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lnner Product Spaces Linear Algebra Notes Satya Mandal November 217 2005 1 Introduction In this chapter we study the additional structures that a vector space over eld of reals or complex vector spaces have So7 in this chapter7 R Will denote the eld of reals7 C Will denote the eld of complex numbers7 and l Will deonte one of them In my View7 this is Where algebra drifts out to analysis 11 De nition Let l be the eld reals or complex numbers and V be a vector space over F An inner product on V is a function 7 V X V 6 l such that 1 am by72 1m72by727 for ab 6 l and 7112 6 V 2 my W for my 6 V 3 am gt 0 for all non zero m E V 4 Also de ne m H This is called norm of m Comments Real Case Assume l lR Then 1 Item means Ly y 2 Also I and 2 means that the inner product is bilinear Comments Complex Case Assume l C Then 1 Items 1 and 2 means that the Loy d2 any E2 12 Example On R we have the standard inner product de ned by my 221 myi Where m 177n E R and y y17yn E R 13 Example On C we have the standard inner product de ned by my 21 E7 Where m 177n E C and y y17yn E C 14 Example Let l R OR C and V For A myLB b2 6 V de ne inner product A73 Zahar 2 De ne conjugate transpose B P Then A7 B traceBA 15 Example Integration Let l R OR C and V be the vector space of all Rivalued continuous functions on 07 1 For f7 9 E V de ne l m 0 fydt This is an inner product on V In some distant future this Will be called L2 inner product space This can be done in any 77space77 Where you have an idea of integration and it Will come under Measure Theory 16 Matrix of Inner Product Let l R OR C Suppose V is a vector space over l With an inner product Let 617 7en be a basis of V Let pm ehej and P p27 6 Then for v 161 mnen E V and Wy1e1ynen E Vwe have U710 Ziy jpij 177nP yz This matrix P is called the matrix of the inner product With respect to the basis 617 7e De nition A matrix B is called hermitian if B B So7 matrix P in 1 is a hermitian matrix Since my gt 07 for all non zero v 6 V7 we have XPYtgt0 for all nonizero XElF It also follows that P is non singular Otherwise XP 0 for some non zero X Conversely7 if P is a n X n hermitian matrix satisfying such that X137t gt 0 for all X e r then X Y X137t for X e r de nes an inner product on lF 2 Inner Product Spaces We will do calculus of inner produce 21 De nition Let l R OR C A vector space V over l with an inner product gtk7 is said to an inner product space 1 An inner product space V over R is also called a Euclidean space 2 An inner product space V over C is also called a unitary space 22 Basic Facts Let l R OR C and V be an inner product over F For v10 E V and c E l we have 1 llcvlll0lllvll7 2 HUHgt0 if va O7 my HvH2 39 3 l v10 l S v w H Equility holds if and only if 10 It is called the CauchySwartz inequality 4 vw S v w It is called the triangular inequality Proof Part 1 and 2 is obvious from the de nition To prove the Part 37 we can assume that v a 0 Write 107v 2107 1 Hvll2 Then 21 0 and w w o H 2 H2 2710 H v v H2v 27w 10 va This establishes We will use Part 3 to prove Part 47 as follows H v 10 H2H v H2 v7w wiv H w H2H v H2 2R lv7wl H w H23 H v H2 2 l WU l H w HZSH v H2 2 H v NH 10 H H 10 H2 H v H H w W This establishes Part 4 23 Application of CauchySchwartz inequality Application of 3 of Facts 22 gives the following 1 Example 12 gives 1 2 1 Zx 12Zy 12 2 1 21 21 for my 6 1R 2 Example 13 gives 1 1S 2 1 239 1212Z l 11239 mm 21 21 21 for my 6 C 3 Example 14 gives traceAB 1 traceAAlZtmceBB12 for AB E 4 Example 15 gives 1 forme lft12dt12 mt 2 ml2 for any two continuous Civalued functions f g on 01 21 Orthogonality 24 De nition Let l be R or C Let V