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# LINEAR ANALYSIS MATH 554

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6 Linear Algebra and AIatrix Analysis Vector Spaces Throughout this course the base eld IEquot of scalars will be R or 1C Recall that a vector space is a nonempty set V on which one de nes the operations of addition for v w E V v w E V and scalar multiplication for oz E IEquot and v E V ow E V which satisfy the following axioms for every 3 y z E V and A u E IEquot 1 atyzatyz 2 atyyat 3 30 E V such thatat0at 4 Va E V there exists 2 E V such that a z 0 written 2 T 5 A3 5M 6 Ma A3 Ag 51 A 103 A3 113 8 Must Auk and 9 13 A subset W C V is a subspace if H is closed under addition and scalar multiplication so W inherits a vector space structure of its own Examples 3T1 115quot g eachatjElE39 7121 37m 2 0 3T1 3 13 C 372 each atjElE39 371 0c 4 IF C 13 C where IF 3T2 lt 00 jl 3T1 C 150 WhOI O 37 supiatji lt 00 7 i 1IE and 0 are clearly subspaces of 150 Vector Spaces 7 371 0c Let 0 lt p lt 00 and de ne 1 1539 372 Ziatjip lt 00 jl Since Eaty lt sciHyng2max c y p 2pmax a p iyip 3 2p x p iyip it follows that p is a subspace of 150 Exercise Show that 7 1539 g quotIE39 if 0 lt p lt q 3 00 5 Let X be a nonempty set then the set of all functions f X gt IEquot has a natural structure a vector space over 1539 de ne f1 f2 by f1 f25 f1 f2a and de ne of by 6 For a metric space X let CX I be the set of all continuous F Valued functions on X CX I is a subspace of the vector space de ned in De ne CbX I C CX I to be the subspace of all bounded continuous functions f X gt IF and let CkX F C C X I to be the subspace of all k times continuously differentiable functions f X gt IF In the case where X IE3quot we simplify this notation to C13 CECE and Ckl respectively 7 De ne 791539 C CR I to be the space of all F Valued polynomials on R 771539 a0 an A A A amatm m 2 0 each aj E Each 1 E 791539 is Viewed as a function p R gt IEquot given by aga1 lmm 8 De ne 791539 C 791539 to be the subspace of all polynomials of degree 3 n 9 Let V E 021539 u u 0 It is to check directly from the de nition that V is a subspace of C205quot For IE3quot 1C one knows that V a1 COSQ a2 sina aha E 1C blew 626 6162 E 1C from which it is also clear that V is a vector space More generally if ulm am1umll A A A alu 1071 is an mth order linear constantcoef cient differential operator then V E C IE39 0 is a vector space V can be explicitly described as the set of all linear combinations of certain functions of the form mic where j 2 0 and r is a root of the characteristic polynomial W amlrm A A A a1quot a0 0 For details see Chapter 3 of Birkho 39 amp Rota Convention Throughout this course if the eld IEquot is not speci ed it is assumed to be 1C 8 Linear Algebra and Allatrix Analysis Linear Independence Span Basis Let V be a vector space A linear combination of the vectors v1 39vm E V is a vector v E V of the form v mm A A A amvm where each aj E 1539 Let S C V be a subset of V S is called linearly independent if for every nite subset L3 of S the linear combination aw 0 i 39 m A A A am 0 Otherwise S is called linearly dependent De ne the span of S denoted SpanS to be the set of all linear combinations of all nite subsets of 3 Note A linear combination is by de nition a finite sum If S set SpanS S is said to be a basis of V if S is linearly independent and SpanS V Faets Every vector space has a basis in fact ifS is any linearly independent set in V then there is a basis of V containing 3 The proof of this in in nite dimensions uses Zorn s lemma and is nonconstructive Such a basis in in nite dimensions is called a Hamel basis Typically it is impossible to identify a Hamel basis explicitly and they are of little use There are other sorts of bases in in nite dimensions de ned using topological considerations which are very useful and which we will consider later b Any two bases of the same vector space V can be put