you generate a new prompt Place the cursor in an input line Simultaneously press the keys TRL and k to insert a prompt above the cursor and TRL and to insert a new prompt below the cursor If you eXamine the Insert menu the EXecution Group corresponds to a new prompt Having toyed with the menu bar observe the Help menu The best ad vice a Maple user can receive is to begin with HelpContents The resulting document that opens is hyperlinked to all sorts ol39inl39ormation about Maple 11 llaple On Line and it is left to the user to navigate through instructions 011 the interface and 011 Maple itself However to get help on a command whose name you know you can use the question mark gt subs Help commands don t need terminal punctuation At the bottom of most help screens is a section of Examples Look there rst Examples can be copied and pasted back into your Worksheet for experimentation The exercises for each section in this manual begin with the instruc tion to restart Maple This means to issue the restart command which clears all variables This command does not erase anything visible in the Worksheet so if a Worksheet has become confused and entangled issuing a restart and then working from the top down reexecuting all the entered commands is a way of beginning at the beginning and retracing the thought process in the Worksheet gt restart gt i We next illustrate some algebraic simplifications gt q 1x 1y gt q1 simplifyq ommands in Maple typically take parentheses around the argument or arguments The letter q makes a handy label because it is easy to find 011 the keyboard Reassigning a new version of a name to itself is actually a violation Lopez s Large Law and should be avoided Thus gt q simplifyq is not a syntax error but is just plain bad worksmanship If parts of the Worksheet are reexecuted which version of q is being referenced Is it the unsimplilied version or the simplified version So use unique names for each meaningful Maple output to avoid confusion when experimenting since that usually requires editing changing reexecuting moving up and down throughout the Worksheet If the same letter has multiple meanings throughout the Worksheet chaos results Maple crashes It is a fact of life that Maple inexplicably and explicably crashes It is exceedingly frustrating to have spent an hour or more on an assignment in Maple only to lose it all because Maple crashed There is only one piece of advice that makes sense here and that is Save gtarly and save often The first time aVVorksheet is saved FileSave or the diskette 11 llaple On Line icon 011 the toolbar lVIaple prompts for a file name and a destination for the saved file Thereafter clicking the diskette icon 011 the toolbar or entering CONTROL 3 from the keyboard saves work to that same file Save early and save often That advice cannot be repeated too frequently lVIaple is both a symbolic and a numeric language Thus there is a difference between 1 and 10 in Maple The first is the exact integer 1 while the second is the decimal version of the number 1 Converting the exact symbolic representation of a number is done with the evalf evaluate floating point command gt q 1sqrt2 gt evalfq Note that Maple immediately rationalizes And note further that Maple can provide many more than the default 10 digits gt evalfq20 The next thing useful to know about Maple is how to reference parts of answers it generates Consider the following solution to a quadratic equation gt qx 23x10 gt solveqx Maple has returned a sequenee of two roots which can be referenced if a tag had been assigned to the solve command Thus the better working strategy is gt q1 solveqx Now the roots can be referenced by the bracket notation gt q11 gt q12 Thus there are three data structures lVIaple uses that are worth under standing Maple uses sequen s lisls and sels In a sequence items are separated by commas A list is a sequence enclosed by square brackets abc A set is a sequence enclosed by curly braces abc The list preserves order and replicas The set does not gt abaac gt abaac i 11 llaple On Line Each structure re ects valid mathematical usage and Maple has com mands for manipulating each data structure proper y Although repetitive tasks can be implemented in Maple by copying and pastng input lines a l39orloop is the appropriate way to repeat similar instructions In this manual the l39orloop is entered as a single input as follows gt for k from 1 to 3 do xk k 2 0d All three input lines are connected to the one prompt by entering all lines but the last with SHIFT ENTER rather than simply ENTER In Release 4 this is now more aesthetic than practical but if the three lines of code above are entered into Maple V Release 4 with just the ENTER key the first line will generate a complaint about 7 incomplete a complaint that disappears when the terminating 0d 7do 7 spelled backwards is entered In previous versions of Maple failure to keep the lines ol39a l39orloop together could lead to terrible consequences if changes were made to individual lines of the loop In Release 4 this is no longer such a problem One advantage of the notation X1 X2 X3 is that such objects can be referenced collectively by gt x v 1 v 3 For large collections of similar objects this turns out to be a handy device for saving repetitive and tedious typing We will be concerned primarily with Maple s functionality in linear al gebra NIaple s code is modularized bundled into related groups called packages There are some 32 packages in Release 4 all of which are present in every properly installed version of Maple The command gt 7packages brings up the list of packages and the names of the packages are hyper linked to more information about the individual package39 The package we will use most is the linalg package tsell39 containing more that 100 commands for manipulating vectors and matrices The linalg package is loaded into Maple via the command gt withlinalg Notice that the terminal punctuation here is the colon which sup presses output This package will be loaded for every exercise set and we will want to suppress the listing of the more than 100 commands made present by this package First we enter the matrix A 11 llaple On Line 7 gt A matrix22 123341 It appears easier to provide the Inatrix command with a single list of entries letting Maple wrap them according to the dimensions given first The alternative is to give the Inatrix command a list of lists the sublists being the rows of the matrix Thus gt matrix 1 2 34 We will use Maple s randnlatrix command to generate matrices at ran dom Maple s random number generator will generate the same sequence of random numbers each time it is initialized by starting or restarting Maple The advantage here is that results are then reproducible even if random matrices have been used There is a way of setting the starting point for the random number generator but that is not used in any of the exercises The help file for rand the random number generator accessed by Yrand will mention the global variable sccd which can be assigned values student s SSN Y thereby making unique assignments for each student That is not done in these exercises gt B randmatrix22 Matrix and vector arithmetic is most easily done by applying evalln evaluate matrix to the desired arithmetical commands gt 2A 439 3B A39Z gt evalm2A 3B A 2 In the first instance merely the names are manipulated by Maple In the second the actual entries of the matrices are manipulated There are times and places for each approach but only the second is used in these exercises Next we address matrix multiplication a process that is known in math ematics to be noncommutative Thus for numbers 23 i but for ma ri V rarely equa s B A Hence Maple distinguishes between the use of for commutative multiplicatio and amp for the noncommutative multiplication of matrices In Re east 4 Maple will warn specifically that AB l39or matrices must be changed to A 55 B In earlier versions the user had to be prescient gt evalmA B gt evalmA amp B evalmB amp A Surprisingly vectors will take more discussion than matrices First of all enter the vector V 22 with the following syntax In a newly 8 11 llaple On Line installed copy of Maple V Release 4 the output will be as you see below gt V vector25 Throughout this manual the output will instead appear as gt V vector25 Why the difference and how do we get Maple to render vectors as columnlike rather than Wowlike And are such vectors column vec tors or row vectors First no matter how we get Maple to print the vector it is always a column vector Second to get your IVIaple session to print the vectors as columns enter the following instructions gt withshare readsharepvac system These two commands cause Maple to load from its Share Library a file called pmc printvectorascolumn the effect of which is to change the way vectors are printed If for some reason this functionality is to be switched off enter the command gt pvac false gt printV gt pvactrne V print V In addition to noting how to turn this display feature on and off observe that it takes print or evalrn to get Maple to display the contents of a vector or matrix For the adventurous from outside IVIaple examine the file structure of the Maple V4 directory There is a subfolder called Shani in which the contents of the Share Library are stored A further subfolder Syslmn contains another subfolder called ch In the PMC folder there is a file I vacmpl a file containing the code for the display feature being discussed If this file can be rendered as a pure text file it can be made into an initialization file so that the code will load automatically every time IVIaple is launched The author of this manual has had this code running as an initialization file in both Release 3 and Release 4 a span of more two years It has worked perfectly and has never given any trouble whatsoever To obtain a text version of the file pvacmpl launch a text editor such as Word etc From within the text editor locate the file pvacmpl and open it Perform a Save As save the file as t x 7 give it the name Mapleini for the PC 011 the Macintosh use MapleInit and for UNIX use mapleinit 11 llaple On Line 9 Be sure that your text editor does not add a hidden txt or other such extension Quit the editor and drop the initialization file into the Lib subl39older ol39 the Maple V4 folder for the PC drop it into the Maple V4 folder for the Mac and experiment With Where to put it 011 a UNIX system Relaunch Maple and you will automatically have launched the pun code To verify that a Maple vector behaves as a column vector no matter What it looks like try the following experiments gt printAV gt evalmA amp V This is exactly the product you should obtain from the product A V done by hand treating V as a column vector Now ask Maple to convert V to a matrix gt VC convertVmatrix gt typeVvector typeVCvector gt type Vmatrix type VCmatrix gt vc11 gt vc12 gt vc21 The matrix VC does not have a second column It has two rows 1139 the vector V is converted to a matrix data structure it gets converted to a 2 x 1 column matrix Maple thinks of the vector V as a column thing even if its default print style is to make it look like a row thing lVIoreover the transpose of V il39 converted to a matrix has all the characteristics of a l x 2 row matrix gt VR converttransposevmatrix gt VR11 gt VR12 gt VR21 The matrix VR does not have a second row Maple thinks of the trans pose of V as a row thing no matter how it prints it In fact it is a great tragedy that the Maple programmers have decided that the default output to the transpose command is gt transposeV 10 11 llaple On Line The evidence has already been presented that Maple understands the transpose It is sad indeed that so fine a program as Maple should have such anomalous behavior when displaying the transpose of a vector V that so obviously has the inherent properties of a column object Incidentally this means that there is no way per se to enter a row vector into Maple You enter a column vector the default vector object then transpose the column vector And yes you live with not being able to see the display of the transpose as a rowlike object aution It is tempting to sidestep this issue of row and column vectors with the belief that instead row and column matrices will be used This is not a good idea There are commands in Maple that specifically demand vectors not matrices For example if you had defined not V but VC a column matrix and wanted to live your Maple life with only matrices you d run afoul o39 gt dotprodVC VC So you cannot live without vectors and if you cannot live without vectors you must then face the issue of row and column vectors Sorry but to reap the benefits of Maple you have to put up with a few quirks Kind of like life in genera There is one final issue to face about linear algebra in lVIaple To perform operations 011 vectors one must map the operator onto the vector For example to simplify a vector use the following syntax gt V vector1x1y 1x1y gt simplifyV Obviously not the right syntax gt mapsimplifyV The operator simplify has to be mapped onto the vector As another example ifV is a function oft and you want its derivative you use the following syntax gt V vectorsint cost gt diffVt Obviously the wrong syntax gt mapdiffVt Additional parameters to the mapped operator go at the end Ifnfortunately there are two exceptions to the rule Map things onto vectors and matrices 12 llaple On Line 11 gt V vectorPi2Pi gt evalf V gt mapevalf V Well that seems to work Why raise that as an exception The com mand evall39 can take an integer as a second argument changing the number of digits returned gt evalfPi20 But if you try that for the vector V it fails gt mapevalfV20 Nonsense is returned The method that works is gt evalfopV20 A second exception is substitution gt V vectorxx 2 gt snbsx1V gt mapsnbsVx1 The syntax that works is gt snbsx1opV Hence the rule is 7 Map all operators except subs and evalf For those don t Inap but use op around the vector If you use Inap don t use op Vhen up is needed you don t use Inap 7 12 On Line The exercises of this section explore the concept of span 7 by visualizing randomly generated members of tie s an of a set of vectors Begin by loading both the linalg and plots packages gt withlinalg withplots 12 12 llaple On Line Exercise 1 Enter into Maple the following four points For most purposes points can be represented as lists a data structure denoted by square brackets gt P111J P211 P311 P411J The plots package makes available a pointplot command that will plot a list of points There are a number of options to this command that will vary the look of the graph and by using the toolbars associated with the graph many characteristics of the plot can be adjusted To obtain help 011 this command enter gt Ypointplot To plot these points enter gt pointplotP 1 01 Exercise 2 l 2 Enter the vectors A l and B a gt Avector 1 1 B vector 2 