be an inner product space over F 1 Supppose v10 E V We say that v and w are mutually orthogonal if the inner product 1710 0 OR equivalently if 1071 0 We use variations of the expression 7mntnally orthogonal and sometime we do not mention the word 7mntnally 7 2 For v10 E V we write 1 l w to mean i and w are mutually orthogonal 3 A subset S Q V is said to be an orthogonal set if vlw for all 1510687 va w 4 An othrhogonal set S is said to be an orthnorrnal set if HoH1 for all UES 5 Comment Note the zero vector is otrhogonal to all elements of V 6 Comment Geometrically7 i l it means i is perpendicular to w 7 Example Let V R or V C Then the standard basis E e1L7 7en is an orthonormal set 8 Example In R2 or C27 we have the ordered pairs 1 my and w 157 are oththogonal Caution Notation Ly here 25 Example Cosider the example 15 over R Here V is the inner prod uct space of all real valued continuous functions on 07 1 Let fnt cos27rnt gnt sin27rnt Then 17f17917f27927m is an orthonormal set Now consider the inner product space W in the same example 15 over C Let fniyn hm ex 27rmt W p Then hmn0717172772777 is an orthonormal set 26 Theorem Let F R or F C and let V be an inner product space over F Let S be an orthogonal set of non zero vectors Then S is linearly independent Therefore cardinalityS S dimV Proof Let v115 71 E S and 1010202Cnvn0 Where Ci 6 F We Will prove Ci 0 For example7 apply inner product gtk7 m to the above equation We get 0101701 0202701 39 Cnvn7v1 0701 0 Since v17v17 0 and v27v1 v37v1 vmvl 07 we get 01 0 Similarly7 Ci 0 for all 239 17 772 27 Theorem Let F R or F C and let V be an inner product space over F Assume 617627 7eT is a set of non zero orthogonal vectors Let 1 E V and UCi 102 2CT T Where Ci 6 F Then forz391r Proof For example7 apply inner product 761 to the above and get U761 0161761 Cl ll 61 HZ So7 01 is as asserted and7 similarly7 so is Ci for all 239 7 28 Theorem Let 1 R or 1 C and let V be an inner product space over 1 Let v1v2 1 be a set of linearly independent set Then we can construct elements 61 62 6 E V such that 1 61 62 6 is an orthonormal set 2 6k 6 Spanv1 vk Proof The proof is known as GramSchmidt orthogonalization process Note v1 a 0 First let vi 61 H v1 H Then 61 H 1 Now lat 02 127 1 1 2 ll 02 02761 1 ll Note that the denominator is non zero 61 1 62 and 62 H 1 Now we use the induction Suppose we already constructed 61 ek1 that satis es 1 and 2 and k S 7 Let we 211 U167 629 6k 39 H we Z1vk7 i H Note that the denominator is non zero 6k H 1 and 6k 1 e for 239 1 k 7 1 From construction we also have Spanv1 vk Span61 ek So the proof is complete 29 Corollary Let 1 R or 1 C and let V be a nite dimensional inner product space over 1 Then V has an orthonormal basis Proof The proof is immediate from the above theorem 28 Examples Read examples 12 and 13 page 282 for some numerical com putations 210 De nition Let l R or l C and let V be a inner product space over F Let W be a subspace of V and v E V An element 100 E W is said to be a best approximation to v or nearest to v in W7 if Hv7w0H S Hv7wH for all wEW Try to think what it means in R2 or C when W is line or a plane through the origin Remark I like the expression 77nearest77 and the textbook uses 77best ap proximation I will try to be consistent to the textbook 211 Theorem Let l R or l C and let V be a inner product space over F Let W be a subspace of V and v E V Then 1 An element 100 E W is best approximation to v if and only if v 7100 T w for all it E W 2 A best approximation 100 E W to v if exists7 is unique 3 Suppose W is nite dimensional and 6176277 n is an orthonormal basis of W Then n 100 21 ekek 191 is the best approximation to v in