into 1 1 correspondence De ne the dimension of V denoted dim V E 0 1 2 U 00 to be the number of elements in a basis of V The vectors 231 en where 0 ej 1 jth entry 0 form the standard basis of IF and dimlE 72 Remark Any IC vectorspace V may be regarded an R vectorspace by restriction of the scalar multiplication It is easily checked that if V is nitedimensional with basis L3 over 1C then v1vniv1ivn is a basis for V over R In particular dimjkg V 2 dimC V The vectors 231232 E I are linearly independent However Spank1232 is the proper subset of I consisting of all vectors with only nitely many nonzero components so 131232 is not a basis oflE OC But m E 0 1 2 is a basis of 79 Now let V be a nitedimensional vector space and U1 vn be a basis for V Any n v E V can be written uniquely v 25mm for some 3 E 1539 So we can de ne a map i1 1 from V into If by v e gt g The 33 s are called the coordinates of v with respect to the a basis vb This coordinate map clearly preserves the vector space operations and is bijective so it is an isomorphism of V with IEquot in the following sense De nition Let V H be vector spaces A map L V gt W is a linear transformation if V Uh vg ElVO102 E I Lozyv1 Olg UQ a1L39v1 a2Lv2 Vector Spaces 9 If in addition L is bijective then L is called a vector space isomorphism Even though every nitedimensional vector space V is isomorphic to Iquot where 71 dim V the isomorphism depends on the choice of basis Many properties of V are inde pendent of the basis eg dim V We could try to avoid bases but it is very useful to use coordinate systems So we need to understand how coordinates change when the basis is changed Change of Basis Let V be a nite dimensional vector space Let L3 vn and 151 wn be two bases for V For v E V let a 31 3 T and y y1ynT denote the vectors of coordinates of v with respect to the bases 81 v139vn and 82 2121 wn respectively So 39v 231735 Ug 23 ijj Express each wj in terms of v1vn wj 231 aij39v an am at E Let A f E 15 Then 1m arm n n n n i1 j1 i1 j1 so 3 2 agjyj ie a Ay The matrix A is called the change of basis matrix Notation HornJohnson uses iMmm 39 to denote 15mm set of m X 7 matrices with entries from 1539 HJ writes 51 for 3 MB for y and 51mg for A so a Ay becomes MB 5 l1 lB2 ller Similarly we can express each by in terms of 151 wn 39vj 231 bijw ij E 511 5m Let B E E 15 Then y Bat We obtain that A and B are invertible bnl brm and B A Formal matrix notation Write the basis vectors v1 4 A and w A A formally in rows Then the equations wj 2 lflgj39vg become the formal matrix equation v1 4 A using the usual matrix multiplication rules In general v1 4 A and wl are not matrices although in the special case where each 39Uj and wj is a column vector in 1539 we have W VA where V W E 15 are the matrices whose columns are the 39vj s and the wj s respectively We also have the formal matrix equations v v1 and v U21 Uny so v1 4 A U21 4 A v1 4 A AvnAy which gives us a Ay before Remark We can read the matrix equation W VA saying the jth column of W is the linear combination of the columns of V whose coefficients are in the jth column of A 10 Linear Algebra and Matrix Analysis Constructing New Vector Spaces from Given Ones 1 The intersection of any family of subspaces of V is again a subspace let W y Y E G A 3 V V l V be a family of subspaces of V where G is an index set then mil 7 is a subspace of 766 V Sums of subspaces If 1 W are subspaces of V then W l W U21 w U21 E W 2122 E 1 is also a subspace and dimW 1 I Vg dimW 1 0 WE dim W l dim 1 We that the sum 1 113 is direct if W 0 W 0 equivalently for each v E W 113 there are unique U21 E 1 and 2122 E W for which v U21 w and we write W l Q W for W 1 More generally if I39V1139Vn are subspaces of V then 1 A A A W n U21 A A A w wj E 1131 3 j g 72 is a subspace We that the sum is direct if whenever wj E HQ and 11le 0 then each wj 0 and we write WK Q Q 11 Even more generally if Ll397 y E G is a family of subspaces of V de