3 Ifsing the pointplot command plot these two vectors as points in the plane Note the additional parameter view which sets a viewing window for the graph gt pointplotABJ view0 30 31 Produce several different linear combinations of A and B Appropri ate syntax lor doing this would be as follows gt eva1m2A 3B b Ifse Maple s random number generator to create several random lin ear combinations ol39 A and B First define a function l39 which generates random l39ourdigit numbers in the interval 41 This is done with Maple s rand function as follows The evalf command changes exact fractions to decimals gt f evalfrand1000O v v 10000100002 Then invoke the function f with the syntax For example create c one random linear combination with the syntax gt cevalmfAfB 12 llaple On Line 13 c Plot enough points in the span of A and B to get a discernible geo metric gure Begin with 100 random linear combinations and using the pointplot command plot them as points in the plane Maple s seq com mand will produce a sequence of similar objects from a pattern provided to it Terminate the command with a colon to suppress the output Then feed the resulting sequence to the pointplot command remembering to enclose the sequence in square brackets since pointplot requires a is of points or vectors gt qseqevalmfAfB k1 t 100 gt pointplot q d The plot in part c is only part of the span the portion being determined by our use of random numbers in the interval ll Rather than redefine the function 139 try multiplying the vector A by 2 creating another plot of at least 200 random linear combinations with A multiplied by 2 Exercise 3 Describe in words the set of points corresponding to the collection of linear combinations de ned by the sum 9 A 13 where both 9 and I lie in closed intervals of the form 22 Plot the resulting set of linear combinations as points in the plane using a different color than used in Exercise 2 See the online help for how to specify color in the pointplot command Exercise 4 In Exercise 9 ol39 the noncomputer problems it was stated that each element x y in the span of z 1 l X l andY 3 satisfies5c3y 2z0 1 Write C a X b Y the expression for the general linear combina tion of A and Y Then show that the components of the vector C satisfy the equation of the plane declared above This is most effectively accomplished as fol ows gt X vector111 Y vector 132 gt C evalmaX bY M l31aple On Line gt q5x3y2z0 gt q1 subsxC1 yC2 zC3 q b Plot as points in R3 a few hundred elements of this span This requires use of Maple s pointplot3d command the 3d analog ofpointplot Begin by forming via seq a sequence of random vectors in the span of X and Y The function f defined for Problem 1 can again be used to provide the random coellicients gt qseqevalmfXfY k1 100 The syntax for pointplot3d is similar to that of pointplot However there are several additional parameters whose use makes for a better graph Since all 3d plot packages attempt to plot a 3d object 011 a 2d sheet of paper or computer screen there is a need for a reference frame that provides the sense of depth Try putting a box around your graph either interactively via the toolbars or via options to the plot command itself Having clear and highly visible labels 011 the axes is equally useful Thus the following syntax could be used to plot the vectors randomly generated above pointplot3d q colorblack axesboxed labelsxyz labelfont TIMES BOLD 14 Observe that in Maple 3d graphs can be rotated 011 the screen by ma nipulation with the mouse lick 011 the plot to make it li Then click and hold down the mouse button dragging the bounding b x that now replaces the graph This bounding box is rotated as the mouse is moved When released the bounding box has a new orientation and the graph is redrawn by clicking the R redraw 011 the toolbar Try to rotate your graph to demonstrate that the plotted points lie 011 a plane 13 On Line Maple Release 4 permits more than one worksheet to be open at the same time On some platforms Maple can be put into the multiple kernels mode in which each worksheet is attached to its own kernel or memory state In this mode variables declared in one worksheet will not be known to any other worksheet opened simultaneous y However the default setting for Maple might be the shared kernel mode in which all open worksheets share the same memory state In this case variables declared in one worksheet have the same value in every other worksheet opened simultaneously The potential here for grave confusion 13 llaple On Line 15 is very high Since in Release 3 only one worksheet could be attached to a kernel this con ict betw e11 multiple worksheets never arose In Release 4 it is essential that this otcha be understood A simple precaution in the shared kernel world is starting every new worksheet with a restart a Maple command which clears memory and resets all variables The use ol39the restart command at the beginning of each new worksheet is highly recommended Here we both restart Maple and load the linalg package gt restart gt withlinalg Exercise 1 4 2 3 Obtain X 0 5 1 1 0 as the general solution to the 0 0 1 equation gr 21 3 z 4 lVIaple s linsolve command the general linear system solver will give the general solution to systems of equations We begin by creating a matrix the coef cient matrix whose entry is the coef cient of the th variable in the i th equation You can either type in the coef cient matrix A directly or use the genulatrix command giving a list of equations and a list of variables as parameters Thus for the preceding linear equation we could enter gt q x2y3z 4 gt A genmatrixq xyz This extracts the coef cient matrix for the system Next the coef cient matrix and the constants 011 the right side of the equations are entered into linsolve We express the constants as a list which in this case has only one entry gt X linsolveA 4 The arbitrary constants that Maple has introduced are 11 and 12 The lead character is the underscore not a minus sign and the numbers 1 and 2 are subscripts To address these constants in Maple use the syntax tl and t2 Setting these constants alternatively equal to 0 and 1 will extract the basis vectors in the general solution After delining u the translation vector it is essential to subtract u from X when extracting the vectors v and w the multipliers of 11 and 12 16 l31aple On Line gt u subst10 t20 01300 v subst 11 t 210 evalmXu w subs t 1 0 t 2 1 evalmXu To get Maple to write the general solution in the vector form given in the statement of the problem adroit use of the evalln command is necessary Unless evalln is applied to the vector v the screen will merely display the name v gt Xg evalmu sevalmv tevalmw With the vectors v and w declared as above or typed in afresh from the pencilandpaper solution form C the general element in the span of v and w gt C evalmav bw b Show that F u C solves the given equation 1 2 y 3 z 4 gt F evalmuC gt q gt subs xF1 yF 2 zF 3 q This shows that the general solution of the given equation is the span of the vectors v and w translated by the vector u c Substitute the vector 7 into the given equation and describe the result What should you conjecture from this result Can you prove it Exercise 2 Repeat Exercise l for the system 1 from Section 13 The essential ques tion to be ultimately resolved is Does the conjecture made in Exercise 1 still hold Can you then prove your answer is correct Begin by entering the system 1 being careful to clear the variable w that was used in Exercise 1 using the command 7 w w all your equations ql q2 q3 and q4 Ifse gennlatrix to form A the coef cient matrix for the system If gt A genmatrixq 1 JD xyzw Ifse linsolve to obtain the general solution of system If using a list for the values on the right hand sides of the equations gt X linsolveA 1012 Extract the translation vector u and the basis vectors v and w as in Exercise 1 Form C the general element in the span ol39v and w 14 llaple On Line gt C evalmav bw b Show that F u C satis es the system If This can be done by repetitive typing by typing once and using copypaste or by a forloop that does repetition automatically gt F evalmu C gt for k from 1 to 4 do subsxF1 yF2 zF3 WFE4J qk 0d c Substitute 7 into each equation in the system If in an attempt to determine if the conjecture made in Exercise 1 is still viable If you still believe your conjecture is true can you prove it 14 On Line Alter restarting Maple and reinitializing by loading the linalg package we examine the reef command for putting a matrix into its reduced row echelon l39orm Consider the matrix A de ned by gt A matrix46 1113oe22260711112 Subject A to the reef operator gt rref A The result just obtained is exact since Maple obtained it by doing ra tional arithmetic There is no truncation error introduced by a conversion of integers to decimal form and there is no roundoff error produced by a numerical algorithm This is the result that would be obtained working with a pencil and paper This is ol39course wonderful We can in princple do every computation with total accuracy At first glance then it seems then we should never again need to round off an answer Unlortuately life is not so simple Suppose for example we want MAPLE to compute 104857720 We enter gt 10485771000000 20 Maple s response is a 121 digit integer divided by an equally large power of 10 Imagine now what would happen if we attempted to perform a calculation which required say addition and multiplication several hundred such numbers The number of digits our computer would need to store would become astronomical and the speed would be reduced to a snail s 1441936J 18 14 llaple On Line pace Furthermore in an actual application the number 1048 would probably represent the result of some measurement which itself might be accurate only to within the given number of digits Thus the vast majority of the digits that our compter is so laboriously computing and saving are totally meaningless The moral is that the perils of numerical computations must be faced Exercise 1 We rst examine how to convert the matrix A to oating point decimal form then look at the same row reduction done numerically instead of symbolically The conversion can be done by tom39crling each element to oating point l39orm Inap the convert operator onto the matrix A via gt A1 mapconvertAfloat Row reduce to reduced echelon form the oating point form of matrix gt rrefA1 For the matrix A there is no difference in the reduced echelon form when working numerically This will not always be the case In fact we can inves tigate Maple s numerics by a stratagem used 011 any numerical calculating device Compute the value of 99 l and successively append 9 s both inside and outside the brackets Eventually there will be enough 9 s so that the numeric calculation will no longer yield 0 That gives an indication of how accurate the computing device is To force Maple to evaluate the ex pressions in oating point l39orm make one of the numbers a decimal For example use l0 rather than just 71 in the numerator ol39 the fraction To see the difference between working numerically and symbolically in Maple change the numerator l39rom l0 to just 1 Then Maple will eval uate the expression symbolically and produce 0 The roundoff error only appears when working with oating point numbers It is possible to vary the number of digits with which Maple computes This is done via the Digits variable as follows gt Digits 12 Test Maple s numeric behavior 011 the oating point calculation above that failed to yie c 0 Notice that with more digits availabl Maple escaped the effect of roundoff in a computation that l39ailed with just the default 10 digits 14 llaple On Line 19 The price one pays for increasing the number of working digits is com putation time since these extra digits are being simulated by the lVIaple sol39tware Reset the number of digits back to the default 10 via gt Digits 10 Exercise 2 Ifse the reef command to find all solutions to the system in Exercise 5g Section 13 First enter the equations of that system all your equations ql q2 q3 qnd qd Next get Maple to write the augmented system matrix The gennla trix command converts the equations into matrix form and the additional parameter ag signals Maple to include the numbers on the right side of the equations Incidentally the parameter can be any character or word that is not already a reserved word in Maple gt Agenmatrixq14xyzwflag Apply the reef command gt A1rrefA To obtain solutions from the rrel39 form of the matrix A apply the process of back substitution Start with the bottommost nonzero row of rrel39A and interpret it as an equation de ning the value of z Solve that equation for the value of z and substitute that value into the equation above Solve the resulting equation for the value of y so determined Substitute both the value of y and z into the remaining equation which is then solved for it Check your work by invoking Maple s builtin backsub command gt backsnbA1 Exercise 3 The rank of a system of equations is the number of equations left after eliminating dependent equations This number does not depend 011 which equations were kept or eliminated Hence it is plausible that the rows in the reduced row echelon form of the system s matrix re ect the distinct equations that would survive an elimination of dependent equations Hence the rank of the system should be the number ol39nonzero rows in the reduced row echelon form of the matrix for the system 20 14 llaple On Line Check this conjecture experimentally by creating 4 x 5 matrices Al A2 A3 and A4 with ranks respectively 1 2 3 4 In particular insure that no matrix has a zero entry A process for creating a random matrix of prescribed rank is based 011 l39orming rows that are themselves linear combinations of other rows Begin by de ning l39 a function returning a random integer in the closed interval 4010 This is done with the rand command gt f rand10 10 We begin by constructing a matrix A1 of rank 1 This requires that the rows ol39Al be linear combinations of a single row Begin by constructing a l x 5 matrix Ml by using the function f in conjunction with the randnlatrix command to guarantee that the random matrix has entries that are integers in the interval 4010 gt M1 randmatrix1 5 entriesf From Ml build a 2 x 5 matrix M2 in which the rows are multiples of the single row in Ml Maple s stack command assembles a rows or vectors into a new matrix making the buildingblocks into the rows of the new matrix gt M2 stackM1 evalmfrowM1 1 From M2 build a 3 x 5 matrix M3 in which the rows are random linear combinations of the rows in M2 A single row in M2 can be referenced by the row command as illustrated below gt M3 stackM2evalmfrowM2 1frowM22 Finally build the required 4 x 5 matrix Al by taking linear combina tions of the rows of M3 gt A1 stackM3evalmfrowM3 1 frowM32 frowM33 Test that Al has rank 1 by invoking Maple s builtin rank command gt rankA1 