ne 276611 span LJW 7 We that the sum is direct if for each nite 39yEG subset G of G whenever 2127 E IV for Y E G and 2766 11 0 then each 21 0 for Y E G equivalently for each 8 E G W33 0 Zyecww EV Direct Products Let y E G be a family of vector spaces over IE3quot De ne V 756 V7 to be the set of all functions v G gt for which Viv E G E V7 We 39yEG write v7 for 39v y and we write v v yw or just v Define v w my and Ow Then V is a vector space over 1539 Example G N 12 each Vn 1539 Then X in 1121 External Direct Sums Let Y E G be a family of vector spaces over 1 Define 63 V7 to be the subspace of X6 V7 consisting of those v for which v7 0 except for 766 7395 nitely many 7 E G Example For 71 0 12 let Vn in P Then 79 G Vn Technicality we should technically assume that for y 8 Vy Vg 0 1120 if not rename the elements of each V7 to make it true the given definition avoids the technicality by using the direct product Facts If G is a nite index set XV7 and QV7 are isomorphic b If each W 7 is a subspace of V and the sum 2 W 7 is direct then it is naturally isomorphic to the external direct sum QW Q 39yEG Quotients Let H be a subspace of V Define on V the equivalence relation v1 N U if v1 U E W and define the quotient to be the set V W of equivalence classes Let U W denote the equivalence class of Define a vector space structure on V W by defining 11v1 11 a2v2 11 mm ag39vg 11 Define the codimension of W in V by codimW dimVW Vector Spaces 11 Dual Vector Spaces De nition Let V be a vector space A linear functional on V is a function f V gt IF for which fozyv1 Olg UQ mf39v1 a2f39v2 for U1 U2 E V aha E 1539 ie a linear transformation from V to the 1dimensional vector space 1539 Examples 1 Let V IE and let f be a linear functional on V Set f f23 for 1 g i g n Then for a munsxny 231 375 E 1539quot M Zane Zia f1f2fnar i1 i1 So the row vector 1quot f is the matrix of f using the standard basis 61 e on I and the basis 1 on 2 Let V 13 C Given N and some f1f2 fN E 1539 we could de ne at for a E 15 However not all linear functionals on I are of this form 3 Let V IE39 Iff E POOP then fora E IE39 fix 3 sup f5 21 ix lt 00 so the sum f 1 fix converges absolutely de ning a linear functional on 1 Similarly if V and f E 1 IF fiat de nes a linear functional on 4 Let X C Rquot and an E X Then 71560 de nes a linear functional on 5 If 00 lt a lt b lt 00 de nes a linear functional on Cab De nition If V is a nitedimensional vector space the dual space of V is the vector space V of all linear functionals on V where mfl a2f239v mfl a2f239v Remark When V is in nite dimensional the set of all linear functions is often called the algebraic dual space of V it depends only on the algebraic structure of V We will be more interested in linear functionals related also to a topological structure on V After introducing norms which induce metrics on V we will de ne V to be the vector space of all continuous linear functionals on V When V is nite dimensional with any norm on V every linear functional on V is continuous Dual Basis in Finite Dimensions Let V be a nite dimensional vector space and let v39v be a basis for V For 1 g i g 71 define linear functionals f E V by 65 1 for i j 0 for i g Let U E V and let a 31 atnT be the vector of coordinates of v with respect to the basis L3 v ie v 2313 Then f39v 37 ie f maps v into its coordinate at of v Now if f E V let 1 f39v then 1 23803 2am meX UL i1 i1 i1 12 Linear Algebra and AIatrix Analysis so f 231 af This representation is unique exercise so f1 is a basis for V called the dual basis to L3 We get dim V dim V f1 If we write the dual basis in a column 3 and the coordinates a1 of f fn f1 23 lgfg E V in a row then f a A an 3 The de ning equation of the dual fn basis is matrix multiply evaluate f1 1 O a s mwa a 1 fn O 1 Change ofBasis and Dual Bases Let 211 be another basis of V related to L3 by the changeofbasis matrix A ie 2121 v1 Leftmultiplying by A 1 and rightmultiplying by A gives f1 A l I fn Therefore 91 f1 91 A l satis es g 11 A A I and so g1gn is the dual basis to u1wn If 614446 are the coordinates of f E V with respect to 11 gn then 91 f1 f1 fawa s awwar s mwa s an fn fn so 11 bnA1 a1 an ie 11 5 a1 anA is the transformation law for the coordinates of f with respect to the two dual bases f1fn and 11 gn Linear Transformations Examples 1 Let T Iquot gt 15quot be a linear transformation For 1 g j g 71 let 3 4 Vector Spaces If a E 15 TEajej Eathej ie 3 3T1 til tin 371 tr in i 377i tml trim 37 So every linear transformation from IEquot to 15quot is given by multiplication by a matrix in 15mm One can construct linear transformations C IF0C gt IF0C by matrix multiplication Let tn t1 T t21 be an in nite matrix for which each row has only nitely many nonzero entries In 3T1 forming T3 for a E 13 C each entry in Ta is given by a nite sum so Ta makes sense and T clearly de nes a linear transformation from I to itself The shift operators x1atgT r gt UI 1JIQT and 1372T r gt xgx3T are examples of linear transformations on I that can be written in this form However not all linear transformations on I are of this form If sup itgji lt 00 and a E 5 then for each i itgjatji 3 sup itgji It follows 2kj 213 that matrix multiplication Ta de nes a linear transformation T I gt 5 C There are many linear transformations arise on function spaces eg Let k E Ch d X hu b where hub h d are closed bounded intervals De ne the linear transformation L Chub gt Chu d by L is called an integral operator and Mr is called its kernel b Let in E Chub Then de nes a multiplier operator L on Chub c Let g hud gt hub Then L Chub gt Chi de nes a composition operator d u e gt u de nes a di erential operator L C hu b gt Chub 14 Linear Algebra and AIatrix Analysis Matrices and Basis Transformations Suppose V W are finitedimensional with bases L3 vn 151 mm respectively and suppose L V gt W is linear For 1 g j g 71 we can write L39vj t wi The matrix tn tm T g g EIWquot tml tmn is called the matrix of L with respect to the basis 81 L3 vn 82 215 wm HJ 3T1 writes T BglLlsi If v E V f are the coordinates of v with respect to 81 and 37 91 are the coordinates of Lv with respect to 82 then ym m n m n Zygwg Lv L Z Ztgjirj wt i1 j1 i1 j1 so for 1 g i g m y Zltgjj ie y Tat Also U1 UnT Now let 8 and B 211 min be different bases for V W respectively with changeofbases matrices A E 15quot B E 15quot 7 A A v1 and A A ml 4 A AwmB Then ml 4 A 421 TA A A AunB TA so the matrix of L in the new bases is sglLls ETTA s l1l5252 lle 51 lIlsq In particular if W V 82 81 and 8 2 1 then B 4 so the matrix of L in the new basis is A TA This matrix is said to be similar to T The transformation T r gt A TA is said to be a similarity transformation of A A similarity transformation of a matrix corresponds to the representation of the same linear transformation with respect to different bases Linear transformations can be studied abstractly or in terms of matrix representations For L V gt W the range RL null space NL or kernel kerL rank L dim RL etc can be defined directly in terms of L or in terms of matrix representations If T E FMquot is the matrix of L V gt V in some basis it is easiest to define det L det T and trL tr T Since det 4 TA detT and tr 4 TA tr T these are independent of the choice of basis Vector Spaces 15 Vector Spaces of Linear Transformations Let V W be vector spaces If L V gt W L2 V gt W are linear de ne mL 1ng V gt W for aha E IEquot by mL a2L239v aL39v a2L2v so the space of all linear transformations from V to W is naturally a vector space over IE3quot If V W are nitedimensional we denote this vector space by 8V W in the in nitedimensional case we will use this notation to mean all bounded linear transformations to be de ned from V to W with respect to norms on V W Remark For normed linear spaces a linear operator is a bounded linear operator i 39 it is continuous i 39 it is uniformly continuous When V W are nite