The process for constructing random matrices of rank 2 is similar The only difference is that we start with a random 2 x 5 matrix produced using the randnlatrix command instead of a l x 5 matrix Similarly for a rank 3 matrix we would begin with a random 3 x 5 matrix and for a rank 4 matrix we would begin with a random 4 x 5 matrix The rank of the matrices Al A2 A3 and A4 can be corroborated by reducing each to reduced echelon form The following loop implements the required calculations 14 llaple On Line 21 gt for k from 1 to 4 do rrefAk 0d In each case the number of distinct nonzero rows exactly matches the known rank of the matrix These row reductions are exact without round olf error since Maple computes symbolically unless told otherwise How ever all computing devices when computing with oating point numbers can experience dif culties attributable to roundoff and truncation errors Examine this issue in Map e gt for k from 1 to 4 do rrefmapconvertAkfloat 0d For the matrices created in this session remember we are using a ran dom process rrel39A3 is wrong The n ef command declares that a small number which ought to be seen as zero is not zero Hence it suggests the rank of A3 is four ne defense against such numeric errors is the gausselinl command which row reduces a matrix but does not make the diagonal elements 1 By not dividing by the diagonal elements this command is less likely to err in numeric computations Let B3 be the oating point version of A3 obtained by mapping the process of conversion to oats onto gt B3 mapconvertA3float Now apply gausselinl gt q gansselimB3 The small entries in the fourth row should be taken as 0 s These are the numbers that rrel39 sees as nonzero leading to errors In Maple we can apply the fnorrnal command to set to zero numbers smaller than a given tolerance As with all operations applied to matrices and vectors the fnorrnal command is mapped onto the matrix 01 gt mapfnormalq 10 Exercise 4 Row reduce the transposes ol39 the matrices Al A2 A3 and A4 constructed in Exercise 3 The Maple command for the transpose is 7 transposeA What do you notice about the rank of the resulting matrices Exercise 5 Define vectors X Y and Z as indicated below gt X vector12543 Y 2 V6Ct0r613982103 Z vector51219241 22 1311aple On Line Determine whiCh ol39 the veCtors U and V below is in the span of X Y and Z Solve a determining system of equations by using the 139139ef command gt U vector52341468J v vector22o22o34 The question requires solving a X b Y C Z U and a X b Y C Z V for constants a b and C Both sets of equations Can be solved at the same time if the following augmented matrix is formed gt q angmentXYZUV Row reduCing via the rrel39 Command gives solutions to both systems of equations at the same time b Imagine that you are the head of an engineering group and that you have a Computer teChniCian working for you who knows absolutely nothing about linear algebra other than how to enter matriCes and Commands into Maple You need to tell your teChniCian how to do problems similar to part above SpeCiliCally you will give the teChniCian an initial set of three veCtors X Y and Z from R3 You will then provide an additional veCtor U and you want the teChniCian to determine whether U is in the span of X Y and Z Write a brief set ofinstruCtions whiCh will tell your teChniCian how to do this job Be as expliCit as possible Remember that the teChniCian Cannot do linear algebra You must provide instruCtions 011 how to ConstruCt the neCessary matriCes what to do with them and how to int gtrpret the answers The final output to you should be a simple W s or No You don t want to see matriCes C ne of your assistant engineers Comments that it would be easier for the teChniCian in part b to use Maple s rank Command rather than l39r39ef What does your assistant have in mind 1 5 On Line Alter Clearing Maple s memory by issuing a restart Command and re initializing by loading the linalg paCkage enter the matrix A and the veCtor gt A matrix34 12135722134 4 3J gt X vector1324J Obtain the produCt B A X You should Consult On Line SeCtion 11 for a disCussion of matrix produCts in Maple 15 llaplc On Line 23 gt BevalmAampX Exercise 1 The matrix multiplication A X just obtained represents a linear combi nation of the columns of A with coef cients taken from the vector X Implement this notion and show the result is the vector 13 found in the Introduction olumns of A can be referenced with the col command and elements of the vector X can be referenced as Xk Hence the brute force way of obtaining the required linear combination would be with the following syntax gt evalmcolA1X1 colA2X2 colA3X3 colA4X4 Since the columns of A are referenced in numerical order with an ill dex that is repeated when referencing the components of X it should be possible to form the same linear combination of columns with some sort of 39 xactly the mathematical sigma notation Zkl 1 k 36 There is one syntactical quirk to overcome however The col command requires a value for the index before the sum command can provide it so naive use of the notation will a E E S O I U H O a z a v 7 D 3 v 7 E w p v S E w a E E E S 5 Cr z E E 9 Cr result in a syntax error The trick is to put single forward quotes on the col command thereby preventing it from demanding priority in getting a value of the index before the sum command is ready to provide it gt evalmsum colAk Xk k1 4 Exercise 2 Solve the system A X B for X Keep in mind that B was formed by multiplying A against X This exercise seeks to determine whether or not X can be recovered from B One method of solution consists of row reducing the augmented matrix AB then using back substitution implemented in Maple via the backsub command gt C rrefaugment AB gt X1 backsubC By inspection determine a value of the parameter 11 that makes the general solution in X1 become precisely X Remember this parameter is a subscripted quantity and can be addressed in Maple via the syntax tl 24 1311aple On Line Exercise 3 Another method for finding the general solution first obtained in Exercise 3 is predicated 011 finding a basis for the null space ofA This basis can be found via the Maple command nullspace as shown below gt q nullspaceA Observe that the nullspace command returns a set of vectors Here there is but one member in the set a single vector that can be addressed via the syntax gt Zq1 Verification that Z is indeed in the null space of A resides in the product A Z gt evalmA amp Z The general solution for the system A X B is then Xg X t Z where t is an arbitrary parameter Form Xg and show that it satisfies the equation A Xg B By inspection determine a value of the parameter t in Xg for which Xg becomes exactly X1 the first form of the general solution found above Exercise 4 Enter the matrix A and the vector 13 as shown below A matrix461761327641946332579552461061996689 36321eo3271o4 gt B vector1745589 Determine the rank ofA From this information determine how many free variables the system A X 0 wi 1ave b How many spanning vectors will the null space of A contain c Using Maple s nullspace command find a spanning set for the null space of A Since this command returns a set of vectors extract all the vectors from this set naming them wl w2 etc gt q nullspaceA gt for k from 1 to 4 do wk quJ 0d d By inspection find a vector X satisfying the equation A X B Verify that your guess indeed satisfies the equation e If F is a general linear combination of the vectors W1 w2 W4 show that C X F is still a solution to the equation A X 13 Note 15 llaple On Line 25 You might need lVIaple s print command as well as evalrn to force lVIaple to display the results of your computations 139 Explain the statement 7 The general solution to A X B is the vector Xo W Y where W is the matrix whose columns are W1 W2 W3 and W4 and Y is any vector in R4 Hint omputationally it would be useful to form the matrix W with the augment command form the vector Y with four parameters for components and to find the general solution to A X 13 via the linsolve command This solution should match Xo W Y g Find a basis for the null space of A by solving the equation A X 0 for the general solution X This is easily done in Maple by using the linsolve command This command takes as arguments the matrix A and a vector or list of zeros as the righthand side values Maple will deliver a linear combination of the vectors W1 W2 W4 that were found by the nullspace commanc h Find a basis for the null space of A this time solving the system A X 0 by augmenting A with a column of zeros and using the l l ef and backsub commands The solution will not be readily recognized as a linear combination of the vectors W1 W2 W4 i Verify that the basis found in part h is equivalent to the basis W1 W2 W4 show that the set of equations 1 W1 6W2 t W3 1 W4 v16 has a solution for each mg in the basis found in part h This is most easily done by augmenting W with the general solution found in part h and using the rrel39 command to show the equations are consistent for any values of the parameters in that general solution 15 llaplc On Linc Chapter 2 Dimension 21 On Line Restart Maple to clear its memory of all variables and reinitialize by loading the linalg package Exercise 1 Given the matrix A entered below gt A matrix43123451321111 nd the reduced row echelon form of A How can you tell just from this reduced form that the columns of A are independent Relate your answer to Theorem 1 Exercise 2 Let A be a random matrix with more rows than columns State a gen eral rule for using rrel39A to decide whether or not the columns of A are independent Demonstrate your condition by producing a 5 x 4 matrix A with no nonzero entries and with independent columns and by b producing a 5 x 4 matrix with no nonzero entries and with dependent columns In each case obtain rrel39A Prove your condition using Theorem Exercise 3 Let A be the matrix entered below 27 28 22 llaplc On Linc gt A matrix5e126814324185131491oo32112 14571111199 Ifse the rref command to find the pivot columns of39 A Write them explicitly as columns Then express the other columns of39 A as linear com binations of39 the pivot columns See Example 4 in the text You should discover that the first three columns of A are the pivot columns Exercise 4 If Ak represents the kth column of39 the matrix A defined in Exercise 3 form the matrix B whose columns are the columns of39 A in the following order B 1 4 Ag Al Ag Ag A5 This is most easily done by using the augment and col commands Find the pivot columns of39 B by using the rref command Do you obtain a different set of39 pivot columns Use rref39B to express the other columns of39B as linear combinations of39 the pivot columns Could you have derived these expressions from those in Exercise 3 If so how Exercise 5 Find a matrix 7 whose columns are just those of39 A listed in a different order such that the column of39 T which equals A5 and the column which equals Al are both pivot columns Is it possible to find such a T where Ag is a pivot column as well If so find an example If not explain why it is not possible 22 On Line Restart lVIaple to clear its memory of39 all variables and reinitialize by load ing the linalg package In addition use the command 7 withstudent to load the sludcnl package in order to access its equate command Exercise 1 Let A be the matrix entered below gt A matrix563912639839143241398539139314 399 3910 0 3 392 391 12 39 14571111199 22 llaple On Line 29 Part a Find the rank of A via the rank command Using only the value of the rank explain why the statements 139 and ii given below are true i The reduced form of the augmented matrix for the system A X 0 has three free variables Recall that in a previous 011 Line section it was noted that the rank is the number of nonzero rows in the reduced form ii The null space of A has dimension 3 Hint How many spanning vectors are there in the general solution to A X 0 Part b Show that each of the vectors X1 X2and X3 given below satisfy A X 0 gt X1 vector51310231 X2 vector3611125 X3 vector479516 Note Since verifying that A XC 0 is a repetitive task it can be done in a forloop Part By computing the rank of the matrix X1 X2 X3 prove that X1 X2 and X3 are linearly independent Recall that the maximal number of linearly independent columns equals the rank Part 1 How does it follow that the dimension of the null space of A is 3 How does it follow that the XC constitute a basis for the null space Part e Using Maple s nullspace command find a basis for the null space of A Express each vector in this basis as a linear combination of the Xk s from part d Hint Given two bases for this null space showing that they are equivalent requires showing that any vector in one can be found as a linear combination of the vectors in the other Hence a set of equations of the form a X1 b X2 c X3 wk must be solved for each k l 2 3 This can be done by forming the augmented matrix X1 X2 X3 W1 W2 W3 and row reducing In row reduced form this matrix will indicate whether or not these equations are solvable and if so how to express the nonpivot 30 23 llaple On Line vectors in terms of the pivot vectors See Theorem 1 and Example 4 in Section 21 Part f In part e what made us so sure that the rst three columns would be the pivot columns Why couldn t for example the pivot columns be columns 1 3 and 4 Hint Think about what this would imply for the reduced form 0139 XL X2 X3 Part g In part you expressed the vectors wk in terms of the vectors X16 In this part now express the XC in terms of the wk This would complete the demonstration that the XC and the wk are equivalent spanning sets for the null space of A Hint Use the technique in part e Part 11 a Find by inspection a vector T which solves the equation A X H gt gt gt Part 7 Let 8 be an arbitrary element of R3 and let Z T r X1 s X2 1 t X3 where T is as found in part h ompute A Z Explain why you get what you get Find constants u v and w such that Z T u w v w2 w W3 What theorem does this demonstrate Note An ef cient eay of doing this is to use the equate command from Maple One might first enter the two proposed expressions for Z as follows gt q1 evalmaugment X 1 v v 3 ampvector rs t q2 evalmaugmentwv 1 v v 3 We can equate these two vectors with the equate command from the sludcnl package and then solve for the required constants using the following syntax gt 13 equateq1q2 gt 14 solveq3 uvw q5 solveq3 rs t 23 llaplc On Linc 31 23 On Line Restart Maple to clear all variables and reset its memory then initialize by loading the linalg package Exercise 1 Using Maple s randnlatrix command construct M a random 3 x 5 matrix What do you expect for the rank of M heck using the rank command Is it conceivable that the rank could have turned out otherwise Why is it unlikely Finally retain this matrix for use in the other exercises of this section Exercise 2 Form two different random linear combinations of the three rows of