dimensional normed linear spaces every linear transformation from V to W is continuous and is therefore a bounded linear transformation If V W have dimensions 71 m respectively then 8V W is isomorphic to Fm so it has dimension nm When V W we denote 8V V by 8V Since the composition 1 I O L V gt U of linear transformations L V gt W and AM W gt U is also linear 8V is naturally an algebra with composition the multiplication operation Projections Suppose W W are subspaces of V and V W Q W g Then we W and W are complementary subspaces Any v E V can be written uniquely v 212 102 with w E W 2122 E W 2 So we can de ne maps P V gt W P2 V gt W by Pv w ng wg It is to check that P P2 are linear We usually regard P P2 mapping V into itself W C V W C V P is called the projection onto W along W similarly P2 is the projection onto W along W It is important to note that P is not determined solely by the subspace W C V but also depends on the choice of the complementary subspace W 2 Since a linear transformation is determined by its restrictions to direct summands of its domains P is uniquely characterized as that linear transformation on V which satis es 13 I and HEW 0 W W It follows easily that P 13 P P2 P P2 1 and 13132 P2P 0 In general an element 1 of an algebra is call idempotent if 12 1 If P V gt V is a linear transformation and P is idempotent then P is a projection in the above sense it is the projection onto RP along This discussion extends to the case in which V W Q A A A Q W m for subspaces W We can de ne projections P V gt W in the obvious way P is the projection onto W along W Q Q W Q W Q Q W m Then PfP for1 i m PAPm1 and PPjPjPUfori j If V is nite dimensional we that a basis w mp 3911 uq for V W Q W is adapted to the decomposition W Q W if u mp is a basis for W and u uq 16 Linear Algebra and Allatrix Analysis is a basis for 1 With respect to such a basis the matrix representations of P1 and P2 are I 0 and gwheretheblockstructureis po pxq 0 0 1X12 1X0 P I abbreviated p it 1 I Invariant Subspaces We that a subspace Wquot C V is invariant under a linear transformation L V gt V if LUV C W If V is nite dimensional and wlg mp is a basis for W which we complete to some basis wlg wp 711 uq of V then W is invariant under L ifl39 the matrix of L in this basis is of the form P I p 1 l0 ie block uppertriangular We that L V gt V preserves the decomposition 1 Q A A A Q W m V if each W is invariant under L In this case L de nes linear transformations L WE gt WE 1 g i g m and we write L L Q A A A Q Lm Clearly L preserves the decomposition ifl39 the matrix T of L with respect to an adapted basis is of block diagonal form T1 0 T2 0 Tm where the are the matrices of the Li s in the bases of the HTS Nilpotents A linear transformation L V gt V is called nilpotcnt if L quot 0 for some 7quot gt 0 A basic example is a shift operator on 1539 de ne 3231 0 and Sci 1 for 2 g i g 71 The matrix of S is 0 1 0 33quot 1 EIBWquot 0 0 Note that 3 shifts by m SW6 0 for 1 g i g m and 37 114 for m 1 g i g 72 Thus 3quot 0 For 1 g m g 71 1 the matrix 307 of 3 is zero except for 1 s on the mth Vector Spaces 17 superdiagonal ie the ij elements for j i m 1 g i g 72 m are 1 s 0 Hi 0 1 0 1 m 1 element 1 71 m 7 element 3 Snm 0 U 0 Note however that the shift operator on IE Se 411 i 2 2 3231 0 is not nilpotent Structure of Nilpotent Operators in Finite Dimensions Let V be nite dimensional and L V gt V be nilpotent We will show that there is a basis for V in which L is a direct sum of shift operators This decomposition results from a direct sum decomposition of both the domain and range of L The basic idea is to build the decomposition by using the structure of the subspaces NLk in the domain of L and the structure of the subspaces RLk in the range of L This is a key step in showing that every matrix is similar to a matrix in Jordan form Since L is nilpotent there is an integer r such that LT 0 LT The proof proceeds by successively considering the subspaces RL quotj starting with j 1 and decomposing along these subspaces Let v1 39ng be a basis for RLT and for 1 g i g 1 choose 21 E V for which v LP wi Observe that 1 V NLT NLT1 G3 spanw1wg1 We claim that the set 81 LT 1u21L quot 2w1Hum LT1u2g LT2wg wg LT1ug1 LTQWI 11 is linearly independent note the cyclic nature of the decomposition already appearing in the description of 81 Indeed suppose 1 r 1 2 ngLkwi 0 i1 k0 Apply LT l to obtain 1 E mmeg 0 21 1 Zng U530 so 300for 1 sigh i1 18 Linear Algebra and Alfatrix Analysis l39ow apply L to the double sum to obtain 1 1 1 0 E any 11 011 21 21 so a 0 for 1 g i g 1 Successively applying lower powers of L shows that all a 0 Observe that for 1 g i g 1 spanL quot w L 111 w is invariant under L and L acts by shifting these vectors It follows that on span 81 L is the direct sum of 1 copies of the 7quot X 7quot shift 3 and in the basis LT 1U21LT2U1 39u1 LT1U2 L quot2u2g U2 LT1Ug1LT2Ug1 ng for span81 L has the matrix S U 0 S In general span81 need not be all of V so we aren t done We know that LT 12111 LT1Ug1 is a basis for RL quot and that 2 LT 11121LT1Ug1LT2U1LT2Ug1 are linearly independent vectors in RLT2 indeed we showed that all of the vectors in 81 were linearly independent and L quot w LT 2Lw e Rm 1 g i g 51 lomplete 2 to a basis of RL quot2 if necessary by adding vectors l a As before choose 3917ng for which LT Q mi aj 1 31 E 52 We now further re ne the direct sum decomposition in 1 using the vectors 3917ng and the subspace NLT 2 to construct a direct sum decomposition of N LT 1 However the span of the vectors 3917ng may not be contained in the subspace NLT which would make this re ned direct sum decomposition impossible to construct We get around this problem by replacing the vectors 3917ng 1 g j s 2 by vectors in NL quot This is done by projecting each 3917ng onto NLT along RLT in the following Recall that the vectors v L112 1 g i g 1 were chosen to form a basis for RL quot Now since LT l Ehihj E RLT1 there exist coef cients 15 E IEquot 1 g i g 11 gj g 2 such that 1 LT1U2gj ZaszT w 11 1 2 4 Ul 1j Ul 1j E hill and uj LT wg j 1 g j 3 2 i1 Vector Spaces 19 Replacing the j s by the uj s still gives a basis of RL quot2 above exercise Clearly LT1Ug1j U for 1 31 s 2 Observe that now we have the direct sum decomposition NLT1 NLT2 G3 spanLw1ng1wg11wg1g2 re ning that given in We also have a basis for RL quot2 of the form LT 11121LT1Ug1LT2U1LT2Ug1LT2Ug11LT2Ug1g2 for which LT lwh i 0 1 1 E 52 We now repeat the argument given above for the set 81 but for the set 52 LT 2w 1l LT3U 1M WW1 LT QUl 1QaLT3Ul 12a w 12g LT2Ug1g2 LT3Ug1g2 U2g1g2 This gives that 8 U 82 is linearly independent and L acts on span82 a direct sum of 2 copies of the 7quot 1 X 7quot 1 shift 34 We can continue this argument decreasing 7quot by one each time and end up with a basis of RL0 V in which L acts a direct sum of shift operators 1 2 7 r A r A r quot LSrewesres ewes ewes ewesl N0t02310 13m Remarks 1 For 1 g j let kj dimJL1 It follows easily from the above that 0 3 k1 lt k2 lt 44ltkrzkr1 k2471 and thusrgn 2 The structure of L is determined by knowing 7quot and 1 T These in turn are determined by knowing k1 kn Exercise express 1 T in terms of k1 kn 3 General facts about nilpotent transformations follow from this normal form For ex ample if dim V 71 and L V gt V is nilpotent then i Lquot 0 ii m 0 iii det L 0 iv det I L 1 v for any A E 1539 det AI L Aquot 20 Linear Algebra and AIatrix Analysis Dual Transformations Recall that if V and W are nite dimensional vector spaces we denote by V and 8V W the dual space of V and the space of linear transformations from V to W respectively In the in nite dimensional case we will reserve this notation for the bounded linear functionals and transformations with respect to norms on V and W So we now introduce the notation V for the algebraic dual of V all linear functionals on V and CV W for the space of all linear