M then append these two rows to M thereby creating a 5 x 5 matrix It helps to use rand to define a function f as a generator for the random coellicients needed for the linear combinations Then the sum and row commands simplil39y constructing the linear combinations of the rows of M Finally the stack command appends rows to the bottom of M See Exercise 3 from the On Line exercises for Section 14 Using both rank and rref test the rank of the enlarged matrix What is the maximal number of linearly independent columns in the new matrix Exercise 3 For the 5 x 5 matrix created in Exercise 2 nd a set of columns which forms a basis for the column space Express the other columns as linear combinations of these columns Ifse the technique of Example 3 in Section Exercise 4 In this exercise you will explore Maple s ability to obtain the reduced row echelon l39orm numerically In Exercise 3 the reduction is found symbolically using exact arithmetic and the result suffers no loss of accuracy from round off error A oating point evaluation of this exact answer will serve as the target answer that we will expect Maple to deliver numerically First apply the evalf command to the reduced row echelon matrix found in Exercise 3 Next by mapping the converttolloat operation onto it convert the 5 x 5 matrix of Exercise 2 to oating point l39orm See 32 23 llaple On Line Exercise 1 in the On Line exercises for Section 14 Then obtain the reduced row echelon form of this numeric matrix Observe that the result is wildly wrong In fact it shows the matrix to be of rank 4 whereas the matrix is known to have rank 3 gt mm mapconvertMMfloat gt mm1rref mm The reason for the error can be seen by row reducing the matrix to an uppertriangular form without dividing the diagonal elements to make them l s This will prevent division by possible small numbers This row reduction can be accomplished via the gausselinl command gt q gausselimmm So although the work was done numerically the only dif culty that has surfaced is the small positive value 011 the main diagonal a value that in exact arithmetic would be zero To get Maple to render such small values as zeros map the fnorrnal command onto the matrix The fnorrnal command takes as additional parameter an integer specifying the number of digits to which the rounding is to be performed gt rref mapfnormal q 9 This result now matches what was obtained when the exact solulion was converted to oating point form Exercise 5 Find a basis for the column space of the 5 x 5 of Exercise 2 by row reducing its transpose and invoking the NonZero Rows Theorem Express each of the basis columns found in Exercise 3 as linear combinations of the basis columns found by this technique Exercise 6 Create four random 4 x 4 matrices of rank 1 2 3 and 4 respectively See Exercise 3 from the On Line exercises for Section 14 Determine the null space and the dimension of the null space called nullin in some texts for each matrix Relate the rank the nullity and the number of rows in the matrix Chapter 3 Transformations 31 Section 31 On Line Restart Maple to clear all variables then load the linalg and plots packages The exercises in this section deal with transformations in the plane Hence constructing several different types of graphs needed in this section requires the plots package Exercise 1 Create a 2 x 11 matrix F whose 11 columns are the coordinates of certain points in the plane These points are the endpoints of line segments that constitute a letter of the alphabet Because it might be tedious to con struct the letter 7 7 with line segments feel free to select some more easily constructed letter such as T Sketch the letter of your choice on a piece of paper using a minimum number of arcs and a maximum number of line segments Pick an endpoint of some segment as the origin and assign coordinates to each of the other endpoints Sketching the letter 011 a sheet of graph paper might make this easier Enter into the matrix F the coordinates of the endpoints of the segments making up your letter Each column represents an endpoint Start at an extremity and use contiguous columns to represent points connected by a line segment If your letter requires you to retrace a segment this happens with the arms in letters like E and F list the coordinates of any endpoints in the order in which they are traversed For example a recognizable letter F can be represented by the matrix 33 34 31 llaplc On Linc 0 0 2 0 0 g 0 2 2 2 l l where the origin has been taken at the base of the vertical stroke Save the worksheet in which you have entered your matrix F since it will be used in the remaining exercises Exercise 2 Since you will need and want to plot your letter create the following Maple function that takes as input the name of the matrix of your letter and returns a plot of the letter The effort required to type in this function will more than pay for itself as you experiment with these exercises x gt pointplot seqconvert colx k list k1 t t coldimx styleline axesboxed scalingconstrained Obtain a plot of your letter by applying the function f to the variable F associated with the data for your letter gt fF Exercise 3 Example 1 iron g iction 31 contains a matrix M which represents a 7 shear along the xa 39 all this shear matrix Say and apply it to the letter stored in F by forming the matrix product Sag F Plot the image of the sheared letter Exercise 4 Construct the rotation matrix corresponding to a counterclockwise rotation of 20 degrees Apply this rotation matrix to your letter and plot the result Since you will need other rotations it will be more ef cient if you build R a rotation matrix generating function that accepts as input the num ber of degrees counterclockwise through which the rotation is to take place and returns the matrix for this rotation Maple s arrow notation for building functions is appropriate here An appropriate syntax would be gt R x gt matrix22cosxsinxsinxcosx gt R20 R20Pi180 31 llaplc On Linc 35 Rotate your letter by multiplying it by R20 and plot the rotated letter by invoking the plotting function f built in Exercise 2 Exercise 5 Create another letter reduce it to a matrix ofcoordinates and store it in an appropriate variable For example the letter E can be created by a simple modification of the matrix representing the letter F and its matrix stored in the variable E Plot the new letter Then in anticipation of plotting the combination F E determine a way to plot the result of shifting E three units to the right all your shifted letter TE Hint Since the translate of E three units to the right would have each xcoordinate increased by 3 you want to find a simple way to add 3 to each element in the first row of the matrix E The Maple syntax 337 writes a sequence of seven 3 s separated by commas Hence the following matrix has row of threes and a row of zeros gt T matrix27 3707 Next plot the combination FE The utility function f which we built for graphing letters is not sophisticated enough to accept multiple inputs the way the standard plot functions in Maple will Hence create plots of each letter and combine the resulting graphics objects with the display com mand Assign the plot of each letter to a variable being sure to terminate each command with a colon Then invoke display gt p1fTE gt p2fF gt displayp1p2 Exercise 6 Let S be the transformation of R2 to itself wherein 3X is a shift of X one unit to the right Show graphically that S is not linear Specifically use the letter created in Exercise 1 to show that 32 X 9E 2 3X Exercise 7 For each of the following find a matrix M for which multiplication would accomplish the indicated transformation In each case validate your matrix by applying it to the letter created in Exercise 1 36 32 llaple On Line Part a Ma ips a letter upside down Part 1 Mb ips a letter lel39ttoright Part 6 Mc rotates a letter by 20 degrees counterclockwise Part d Md is a shear along yaXis Exercise 8 Plot the effect of each of the following transl39orinations applied to the letter created in Exercise 1 Part a A shear along the XaXis followed by rotation of 20 degrees counterclockwise Part 1 Rotate 20 degrees counterclockwise then shear along the XaXis Part c Shear along the Xaxis followed by shear along the yaXis Part d Shear along the yaXis followed by a shear along the Xaxis 32 llaple On Line 3 32 On Line Restart Maple to clear its memory of all variables then reinitialize by loading both the linalg and plots packages These exercises will continue the study the geometric aspect of transfor mations in R3 For this work it will be useful to again define the function f used in On Line Section 31 In that section the function took in a matrix representing a plane figure and returned a plot of the figure Exercise 1 In place of the letters of the alphabet that were used in the exercises of Section 31 use line segments to create the outline of a car Draw the outline 011 a piece of paper even a sheet of graph paper pick for the origin the endpoint of one segment and find the coordinates of the endpoints of all the line segments Enter these coordinates as columns of a matrix 7 Hence you might have a matrix such as the following gt c matrix01325600 1 52 20 O representing the car whose shape is given by the following graph gt f C For the reader who believes this car looks more like a flatiron we give the data points for a car that is signi cantly better looking The interested reader can enter the data into an appropriate matrix and run the experiments in these exercises with the improved image 6 9510 1 104e1 1612 1125 2237 2121 2336 2513 2161 2161 3187 3158 3172 3553 3912 3681 1611 3756 3783 3783 6373 3783 6716 3681 6891 3526 3388 3191 7366 2961 7169 2763 7876 2661 2566 8391 2368 8172 2672 8197 1612 1283 8497 1020 20e2 1053 Transform your plane image into a 3d object by altering the matrix T as follows Add a middle row of zeros by first adding a third row of zeros then swapping the second and third rows This can be done in Maple by first stacking T with a row of zeros then using the swaprow command to interchange the new row of zeros with the original row 2 gt d swaprowstackC O6 23 Since several 3d plots will be required it is very useful to define a function f3 that will take in a matrix representing a 3d object and return a 3d plot of the object so represented 38 32 llaplc On Linc gt f3 ugt pointplot3dseqconvertcoluk list k1 v coldimu styleline axesboxed scalingconstrained colorblack labelsxzy labeliont TIMES BOLD 14 Having defined the function 13 apply it to the matrix d which represents a first version of39a 3d car Since this is a 3d plot it can be rotated in Maple by clicking 011 the image and then using the mouse to 7grab and 7rotate the bounding box flicking 011 the R in the toolbar redraws the graph gt f3d Exercise 2 Further dimension can be added to the image of the car by adding 14 to each element in row 3 of the matrix d then augmenting the matrix d with the altered version e One way to do this is to assemble via stack the first row of d the altered second row and then the third row gt e stackrowd 1 evalmrowd214 rowd3 gt F augmentde Test the ef cacy of these improvements by plotting using the function Exercise 3 Add some substance to the car by sketching in diagonal lines 011 each of the narrow faces This requires alternating the columns of the augmented matrix P so that the first column comes from d the second from matrix e etc In effect build a new matrix FF by augmenting pairs of columns of the form clC ck This is accomplished in Maple via the syntax gt FF augment F seqop coldk cole k k1 v v coldimC The validation of the manipulation is in the plotting gt f3FF Exercise 4 Some of the transformations that will be applied to the car include rota tions To keep the rotated car in the viewing window it will help to move the origin to the center of the car Deduce the coordinates of this center and move the origin to that point by subtracting appropriate constants 33 llaplc On Linc 39 from the first and third rows of the matrix FF Remember that the first row records xcoordinates the second row zcoordinates and the third row ycoordinates Form this new matrix by altering the appropriate rows and reassembling them into a new matrix gt FFF stackevalmrowFF 13 rowFF2 evalmrowFF332 Next rotate the figure counterclockwise 30 degrees about the xaxis then rotate that image 20 degrees counterclockwise about the zaxis This is most easily done by building functions that yield the appropriate three dimensional rotation matrices Exercises 10 11 and 12 in Section 31 then invoking the functions for the required angles See Exercise 4 in the On Line section for Section 31 Exercise 5 Obtain a single matrix whose action under multiplication reproduces the two successive rotations implemented in Exercise 4 Validate your single matrix by again plotting the rotated car Exercise 6 What image would you see if you transformed the matrix FFF by a rank 2 transformation Create a random rank 2 matrix and test your guess After printing graphs of the transformed and untransformed images attempt to label several points where the transformation is manytoone The entries in the transformation matrix should be random numbers in the range 11 lest the scale of the car be altered completely Exercise 7 What image would you see if you transformed the matrix FFF by a rank 1 transformation Create a random rank 1 matrix and test your guess Again be sure to restrict the entries in your random matrix to the range 11 33 On Line Restart Maple to clear its memory of all variables then reinitialize by loading the linalg plots and 110110015 packages 40 33 llaple On Line Exercise 1 Create M a random 2 x 3 matrix with rank 1 See Exercise 3 in Section 14 If M is interpreted as the matrix of a transformation acting 011 R3 what should the dimension of the image of this transformation be Verify this by creating 100 random points in R3 transforming them under M and plotting the points Next generate P a matrix containing 100 random points in R3 A moment s thought about how the transformed points will be plotted will determine the optimum strategy for generating and plotting the points If the points are stored as columns of a matrix they can be plotted by apply ing the pointplot command to the matrix Hence let P be of dimension 3 x 100 so the product M P will be 2 x 100 reate P by juxtaposing via augnlent 100 vectors generated by randvector Terminating the com mands with a colon signals Maple not to print the rather large outputs to the screen Finally anticipating Exercise 2 where this plot will be re quired store the plot data structure in a variable say pl so later other images can be superimposed 011 it gt P augmentseqrandvector3 k1 v v 100 gt S evalmMampP gt p1 pointplotS gt p1 Exercise 2 The plot generated in Exercise 1 should show the span of any nonzero column of M Demonstrate this by choosing a column of M and plotting 011 the