transformations from V to W Let L E CV W We define the dual or adjoint transformation L W gt V by 9Lv for g E W 39U E V Clearly L r gt L is a linear transformation from CV W to W V and L o AMY AF 0 L if AM E CU V When V W are nite dimensional we can choose bases for V and W along with corre sponding dual bases Using these bases we can represent vectors in V W V W by their coordinate vectors aa14an and bl1 bm respectively The linear operator L E CV W can then be represented by a matrix T E 15m for which y T Hence given g E W having coordinates b bl A A Abm with respect to the dual basis we get so Lg has coordinates a1 4 A b A A AbmT Thus L is represented by leftmultiplication by T on column vectors and L is represented by rightmultiplication by T on row vectors Another common convention is to represent the dual coordinate vectors also columns taking the transpose in the above gives 11 51 TT an bm that is L can also be represented through leftmultiplication by T T on column vectors TT is the transpose of T TTj tj We can take the dual of V to obtain V There is a natural inclusion V gt V if v E V then f r gt de nes a linear functional on V This map is injective since if v 0 there is an f E V for which 0 Proof Complete to a basis for V and take f to be the rst vector in the dual We identify V with its image so we can regard V C V If V is nite dimensional then V V since dim V dim V dim V If V is in nite dimensional however then there are elements of V which are not in V Vector Spaces 21 If S C V is a subset we de ne the annihilator Si C V by SL E V Vv E S 0 Clearly Si spanSi Now Sii C V and if dim V lt 00 we can identify V V above Proposition If dim V lt 00 then Sii spanS Proof It follows immediately from the de nition that spanS C Sit To show Sii C spanS assume VVLOG that S is a subspace We claim that if H is an mdimensional sub space of V and dim V 71 then dim I39Vi codimI39V 71 m choose a basis 211 for IV complete it to a basis 211 u2m1 wn for V then clearly the dual basis vec tors fm1 are a basis for W so dim I39Vi n m Hence dim Sii n dim Si 72 72 dim S dim S and we know 3 C Sn E In complete generality we have Proposition Suppose L E CV IV Then RLi Proof Clearly both are subspaces of W Let g E W Then 9 E 4 Lg 0 4 V UEV Lgv0lt rV39vEV 9Lv0lt 29ERLL II We are often interested in identifying RL for some L E CV IV In the nite dimensional case this amounts to determining those 21 E II for which El39v E V satisfying Lv w choosing bases of V II and coordinate vectors a E 1539 y E 15quot for v w and letting T be the matrix of L this amounts to determining those y E 15quot for which the linear system Ta y can be solved Combining the two Propositions above we see that if dim H lt 00 then RL NLi Thus El39v E V satisfying Lv w i 39 0 for all g E W for which Lg 0 In terms of matrices Ta y is solvable i 39 yl bl bm 0 gm 51 for all i A A Abm for which i AAAbmT 0 or equivalently TT 3 0 These are often called the compatibility conditions for solving the linear system T a y Bilinear Forms A function 99 V X V gt Iquot is called a bilincar form if it is linear in each variable separately Examples 22 Linear Algebra and Allatrix Analysis 1 For any matrix A E 15 the function 23 291 a agyj is a bilinear form In fact all bilinear forms on IEquot are of this form 376Zyjcj 231 291 3yj99c 6 just set 125 9923 6j Similarly for any nitedimensional V we can choose a basis v1vn if 99 is a bilinear form on V and v Zayvi w Zyj39vj then 23 29 atgyj9939v atTAy where A E 15 satis es 15 99v A is called the matrix 0f99 with respect to the basis vb A 3 V One can also use in nite matrices ajgtj21 for V If long convergence con ditions are imposed For example if all iawi 3 M then lijxg yj de nes a bilinear form on 1 since iagjatgyji E MQXl Sim 7 ilarly if Erlng lt 00 then we get a bilinear form on 5 C 3 If fg E V then is a bilinear form 4 If V Ca b then i for k E Cm b X 96 ii for h E Cm 6 iii for 30 E 96 UC g are all examples