graph from Exercise l 100 random points in the span of this column Define the function f which generates a random integer in the closed interval 500500 Then using the seq command give to pointplot a sequence of 100 random multiples of the first column of M olor the points red and assign the plot data structure to a variable say p2 After viewing the graph merge it with the plot from Exercise 1 by use of the display command Assign this composite graph to a variable say p3 for use in Exercise 3 gt f rand500 500 gt p2 pointplotseqevalmcolM1fk1400 Color red p2 gt p3 displayp1p2 p3 34 llaple On Line 11 Exercise 3 Using information and insights from Exercises 1 and 2 find a specific vector 13 in R2 for which the equation M X B is no solvable and b a vector C in R2 for which the equation M X C is solvable Indicate these vectors 011 the composite graph produced in Exercise 2 Verify your answers by computing via l l ef the reduced row echelon forms of the augmented matrices M3 and MC Exercise 4 Plot 100 random elements from the null space of M Maple s nullspace command will provide a basis for the null space of M This basis is returned as a set of vectors which can be extracted from the set with the bracket notation The seq command can be used to generate a sequence of 100 random linear combinations of these basis vectors a sequence Which can then be plotted in R3 with the pointplot3d command The 3d graph so generated can be rotated onscreen by grabbing and rotating the bounding box It should then be possible to observe the nature of that portion of the span so generated Finally explain how this plot relates to the RankNullity Theorem Exercise 5 Randomly generate a rank 2 matrix M of dimension 3 x 3 Repeat Exercises 1 through 4 suitably modified to account for the different dimensions Speci cally this means you are to generate M Then in imitation of Exercise 1 the matrix P containing 100 random point in R3 and plot the product 1V1 P In imitation of Exercise 2 plot 100 random linear combina tions of the columns of M In imitation of Exercise 3 find vectors B and C for which the systems M X anc C are not solvable and solvable respectively Verify your choices of B and C computationally Fi nally in imitation of Exercise 4 plot 100 randomly chosen elements from the null space of M When you are done don t forget to relate your findings to the Rank Nullity Theorem 42 34 llaple On Line 34 On Line Restart Maple to clear its memory of all variables and reinitialize by loading the linalg and sludcnl packages Maple contains a variety of solvers l39or equations of various types For example the standard symbolic solver for one or several equations linear and nonlinear alike is solve The standard numeric solver for such equations would be fsolve oating point solve Differential equations are solved by dsolve difference equations are solved by L39solve recursive solve and Diophantine equations are solve by isolve integer solve In the linalg package if a set of linear equations is captured in the matrixvector format A X B then linsolve can be used Both solve and linsolve return general symbolic solutions when applicable In fact linsolve will even return solutions in terms of arbitrary parameters Other approaches to the solution of linear systems include the use of gausselinl followed by backsub or reef also followed by backsub Maple s com mand structure is rich enough that nearly any undergraduate mathematics that can be articulated in standard mathematical notation can probably be implemented in the context of Maple s builtin commands There are two cautions to observe when using Maple to solve linear sys tems First if the calculation is done in oating point arithmetic lVIaple is as liable to roundolf and truncation errors as any other numeric utility Second when working symbolically exact expressions for numbers can get dauntingly large thereby consuming memory and time Hence Maple can not solve symbolically systems as arge as some strictly numeric utilities can solve by working in oats Exercise 1 Let A be the matrix from Exercise of Section 34 Solve the system A X B where B 32 then convert the answer to oating point l39orm Next convert both A and B to oats by mapping the convert oat operator onto them Resolve the system and compare the two oating point results See Exercise 4 in Section 23 34 llaple On Line 13 Exercise 2 In many applications of linear algebra numerical data comes from mea surements which are susceptible to error Suppose the vector 13 in Exercise 1 was obtained by measuring a vector Ba whose actual value is Ba 321 Compute the solution to the equation A Xa Ba 440 lVIeasure error is the absolute value of the difference between the com puted value and the actual value It can be computed in Maple with the following command gt e mapabsevalmXXa Which component of the solution X computed in Exercise 1 has the largest error It might hepl to convert your answer to lloting point form In terms of the magnitude of the components of the inverse Ill l explain why this is the largest to be expected Maple computes the inverse of a matrix with the inverse command Which component of Ba would you change in order to produce the greatest change in Xa Why Back up your answer with a numerical example or with a symbolic calculation where the in Ba are l 39 39 appearing in each component How much error could you tolerate in the measured values of the components of B if the absolute value of the error in any entry of X is to be at most 0011 Exercise 3 Let A and B be as de ned below gt A matrix33112131213141314151 gt B vector834632 Find the solution to A X B As in Exercise 2 suppose the vector 13 was obtained by measuring a vector Ba whose actual value is Ba 8290 11 4607 Solve the equation A Xa Ba What is the percentage error in the least accurate entry of X How much error could you tolerate in the measured values of the components of B if the absolute value of the error in any entry of X is to be at most 001 44 34 llaple On Line Exercise 4 Exercises 2 and 3 demonstrate that the process of solving a system of equa tions can magnil39y errors in disastrous ways One quantitative measure of the inaccuracy of a calculation is the ratio of the percentage error in the final answer to the percentage error in the input data But what do we mean by the percentage error in a vector such as X in Exercises 1 and 2 in which every component might have errors of different magnitudes For vectors in R3 this question has a geometric meaning Think of X1 and X2 as representing points in R3 The distance d between these points is one measure of the error 1139 X1 1 yl z f and X2 12 12 22 then 4m 2 Lil WV 21 22V In Maple this can be computed as norm X1 X2 2 The additional 72 represents the 2norm 7 wherein differences are squared and the square root of the sum taken If X1 is the computed answer and X2 is the actual answer we define the percentage error to be P 100 norm Xi X2 2norm X1 2 Let 13 Ba X and Xa be as in Exercise 2 Ifse the given formula to compute the percentage error in 13 ii the percentage error in X and iii the ratio of the percentage error in X to that in B This is the inac curacy ol39 the calculation of X from B Assuming that accuracy is desired do we want the number I to be large or small Explain b Compute the inaccuracy of the computation of X from B in Exercise c For each n x n invertible matrix A there is a number condA the condition number of A such that the inaccuracy in solving the system A X B is at most condA regardless of B and regardless of the amount of error in B This means that if say condA 20 and the error in B is 001 then the computed value of X will have at most 20 x 001 02 error In general 1 g condA This says that we cannot expect the answer to be more accurate than the input data Maple has a builtin command for the condition number Since the condition number is constructed from a norm 7 of the matrix A we need to specify the version of the condition number we want according to the type of norm we want used Hence we will use the syntax condA2 In addition the 2norm of a symbolic matrix can be a very large and complex expression For this reason we will only compute the condition number based on the 2norm of matrices of oating point numbers 35 llaple On Line 15 Compute the condition numbers for the coelicient matrices in Exercise 1 and 3 Use this to explain the difference in accuracies obaine in these exercises Matrices with large condition numbers are called illconditioned If the coellicient matrix of a system is illconditioned then we must be extremely suspicious of answers obtained by solving the system since any slight error in the input data can make the solution very inaccurate Notice that these inaccuracies are not related to roundoff error Illconditioning is intrinsic in the matrix and not in the method of solution 35 On Line Restart Maple clearing its memory of all de ned variables Then re initialize by loading the linalg and sludcnt packages These exercises will explore LUdeconlp Maple s builtin command for obtaining the LL39 decomposition of a matrix The LUdecomp command can return seven different items namely L L39 a factorization of L7 into 391 and R a permutation matrix P the determinant of Fl and the rank of A We will not need all of these outputs and will restrict explorations to just L L39 and P The syntax for a multireturn Maple function is tedious Each variable that is to have a return assigned to it must be included in the command surrounded by single quotes Since the actual return of the LUdeconlp command is L39 we will not need to make a specific request for L7 to be returned Thus to obtain L L39 and any permutation P needed to complete the decomposition the appropriate syntax would be u LUdecomptA L l P p Of course if only L and L7 were desired then the command could be shortened to u LUdecompA L 1 Exercise 1 Enter the Matrix A from Example 1 ol39 the text then obtain the LU de composition by using the LUdeconlp command Assign the output of this command to the variable u and let 1 be the lower triangular factor This 46 35 llaple On Line matrix will not need a permutation P so it can be omitted from the com mand Finally verify that A l u You may need to use either print or evalrn to view the contents of 1 Exercise 2 Before examining the consequences pivoting has 011 the LL39 decomposition we 39tudy the notion of a matrix factorization When we demand that A be factored into the product LL39 with L being a lowertriangular and I being an uppertriangular matrix we are asking for the solution of a set of equations For example if A is a given 3 x 3 matrix then nding matrices L and If such that ALL39 is equivalent with solving a system of nine equations where each equations is obtained by setting one entry of A equal to the corresponding entry 0139 LL39 In this exercise we investigate this system for the matrix A of Exercise 1 Begin by forming L and 1 3x3 matrices of indeterminates Li and Uij lVIaple s Inatrix command with appropriate il39statements as an option will create the desired matrices gt L matrix33 ijgt if iltj then 0 else Lvivj ii U matrix33 ijgt if igtj then 0 else Uij i lVIultiply L and 17 forming a template of indeterminates which we then demand reduce to the entries ol39A l39rom Exercise 1 above This gives a set of nine equations in twelve unknowns since there are six unknowns in each of L and If The solution of this set of equations will not be unique suggesting that we can impose an additional three conditions 011 the factorization gt LU evalmL amp U The product LU and the matrix A can be equated via the equate com mand from the sludcnl package This command returns a set of equations formed by equating corresponding entries of each matrix gt q equateLUA Typically the solve command needs a set of equations and a set of variables If no variables are suggested to Maple it will attempt to deduce what the unknowns ol39 the problem actually are That is convenient here since it would be tedious to enter the names of all the variables in these equations gt q1 solveq It is clear that we did not get a unique solution for the entries ol39L and If arel39ul inspection shows there are three indeterminates in the answer 35 llaple On Line 17 This becomes more evident if we substitute these solutions into the matrices L anc L39 gt L1 snbsq1opL U1 snbsq1opU Exercise 3 Change the definition of the matrix L used in Exercise 2 Since there are three free parameters in the solution for the factors L and 1 choose to have the diagonal elements of L all be l s This can be done with an appropriate il39statement in the Inatrix command that defines L The matrix I will e the same as used in Exercise 2 Repeat the formation of nine equations in nine unknowns obtaining unique l39actors L and If for the matrix A in Exercise 1 Display the resulting matrices L and If and show that they are exactly the factors produced by the LUdeconlp command in Exercise 1 Exercise 4 In this exercise you will explore the concept of a permutation matrix P whose rows are a permutation of the rows of the identity matrix It a matrix A is multiplied by P the rows of PA will be permuted in the same way that the rows the identity were when forming P Let P be a permutation of the rows of the 4x4 identity This can be done in Maple by stacking a sequence of rows from the identity Create A a random 4x4 matrix and examine A PA and P gt H d diaglt14 gt P stackseqrowldk k 2431 gt A randmatrix44 gt PA evalmP amp A V printAPAP Finally observe that for a permutation matrix P the inverse is the transpose gt printinverseP transposeP 48 35 llaple On Line Exercise 5 We wish to study the effect of pivoting 011 the L1 decomposition For this we need a matrix A that forces a pivot If the LUelement of A were to be zero a pivot would have to be performed immediately since the rowreduction process by which we obtain L and If is basically gaussian elimination Create A a random 4x4 matrix and then reassign its 11 element the value zero Ifse the LUdecornp command to obtain the L1 decomposition this time including parameter P p so that a permutation matrix p will be returnec To have Maple display p l and u sidebyside use the print command gt u LUdecompAL l P p printplu Use the print command to display sidebyside the product lu the matrix A and the product plu You should observe that the product lu does not reproduce A That is the effect of pivoting Whenever rows must be interchanged during the factorization these interchanges are recorded in the matrix P The resulting LU factorization is then a factorization of Pt 1 A not A resulting in the equality Apt l A LIT Thus A PLU not just LU Chapter 4 Orthogonality 41 On Line Restart Maple to clear its memory of all de ned variables and reinitialize by loading the linalg plots sludcnl and 110110015 packages Given the vectors Ql and 2 2 we will generate a coordinate grid corresponding to a space in which Ql and 2 are the ba sis vectors VVe will plot the skewed grid lines in red atop a standard coordinate system in black gt Q1 vector 1 1 Q2 vector1 2 The vector equation of a line through the tip of the vector Ql and parallel to the vector 2 is given by Rtt 1 t 2 To form parallel lines through the tip of 2 1 3 1 etc we need to form the vectors k Ql t 2 with k an integer in some interval Alternatively the equations of lines parallel Ql through the tip of 2 2 2 etc we for the vectors k 2 t 1 In every case we let t the parameter of the line range over an interval bJ Maple s seq command can be used to generate an appropriate sequence of vector representations of the grid lines Plotting them in red will product the desired grid A template for the vector form of each set of grid lines is obtained as gt P1 evalmkQ1tQ2 P2 evalmkQ2tQ1 Sequences of equations of the skew grid lines are formed with the seq command 49 50 41 llaple On Line gt 51 seqP11P12t55k55 2 seqP21P22t55k The plot of the grid lines is assigned to the variable fl so that it can be used again in one more activity We include the scaling parameter in the plot command A 11 scaling can also be imposed interactively from the toolbar A view window is also set in the plot command gt f1 plotsl 52 colorred scalingconstrained View 6 66 61 ii For a finishing touch we use the arrow command from the 110110015 package to draw Ql and 2 the basis vectors of the skew coordinate system We plot the first in green and the second in blue superimposing both 011 the skewed red grid gt a1 arrow00 1 2 2 4 2colorgreen a2 arrow 00 1 1 2 display f1 al a2 Exercise 1 Modify the commands in the Introduction to produce red grid lines corre sponding to the basis vectors Ql 2 and 2 2 In addition produce a graph showing the skewed grid lines and the basis vectors in green and blue Assign this graph to a variable so that it can be reused in Exercise 4 Exercise 2 The curve defined implicitly by the equation 1 2 L2 l is a hyperbola that will be plotted in Exercise 3 Here show that in the basis of Exercise 1 with coordinates u and v the equation of this hyperbola is 411 v 1 Begin by deducing the equations of the transformation The point ma trix M is constructed with columns Ql and 2 Letting X and U the transformation equations are X M U The vectors X and M U can be equated with the equate command from the student package The equation X 1V1 U defines the change of basis from rycoordinates to tuxcoordinates Once these equations are obtained Maple s subs command can be used to impose this change of coordinates 011 the equation of the hyperbola An appropriate syntaxt might be gt q1 equateX MU 42 llaple On Line 51 gt 12 x 2 y 24 1 gt 13 snbsq1q2 gt 14 simplifyq3 Exercise 3 Use Maple s iniplicitplot command from the plots package to obtain a graph in the xycoordinates system of the original hyperbola Assign the graph to a variable so that in Exercise 4 it can be reused Be sure to use 11 scaling so that no distortion is introduced by the computer screen gt g3 implicitplotltq2x10 v 10y10 v 10colorblack scalingconstrained g3 Exercise 4 Use Maple s display command to superimpose on the skewed grid lines of Exercise 1 the graph of the hyperbola l39rom Exercise 3 Then use Maple s digitizer click 011 the graph click 011 a point in the graph read the coordi nates in the window at the topleft ol39 the graphics toolbar to approximate the coordinates of some point 011 the hyperbola By counting and esti mating infer the corresponding uvcoordinates Show that to within the accuracy of the digitizer your coordinates satisfy the equation of the hy perbola in the rysystem and in the uv system 1 o H 1 gt4 E i the point 24 appears to be almost 011 the hyperbola The digitizer gives 2 35 The same point appears to be 2 01 in the skew grid Hence gt snbsx2y35q2 gt subs n2v 1 q l 42 On Line Restart Maple clearing its memory of all de ned variables Then re initialize by loading the linalg and sludcnl packages You are working for an engineering rm and your boss insists that you find one single solution to the following system 2x3y4z3w 129 52 42 JIaple On Line 4c7y Gz 839w 7l 6clOy 2z 5w59 You object noting that The system is clearly inconsistent the sum of the first two equations contradicts the third b You need at least four equations to determine four unknowns uniquely Even if the system were solvable you couldn t produce just one solution The boss won t take no for an answer oncerning the boss points out that the system was obtained from measured data and any inconsis tencies can only be due to experimental error Indeed if any one of the constants 011 the right sides of the equations were modi ed by 1 units in the appropriate direction the system would be consistent oncerning b the boss says Do the best you can We will pass this data 011 to our customers and they wouldn t know what to do with multiple answers After some thought you realize that projections can help with the ill consistency problem The given system can be written in vector format as 2 3 4 3 129 1 4 y 7 z 6 10 8 71 6 10 2 5 59 You realize that this system would be solvable if the vector 011 the right side of the above equation were in the space spanned by the four vectors on the left Using Maple s rank command you quickly compute as 2 the rank of the system matrix 2 3 4 3 A 4 7 6 8 6 10 2 5 showing that these four vectors in fact span a plane all this plane W 129 Your idea is to let Bw be the projection of B 7l onto W 09 Since the system is so nearly consistent Bw should be very close to 13 Furthermore the system A X Bw should certainly be solvable and one of the solutions should be what the boss is looking or Your point b will require some further thought However you do eventually come up with an idea which will be described in the exercises which follow 42 llaple On Line 53 Before going on to the exercises it will be useful to enter the data of this system of equations Enter the matrix A and the vector 13 Convert B to exact rational form and call that vector b Compute the rank of A verifying that it is indeed 2 gt A matrix34 234347396398610 392 395 gt B vector1297159 gt b mapconvertBrational gt rankA Exercise 1 Find an orthogonal basis for the column space ol39A use the Fourier Theorem to obtain Bw the projection of 13 onto W the column space of A Then solve A X Bw expre ing the solution in parametric form Write X as a sum of a translation 7 vector and vectors in the null space of A Show that the translation vector Maple finds is not orthogonal to the null space of A First obtain an orthonormal basis for the span of the column space of A Since the columns of A are not linearly independent A has rank 2 determine via L39L39ef which two columns of A to take as independent and pass those two vectors to Maple s GraulSchnlidt command for orthogo nalization This command returns a list of lists not vectors so map the converttovector operator onto this output You will now have a list of two orthogonal vectors A11 appropriate syntax might be gt q mapconvertGramSchmidtcolA1colA2 vector Next apply the Fourier Theorem to get Bw the projection of 13 onto W the column space of A Work with b the exact version of the vector 13 The formula in the Fourier Theorem is implemented in Maple much as it is written mathematically Reference the basis vectors as ql and q2 compute dot products with lVIaple s dotprod command and norms with Maple s norm command being sure to compute the 2norm Next solve the system A X Bw Maple s linsolve command will yield the general solution This general solution can be put into the form X T a v1 3 v2 where the vectors v1 and v2 are in the null space of A These vectors can be extracted from this general solution by an adroit use of substitution via the subs command 54 42 llaple On Line Exercise 2 bjection b amounts to this If the system is inconsistent there is no solution If the system is consistent there are many solutions Even the use of projections in Exercise 1 has yielded a general solution that is not unique Perhaps your first thought was to report the translation vector as the solution But there is nothing special about this vector and the Translation Theorem says the general solution can be expressed using any particular solution not just the translation vector Your next idea however is sound With T as the translation vector let Tn be its projection onto the null space of A Report to your boss the solution X T Tn Why is this X a solution Ifse Maple s nullspace command to find a basis for the null space of A Then use the Fourier Theorem to obtain the projection of T onto this null space Finally form T Tu and explain why this is a solution Note however that the Fourier Theorem assumes an orthogonal basis for the space into which the projection occurs Hence you will need to apply Maple s GraInSchInidt command to the basis for the nullspace Remem ber though that GraInSchInidt returns a list of lists requiring us to convert the sublists back to vectors an was done in Exercise 1 Exercise 3 Try computing X in Exercise 2 by starting with a solution other than T You should get the same X W 1y It can be shown that the X found in Exercise 2 is the solution of minimal length Exercise 4 Show that of all solutions to A X Bw the one with minimal length is the solution X computed in Exercise 3 For this let ql be the general solution found by Maple in Exercise 1 considered as a twoparameter family of vectors Compute the 2norm and use calculus to minimize this norm The resulting solution should be the same X that was computed in Exercise Hint Use the subs command to repace the free parameters tl and t2 used by Maple with a and b calling the result Xg Now obtain the 2norm of the vector Xg Since Xg is a symbolic vector Maple returns the norm with absolute values thereby making it dif cult to differentiate and set derivatives equal to zero Simplify the norm of Xg adding the parameter symbolic to coax Maple to simplify the absolute values 43 llaple On Line g u gt f simplifynormXg2symbolic Differentiate with respect to a and 7 Use Maple s solve command to solve the system obtained by setting these partials equal to zero as follows Finally substitute these values into Xg and cmpare with X gt fa difffa fb difffb gt qq solvefafbab Exercise 5 Maple would have found the least squares solution of minimal norm with its builtin leastsqrs command Verify that this command yields the solu tion X found in Exercise 2 Note the inclusion of the parameter oplimizc which signals Maple to find the solution of minimal length Without this parameter the leastsqrs command will return the general solution found in Exercise 1 as the vector in 011 gt leastsqrsAboptimize gt leastsqrs Ab Exercise 6 Alter giving the boss your answer you delete all your data except for A and X A month later the customer calls saying WVe know that there must be other solutions Could you please provide us with the general solution 7 Show how the general solution can be reconstructed from A and X with a single Maple command 43 On Line Restart lVIaple to clear its memory of all previously defined variables Then reinitialize by loading the linalg and plots packages In addition enter the following lines of lVIaple code This code written by Dr Mike Monagan of Simon Frasier University in Burnaby British olombia i39anada creates a function that will generate the periodic extension of a function The code rst appeared in the article Tips for Maple Users and Programmers illaplcT I VOL 3 NO 3 1996 published by Dirkmus PE 7 procl39 drange subsT l39 L lhsd D rhsdlhsd procxalgebraic local y y lloorx LD FxyD end end 56 43 llaple On Line These lines of code can be entered into a separate Maple worksheet and that worksheet saved Later it the code is again needed that worksheet can be opened and the lines copied and pasted into the active worksheet There are of course other ways of saving code and making it accessible more easily but some aspects of that process are platl39orin dependent and will not be discussed here Exercise 1 Figure l of Section 66 in the text depicts a sawtooth function called a rasp Plot the first fourth and tenth Fourier sine approximations to this function The rasp is generated by the periodic extension of the function l39x x for x in the interval 11 To get Maple to plot the periodic extension of l39x use the function PE de ned in the Introduction above First be sure to define l39x with Maple s arrow notation thereby making 139 a function not an expression gt f x gt x De ne the function whose name is rasp as the periodic extension hence PE 0139 the function whose name is 139 Do this by applying the PE operator to 139 being sure to terminate the command with a colon since the output will look strange and probably unintelligible The arguments to PE are the function to be extended and the domain of the function being extended gt rasp PEf1 v 1 Obtain a graph of the rasp 011 the interval 33 assigning the plot to a variable for use later The plot option disconl rue signals Maple to observe the discontinuities in the function and tells it not to connect across the uinps gt f1plotraspxx3 3 disconttrue colorb1ack scalingconstrained thickness3 f1 Obtain the Fourier sine series coef cients 0 sintmlrx L L Here I l and l39x x An integral can be entered into Maple with the Int command which stores the integral as an unevaluated syinbol If instead the integral is entered with the int command the evaluation of the integral is immediate Here Int is used so that the integral will be displayed coinpletely gt q 2IntxsinnPix xO v 1 Cl 1 43 llaple On Line To evaluate an integral that has been entered with Int apply Maple s value command to the integral gt q1 value q We wish to simplify this expression For example mm 7r is zero when ever 11 is an integer We will tell Maple that n is an integer with its assume command However that will cause Maple to attach a tilde V to each n it prints thereafter Suppressing the attachment of the tildes can be done either interactively from the Options menu Options Assumed Variables No Annotation or from the command line with the following interface command gt interfaceshowassumed0 Now use the assume command to tell Maple that 71 is an integer gt assume n integer If the Fourier coef cients are now simplified they will appear much like they would if the calculation were done 7 by hand quot Note how in the interest of simplicity we assign the result to the name b and not to something that tries to re ect the dependence on n gt b simplifyq1 The Fourier approximations are simply partial sums of the Fourier se ries The first approximation pl is just 1 sintirx and the tenth one is 210 11 0 mm 726 We can obtain these expressions in Maple by using its sum command 1sumbsinnPix n1 1 p4sumbsinnPix n1 v 10 gt P p10sumbsinnPix n1 v Finally these three partial sums can be plotted with a single plot com mand by grouping the functions in a list olors can be assigned to the functions by