of bilinear forms We that a bilinear form is symmetric if V39v39w E V In the nitedimensional case this corresponds to the condition that the matrix A be symmetric ie A AT or Vij 15 133 Sesquilinear Forms When llquot 1C we will more often use sesquilinear forms 99 V X V gt C is called scsquilivzcar if 99 is linear in the rst variable and conjugatelinear in the second variable ie mun 1211 1199v139u21 1299v 21 For example on C all sesquilinear forms are of the form 99z 23 2 aing39u y for some A E CMquot To be able to discuss bilinear forms over R and sesquilinear forms over 1C at the same time we will speak of a sesquilinear form over R and mean just a bilinear form over R A sesquilinear form is said to be Hermitiansymmetric or sometimes just Hermitian if V39v w E V 99w when llquot R we the form is symmetric This corresponds to the condition that A AH where AH AT ie AH2j A73 is the Hermitian transpose or conjugate transpose of A when llquot C in which case the matrix A E CMquot is called Hermitian or the condition A AT ie A is symmetric when llquot R To a sesquilinear form we can associate the quadratic form We that 99 is nonnegative or positive semide nite if Vv E V 2 0 and that 99 is positive or positive de nite if gt 0 for all v 35 0 in V By an inner product on V we will mean a positivede nite Hermitiansymmetric sesquilinear form Examples Vector Spaces 23 1 IE with the Euclidean inner product 3 23 2 Let V IE and let A E 15quot be Hermitiansymmetric and de ne n n lt5 A Z Z ngTgE i1j1 The requirement that 3 gt 0 for a 0 so that A 4 is an inner product serves to de ne positivede nite matrices 3 If V is any nitedimensional vector space we can choose a basis and thus identify V 2 IE and then transfer the Euclidean inner product to V in the coordinates of this basis The resulting inner product depends on the choice of basis in general there is no canonical inner product on a general vector space With respect to the coordinates induced by a basis any inner product on a nitedimensional vector space V is of the form described in example 2 above A 43 V One can de ne an inner product on 2 by To see from rst principles that this sum converges absolutely apply the nitedimensional Cauchy Schwarz inequality to obtain n n 1i n 0c 0c i1 i1 i1 i1 i1 Now let n gt 00 to deduce that the series converges absolutely The LQinner product on Cm 6 is given by u A 0 V Exercise Show that the inner product de ned in Example 5 above is indeed positive de nite on Cm An inner product on V determines an injection V gt V if U E V de ne E V by aw since 212112 it follows that 0 f w 0 so the map w H is injective The map 21 e gt is conjugate linear rather than linear unless IEquot R since hf The image of this map is a subspace of V If dim V lt 00 then this map is surjective too since dim V dim V In general it is not surjective Let dim V lt 00 and represent vectors in V as elements of I by choosing a basis If 531 L91 39v39w have coordinates f f respectively and the inner product has matrix A E 15 in this basis then i n agijj at i1 j 24 Linear Algebra and Allatrix Analysis It follows that has components 6 231 125 with respect to the dual basis In terms of matrices the map 3911 r gt is represented by bn Lyn An inner product on V allows a reinterpretation of annihilators If W C V is a subspace de ne the orthogonal complement read W quot perp W i E V 0Vw E W Clearly W i is a subspace of V The use of the same notation that we used for the annihilator of W subspace of V is justi ed by the observation that the image of this subspace W i of V under the map V gt V discussed above is precisely the annihilator of W If dim V lt 00 a dimension count and the obVious W 0 W i 0 show that V W G3 W i So in a nite dimensional inner product space a subspace W determines a natural complement namely W i The induced projection onto W along W i is called the orthogonal projection onto W

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