the option color with matching colors being listed in the order of the functions to which they are being ascribed Assigning the plot to a variable allows merging via the display command of the plots package the approximations with the graph of the rasp created above plotp1 p4p10 x3 3colorred green blue gt f2 displayf1f2 58 43 llaplc On Line Exercise 2 1 gr lt 0 4 Let g c l lt t a p1ecew1se defined function btam Fourier sine approximations withil 8 and 20 sine functions The periodic extension of gx is called a square wave Note the ear like peaks which appear in the graph of the partial sums at the discontinu ities of fx These peaks are referred to as the ibbs phenomenon They are quite pronounced even after twenty terms of the Fourier series Their existence shows that it takes a very high fidelity amplifier to reproduce a square wave accurately For this reason square waves are sometimes used to test the fidelity of an amplifier Begin by defining gx as a piecewise function 011 the interval 11 Use Maple s piecewise function which permits the definition a function with multiple formulas Define g as a function by using the arrow notation gt g x gt piecewisexltO1xgt01 Check the behavior of gx by plotting it again using the plot option disconl rue so that jumps in the function are not connected gt plotgxx1 1disconttrue color black Define G as the periodic extension of the function g Use the PE code detailed in the Introduction Again end the command with a clon since the echo will probably not make much sense to the typical student of linear algebra Plot G 011 the interval 33 assigning the plot to a variable for use later As in Exercise 1 the Fourier sine coeflicients 0 are computed by inte grating gx Enter the defining integral using int and gtx The presump tion here is that the interface and assume commands are still operative from Exercise 1 If not reexecute those commands Exercise 3 Obtain the thirtieth partial sum of the Fourier sine series for the function gr x3 39 g gr S 1 then graph the periodic extension of fx and the Fourier approximation Explain why the graphs don t agree See Exercises 6 and 7 of Section 66 Exercise 4 Obtain a Fourier cosine series for the function in Exercise 3 Plot the first fourth and eighth partial sums 44 llaplc On Linc 59 44 On Line Restart Maple to clear its memory of all de ned variables and reinitialize by loading the linalg and plots packages Let be the matrix of a counterclockwise rotation around the r axis and through an angle of radians Let RA be the matrix of a counterclockwise rotation around the y axis and through an angle of radians Let A RAE Since the product of two orthogonal matrices is orthogonal A is or hogonal The purpose of these exercises is a demonstration that A de nes a rotation about a xed axis and through a particular angle See Figure 5 in Section 44 of the text Points 011 this axis remain fixed under the rotation Thus if X is 011 the axis of rotation it will satisfy AX X or equivalently A IX 0 Exercise 1 Construct the matrix A as the product of the matrices and Egg Exercise 2 Find an X 011 the axis of rotation by using the equation A IX 0 Thus X is in the null space of A I Such an X can be found by applying Maple s nullspace command to A I which Maple lets us form via the syntax A l The nullspace command returns a set of vectors so X will have to be extracted from this set Exercise 3 Plot the line segment from X to X If we parametrize this line segment as tX we can plot it with the spacecurve command letting t lie in the interval 11 Assign this plot to a variable so it can be used in a later exercise gt Hi evalmtX gt f1 spacecurvetXt1 v v 1 colorb1ack axesboxed scalingconstrained labels x y z labelfont TIMES BOLD 14 30 44 llaplc On Linc Exercise 4 The plane P through the origin perpendicular to X is called the 7 plane of rotation Since this plane contains the origin it is a subspace of R3 If the vector X is converted to a l x 3 matrix Maple s nullspace command will produce a basis for the plane P the null space of the matrixl39orm of X Why btain this basis nameing its elements N1 and N2 Plot line segments through N1 and N2 as was done in Exercise 3 Join this plot with the one from rom Exercise 3 with display3d from the plots package and assign the merged graphs to a variable for use later in Exercise 4 Be sure to use a 11 aspect ratio so that orthogonal vectors appear orthogonal The conversion of X to a matrix is accomplished by Maple s convert command with matrix 7 as the parameter This will be a 3 x 1 matrix which the nullspace command will reject Apply the transpose operator to produce a l x 3 matrix to give to the nullspace command Maple s nullspace command does not yield normalized vectors After obtaining the basis of the null space normalize the vectors with Maple s nornlalize command Show that the basis vectors are not necessarily orthogonal to each other They are orthogonal to X Use Maple s dotprod command to compute the dot products of vectors Exercise 5 The expectation should be that multiplication by A rotates elements within the plane orthogonal to the axis of rotation To test this hypothesis mul tiply both N1 and N2 by A Then use Maple s angle command to find the angle between N1 and A N1 and between N2 and A N2 A second multiplication by A should rotate A N1 and A N2 by the same amount Verify this Exercise 6 Continue to explore by visual means the idea of rotations in the plane P Create a plot of 15 successive applications of the matrix A to the basis vectors found in Exercise 2 If each application of A rotates these basis vectors through a fixed angle and if they remain in P their collective image should show 7 the plane P Since working with exact expressions can lead to memoryconsuming expression swell it is wise to convert the computations to the numeric 45 llaple On Line 61 domain Map the convert operator with the oal option onto the ma trix A and the vectors N1 and N2 coining new names for these numeric versions For example we might call the floating point versions of these quantities B NNl and NNZ respectively lVIultiplication of NNl by B ktimes produces BkNNl We can pro duce the desired plots using Maple s seq command as follows Finally use display3d to merge these plots with the plot from Exercise 4 gt 51 seqevalmtB k amp NN1k115 2 seqevalmtB k 86 NN2 k1 v 15 gt f4 spacecurveslt0 1colorblue f5 spacecurves2t0 1colorgreen Exercise 7 Determine in the plane P the angle of rotation caused by multiplication by A Ifse Maple s angle command but express the answer in degrees as a oating point number 45 On Line Restart Maple to clear its memory of all defined variables and then re initialize by loading the linalg and plots packages Exercise 1 Imagine that you are an astronomer who is investigating the orbit of a newly discovered asteroid You want to determine what is the closest the asteroid will come to the sun and b what is the furthest away from the sun the asteroid will get To solve your problem you will make use of the following facts Asteroids have orbits which are approximately elliptical with the sun as one focus b In polar coordinates an ellipse with one focus at the origin can be described by a formula of the form H c lasin lzcos9 G2 45 llaple On Line where a b and t are constants You have also collected the data below where 7 is the distance from the sun in millions of miles and 0 is the angle between the vector from the sun to the asteroid and a fixed axis through the sun The data is of course subject to experimental error 0 0 6 18 14 21 32 54 7 32927 3138 31949 31091 32788 37491 36749 Your strategy is to use the given data to find values of a b and t which make the formula for the ellipse to agree as closely as possible with the given data This will involve setting up a system of linear equations in a b and C and solving the normal equation Give the augmented matrices for both the original system and the normal equation You will then graph the given formula and measure the desired data from the graph Note Once you have found values of a b and c you will need to plot the orbit of the asteroid This can be done with the following command where r is the function that de nes the ellipse gt p3 plot rt tO v 2Pi coordspolar Retrialk As stated this is an inherently nonlinear problem which we solve using linear equations There are more accurate techniques based 011 multivariable calculus These techniques are also considerably more complecated than our solution Exercise 2 Maple has a leastsqrs command from the linalg package that is easier and better for solving least squares problems than simply solving the normal equations Use this command to solve the overdetermined linear system developed in Exercise 1 An appropriate syntax is as below where A is the coef cient matrix for the system and F is the vector of constants on the right side of the equations gt leastsqrs AF Chapter 5 Determinants Determinants are extremely useful in many contexts You will for example use them constantly when you study eigenvalues and eigenvectors later in the text In addition you will see them used to write formulas for the solutions to many applied problems In particular determinants are used extensively in the study of differential equations and in the study of advanced calculus Determinants are also used extensively in studying the mathematical foundations of linear algebra omputers however do not generally use determinants for computations Much faster and more ef cient numerical techniques have been found Thus we will not provide any computer exercises for this chapter The reader should be aware however that Maple will compute deter minants The appropriate command is detA Incidentally Maple uses the methods of the next section to compute determinants rather than the methods already described 51 llaplc On Linc Chapter 6 Eigenvectors 61 On Line Restart Maple to clear its memory of all variables and reinitialize it by loading the linalg package In the On Line section for Section 51 we commented that virtually anything you might use determinants for a computer would do otherwise This includes linding eigenvalues Algorithms for computing eigenvalues are very sophisticated and will not be described in this text However we will point out that the techniques do not involve finding the characteristic polynomial and determining its roots In fact the numeric recipes for finding eigenvalues are so good that they are often used to find roots of polynomialsl The exercises in this section explore this idea Exercise 1 0 If A is the matrix b la 1 show by hand that the characteristic polynomial is A2 a A 0 Use this to construct a matrix Al which has 1 A A2 7 A l as its characteristic polynomial Check that your matrix has the correct characteristic polynomial using Maple s charpoly command Note The charpoly command requires that you name the variable to be used in the characteristic polynomial Thus an appropriate syntax might be gt p1 charpoly A lambda Also Maple uses cletM I Ap clet l I as the definition 65 36 61 llaplc On Linc ol39 the characteristic polynomial The relation between the two is that if A is n x n then the characteristic polynomial we compute in the text is 11 Use Maple s eigenvals command to compute the eigenvalues of Al and hence the roots ol39 1 A Note A11 appropriate syntaxt for the solve command is gt solvep1 01ambda Exercise 2 l 0 Compute the characteristic polynomial for the matrix A 0 0 l c b a Ifse this result to obtain the roots of the polynomial le A3 8 A2 17A 10 Test the roots by substitution back into Exercise 3 Let A1 be the matrix obtained in Exercise 3 Obtain the eigenvectors of Al normalizing them so the first element in each is 1 What do you then notice about these eigenvectors Use the pattern you articulate to give a general prescription of the eigenvectors of an n x 11 matrix of the form A Prove your answer The eigenvactors for Al may be computed using the following command gt q eigenvectsA2 Note that there are three lists in q and in each list there are three members The first member of each list is the eigenvalue The second member of each list is the algebraic multiplicity the number of times that eigenvalue is a root of the characteristic equation The third member of each list is a set of eigenvectors Here each such set contains a single eigenvector Hence these eigenvectors can be referenced as follows gt v1 q1 3 1 v2 q2 3 1 v3 q3 3 1 Exercise 4 Find a 4 x 4 matrix A whose characteristic polynomial is A43 A2 5 A 7 Obtain the roots of this polynomial by finding the eigenvalues of the matrix A Check your result with Maple s solve command 62 llaple On Line 6397 The matrix whose characteristic polynomial is is known in math ematics as the companion matrix IVIaple has the builtin command corn panion for finding the companion matrix for a polynomial Maple gener ates a companion matrix that is the transpose of what you might find in some differential equations texts gt p x 4 3x 2 5 7 gt A transposecompanionpx Find the eigenvalues ol39A gt eigenvalsA Maple has expressed the roots of the characteristic polynomial with its Rootlquot7 notation This is a shorthand for what could be large and complex expressions for the exact value of the roots There are several options available at this point First apply the allvalues command to the RootOl39 structure This will return the eigenvalues as exact values containing complicated expressions involving radicals gt q eigenvalsA gt q1 allvalnesq These expressions are too unwieldy to work with onvert them to oating point numbers with the evalf command gt eval ql Another alternative is to include at least one oating point number in the matrix A When eigenvals sees the oat it will compute the eigenvalues numerically gt A1 mapconvertAfloat gt eigenvalsA1 Finally solve the equation 0 numerically by using the fsolve command For polynomial equations this command accepts the parameter complex to indicate that all roots both real and complex are to be found gt fsolve 1 complex 62 On Line Restart Maple to clear its memory of all variables then reinitialize by loading the linalg package G8 62 llaple On Line Create your own eigenvalue problem by constructing a 3 x 3 matrix A with prescribed 39 anc 39 The 39 39 are the diagonal elements in a diagonal matrix D while the eigenvectors are the columns of a 3 x 3 matrix P A simple way to construct the eigenvector matrix P is as a random matrix Hence define l39 a function which generates random integers in the interval 1010 gt f rand10 10 Let P be a random 3 x 3 matrix with entries determined by l39 gt P randmatrix33entriesf Let the eigenvalues be 2 2 and 3 in that order Create a diagonal matrix with these elements 011 the diagonal but assign the matrix to the name d not D The letter 7D in Maple is reserved for one form of the differentiation operator and Maple will not let you assign to it gt d diag223 The matrix A PDPVU will have eigenvalues 2 2 and 3 in that order and will have the columns of P as eigenvectors in corresponding order gt A evalmP amp d amp inverseP Exercise 1 By computing A X for each column X of P verify that each column of P is an eigenvector of A olumns of P can be referenced by Maple s col command learly A X A X must hold for each eigenpair ol39 eigenvector X and eigenvalue A Exercise 2 Verify that the diagonal elements of D are the eigenvalues of A by using Maple s eigenvals command to determine the eigenvalues of A directly from A itself There is however no canonical ordering for the results of this command so Maple need not order the eigenvalues as 2 2 3 gt eigenvalsA 62 llaple On Line 6399 Exercise 3 By applying Maple s eigenvects command to A again verify that the columns of P are the eigenvectors The eigenvects command returns lists with three members in each list These three members are first the eigen value second the algebraic multiplicity of the eigenvalue the number of times the eigenvalue was a root of the characteristic equation and third a set of eigenvectors associated with the eigenvalue in the list Extract the eigenvalues and eigenvectors by adroit use of39 the selector bracket notation See Exercise 3 in the On Line exercises for Section 61 Again there is no canonical ordering for the lists produced or for the eigenvectors associated with an eigenvalue of multiplicity greater than 1 Executing the eigenvects command 011 different occasions can result in a different ordering each time The eigenvectors computed by the eigenvects command may not 7 look like the columns of P The columns of P may be constant multiples of the corresponding vectors or in the case of39 multiple eigenvectors the columns ofP could simply be a different basis for the eigenspace associated with the repeated eigenvalue Compare the third column of P with the eigenvector Maple found for the eigenvalue 3 determining any multiplicative factor needed to make the eigenvectors match exactly To show two bases P1 P2 and v1 by are equivalent you need to show that linear combinations of one set of basis vectors yield the other basis vectors Use the L39L39ef command on a matrix containing as its columns the first two columns of P and the eigenvectors Maple found for the eigenvalue 2 How will this show the equivalence of the bases Exercise 4 For the n x 11 matrix A the characteristic polynomial has been defined in this text as 10111 AI Some texts use clcfAI A thereby making the two definitions differ by a factor of 1 lVIaple s builtin charpoly command for generating the characteristic polynomial uses the latter convention The advantage of Maple s definition is that for an n x 11 matrix the highest order term In Maple compare these two methods for obtaining the characteristic polynomial The charpoly command takes as arguments the matrix A and a variable to be used in the output polynomial Maple computes determinants via the let command and also allows the syntax A A as a short form of A A I Finally typing out the name of the Greek letter A causes lVIaple to print that letter as a Greek letter gt charpolyA lambda 70 63 llaple On Line gt det Alambda Note that the two polynomials are just negatives of each other There is only one degree three polynomial with roots 2 2 and 3 that has A3as its highest degree term What is this polynomial IIint Write it as a product of linear factors and then expand Note that this is the characteristic polynomial ol393A3 as found by Maple 63 On Line Restart Maple to clear its memory of all de ned variables and reinitialize by loading the linalg package xerc1ses complex numbers will appear Maple uses the letter so that the complex number z 2 3 iis entered into IVIaple as z 2 3I It is also useful to remember that if z 2 3 i then 3 2 3139 is the complex conjugate of z Thus the complex conjugate of a real number x is that real number itself since the imaginary part the part with the i is zero lIaple s command for conjugating a number is conjugate Its com mands for extracting the real and imaginary parts of a complex number are Re and Im respectively In purely numeric contexts these commands usually need no additional boosts In symbolic and exact contexts these commands generally work only if an additional evalc evaluate complex is applied T 1us gt Re2 3I Im2 3I but gt Rex Iy Imx Iy thereby requiring gt evalcRex Iy evalcImx Iy Exercise 1 Ifse Maple s eigenvects command to obtain the eigenvalues and eigenvec 1 from Example 2 of section 53 Notice that the eigenvalues are complex conjugates as are the eigenvectors as well tors of the matrix A 1 63 llaple On Line 71 Exercise 2 A11 11 x 11 matrix A with complex entries is said to be Hermitig if the conjugate of the transpose equals A Thus A is Hermitian il39 1 A A moment s re ection reveals that the conjugate of the transpose equals the transpose of the conjugate that is A At Some texts denote the conjugate transpose of A by the symbol Aquot so that Aquot 1 AT and A is Hermitian provided A Give an example of a 3 x 3 Hermitian matrix containing as few real numbers as possible and having no entries zero To verify that your matrix is Hermitian you need to take both the transpose and the complex conjugate Maple has the builtin htranspose command for this Apply it to your matrix A and then separately apply the transpose and conjugate commands In Maple the simplest way to do this is to map conjugate onto the transpose of A gt htransposeA gt mapconjugate transpose A ne remarkable property ol39Hermitian matrices is that their eigenvalues are real Apply lVIaple s eigenvals command to your matrix A in an effort to verify the truth of this claim You will most likely obtain large compli cated 39 for the 39 quot exact 39 39 They might even making it look like the eigenvalues are complex contain the symbol 1 To determine if these eigenvalues are real you want to show that for each the imaginary part is zero After obtaining the exact eigenvalues convert them to oating point numbers by using the evalf command Since eigenvals returns a sequence of eigenvalues you may need to convert this return to a list use before the evalf command works The conversion to decimals of any radicals in your eigenvalues may yield expressions with very small imaginary parts You can instruct Maple to truncate these small numbers to zero by using the fnornial command Instruct Maple to extract and simplify the imaginary part of each eigen value These imaginary parts should reduce exactly to zero Remember to use both evalc and 1111 as well as simplify on each exact eigenvalue Obtain the exact real part of each eigenvalue These expressions will be real but complicated The point of the activity is for you to realize that exact values for the roots of cubic equation are unpleasant expressions to work with Just because Maple is able to provide the roots exactly does not mean that these expressions are always useful or simple Finally apply evall39 directly to the exact real parts of the eigenvalues 72 63 llaplc On Linc Compare the values to what you got when oating the unsimplilied com plex version of the eigenvalue Exercise 3 Change one of the entries of A from Exercise 2 making A nonIIermitian Then recalculate the eigenvalues Are they again real IVIake the change to A in Maple by creating a matrix B via substitution of a new value into the matrix A Reference an element of A say the llelement by All and make a substitution of a new value for such an element into optA not just into A Choosing the new value to be a oating point number will mean that eigenvals will return oating point numbers directly Since Exercise 2 made a thorough study of the complexity of exact calculations work numerically in this exercise Exercise 4 For the matrix A of Exercise 2 compare the action of transpose and htrans pose Is there any dilference if these operators are applied to matrices with just real entries Exercise 5 In Maple let A be a real symmetric 3 x 3 matrix with as many ol39its entries as possible distinct btain B I iA where I is the 3 x 3 identity matrix and i Let 7 I 131 and compute both i iquot and Cquot 7 Can you prove that what you observe is always true Hint Begin with the equality B Bquot 2 I and multiply by 3amp1 on the left and by B t llon the right Notes To generate a random symmetric matrix in Maple add the pa rameter syrmnclric to the randnlatrix command gt A randmatrix33symmetric The matrix B can be obtained in Maple it due note is taken of Maple s usage ol39I for the imaginary unit and due care is taken to distinguish between the written symbols Iand i standing respectively for the identity matrix and the imaginary unit 1 Since an identity matrix is a diagonal matrix with ust Is 011 the diag onal the diag command can be used to create an identity gt B evalmdiag13IA 65 llaple On Line 73 The matrix T likewise requires use of an identity matrix gt C evalmdiag13inverseB2 65 On Line Restart Maple to clear its memory of all variables then reinitialize by loading the linalg plots and 110110015 packages gt restart gt withlinalg withplots withplottools The purpose of this exercise set is to explore the relationship between eigenvalues eigenvectors and the geometry of quadratic forms Exercise 1 Use Maple s inlplicitplot command from the plots package to obtain a graph of the ellipse defined by the quadratic equation 32 1 39 Be sure to use a 11 aspect ratio so that there is no distortion in the scaling Then obtain A the matrix of the quadratic form de ned by this same equation This can be done by typing in A by clever use of partial differentiation or by Maple s hessian command Obtain the eigenvalues and eigenvectors of gt q x 2 4y 2 1 gt implicitplotqx1 1y1 1scalingconstr ained l 0 0 4 By inspection we can write A Alternatively we can note that A m W where l39xy is the left hand side of the defining fyac fyy quadratic equation and subscripts denote partial derivatives Thus gt f lhs q gt fxx difffxx fxy difffxy fyx difffyx fyy difffyy The array of second partial derivatives of the form f aw is called at the hessian matrix and is returned in Maple by the hessian command gt hessiangxy xy 74 65 llaple On Line Hence the matrix of a quadratic form is simply onehalf the hessian matrix The simplest way to get the 12 into the hessian is by including it in the quadratic function Else an evalrn is needed to multiply the hessian matrix by that 12 gt A hessianlhsq2 xy The eigenvalues and eigenvectors can be obtained by use of the eigen vects commanc We next seek to relate the eigenvalues and eigenvectors to the lengths of the semimajor and semiminor axes From the graph these re 1 and 12 respectively The eigenvalues are l and 4 respectively for eigenvectors that have the directions of the ellipse s axes Hence the scale factors by which to multiply the eigenvectors to get the semimajor and semiminor axes are m w1ere 39 2 Exercise 2 Using Maple s irnplicitplot command obtain a graph of the function de fined implicitly by the quadratic equation 2 12 1 y gt 1 Be sure to use an aspect ratio scaled to 1 1 It may also be edil39ying to increase the number of points used in the plot by adding the parameter numpoinls 1000 to the irnplicitplot command Assign the plot to a variable so that it can be reused in subsequent exercises Exercise 3 For the quadratic equation in Exercise 2 form the matrixA ol39 the quadratic form de ned by the equation Ifse Maple s hessian command and check the result by taking partial derivatives of the left hand side of the deliri ing quadratic function Obtain the eigenvalues and eigenvectors of A via Maple s eigenvects command Extract and give unique names to the eigenvalues and eigenvectors btain oating point approximations for the eigenvalues and normalize the eigenvectors to have length l with Maple s norInalize command Apply Maple s radsinlp command to the 2norm computed via IlOl IIl of each normalized eigenvector to verify that each indeed has length 1 Use the arrow command from the 110110015 package to superimpose the normalized eigenvectors 011 the graph of the ellipse De ne each arrow separately The arrow command takes five parameters a list of coordinates for the tail here the origin a list of coordinates for the point here the entries in the normalized eigenvectors then three sizing parame ters which experiment shows are well chosen as 05 l and 05 Color each 65 llaple On Line 1 Cl arrow differently Then use the display command from the plots package to merge the graph of the ellipse and the two arrows into one graph To convert a vector to alist use the convert command with option 13951 Thus if the eigenvectors are named V1 and V2 you might eneter gt a1arrow 0 0 convert V1 list v 05 v 1 05 colorgreen gt a2arrow 0 0 convert V2list v 05 v 1 05 colorblue gt display p1 a1 a2 scalingconstrained Print a copy of this final graph of the ellipse with the normalized eigen vectors Exercise 4 n the plot printed in Exercise 3 draw in the axes determined by the eigenvectors On these axes put tick marks every quarter unit noting that each eigenvector is one unit long Use a ruler to guarantee the accuracy of your tick marks According to the general theory the ellipse should cross the new axes at points whose coordinates in the system determined by the eigenvectors are 30 lrl0 Virgil amp0 where M and A2 are the eigenvalues ol39A Veril39y this by estimating the appropriate coordinates from your graph Exercise 5 Verify in general that the scale factors 7 k l 2 indeed convert nor malized eigenvectors into vectors of precisely the length of the semimajor and semiminor axes Begin by obtaining oating point values of l k xT 39 l 2 Then scale the normalized eigenvectors by these factors Next use Maple s arrow command to build a plot of the ellipse and the newly scaled eigenvectors A plot of the ellipse and these new basis vectors should show that with this scaling the vectors coincide precisely with the semimajor and semiminor axes Exercise 6 Find the equation of an ellipse centered at the origin and for which the major axis is 4 units long and lies along the line determined by the vector 3901 and for which the minor axis is 2 units long Hint If you 76 65 llaple On Line can gure out the eigenvalues and eigenvectors you then can find matrices Q anch for which A QDQf Where A is the symmetric matrix for the quadratic form corresponding to the ellipse Graph the ellipse and the scaled ei 39 1vectors for A demonstrating that the proper scaling of the eigenvectors gives them the lengths of the semimajor and semiminor axes Since the axes of the ellipse are orthogonal you need to get a vector orthogonal to the vector L1 This can be done in the plane by interchanging the x and ycoorclinates and negating one component of the resulting vector Even a casual inspection reveals Why this works

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made $280 on my first study guide!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.