W The textbook mixes up orthogonal and orthonormal and have a condition the looks complex We assume orthonormal and so ei H 1 Proof To prove 1 let 100 E W be such that v 7 100 T w for all it E W Then7 since it 7 100 6 W7 we have H viw H2H viwowow H2H viwo H2 H woiw H2le viwo H2 Therefore7 100 is nearest to v Conversely7 assume that 100 E W is nearest to i We will prove that the inner product i 7 100710 0 for all it E W For convenience7 we write v0 v 7 100 So7 we have We Hz llviw H2 Eqn71 9 for allw E W Writev7w v7w0w07w v0w07w Since any element in 10 van be written as 100 7 w for some 10 E W Eqn l can be rewritten as H v0 HZSH v0 10 H2 for all 10 E W So we have H v0 HZSH v0 10 H2H v0 H2 2113 lv07wlHwH2 and hence 0 S 2Rev0w w H2 Eqn 7 H for all 10 E W Fix w E W with w a 100 and write 739 v07w0 i w wo 7 w 7 H 100 i w H2 Since 739 E W we can substitute 739 for w in Eqn ll and get l 007100 w l2 0lt7 Hwo wHZ Therefore v0w0 7 w 0 for all 10 E W with 100 7 w a 0 Again since any non zero element in W can be written as 100 7 w it follows that v0w 0 for all 10 E W So the proof of 1 is complete To prove Part 2 let 100101 be nearest to 1 Then by orthogonality 1 we have H100 7 101 H2 wo 7 101100 7 101 wo 7 w1w0 7 v v 7 101 0 So Part 2 is established Let 100 be given as in Part 3 Now to prove Part 3 we will prove 10071 l e for 239 1 n So for example wo 7 v61 10061 7 v61 v616161 7 v61 0 So Part 3 is established 10 212 De nition Let l R or l C and let V be an inner product space over F Let S be a subset of V The otrhogonal complement SL of S is the set of all elements in V that are orthogonal to each element of S So SL UEVZ UJ LU for all 1063 It is easy to check that 1 SL is a subspace of V 2 0L V and 3 VL 0 213 De nition and Facts Let l R or l C and let V be an inner product space over F Let W be a subspace of V Suppose v E V we say that w E W is the orthogonal projection of v to W if 10 is nearest to v 1 There is no guarantee that orthogonal projection exists But by Part 2 of theorem 211 when it exists orthogonal projection is unique 2 Also if dimW is nite then by Part 3 of theorem 211 orthogonal projection always exists 3 Assume dimW is nite De ne the map 7139W V a V where 7rv is the orthogonal projection of v in W The map 7139W is a linear operator and is called the orthogonal projection of V to W Clearly 7 WW So 7139W is indeed a projection 4 For 1 E V let Ev v 7 7139Wv Then E is the othrogonal projection of V to Wt Proof By de nition of 7fw we have Ev v 77rWv E WL Now given 1 E V and 10 E WL we have 1 7 Evw WWvw 0 So by theorem 211 E is the projection to WL Example Read Example 14 page 286 11 214 Theorem Let l R or l C and let V be an inner product space over F Let W be a nite dimensional subspace of V Then 7139W is a projection and WL is the null space of 7139W Therefore V W 619 WL Proof Obvious 215 Theorem Bessel s inequality Let l R or l C and let V be an inner product space over F Suppose v1v2 vn is an orthogonal set of non zero vector Then for any element 1 E V we have TL H W l2 2 H H2 3 llvllz k1 1 Also the equality holds if and only if n v Z Ulvkivk H vk H Proof We write 1 6k Hvkll and prove that 3 H0760 l2 S H v H2 191 Where 61 62 en is an orthonormal set Write W Span6162en By theorern 211 n 100 Zvekek 191 is nearest to v in W So 1 7 100100 0 Therefore H2 H we H2 H v H2H 0 i 100 we H2H 0 i 100 So TL H v H2 2 H wo H2 2 l m V 191 12 Also note that the equality holds if and only if v 7 100 HZ 0 if and only if v wo So7 the proof is complete If we apply Bessel7s inequality 215 to te example 25 we get the following inequality 216 Theorem Application of Bessel s inequality For and Civalued continuous funtion f on 01 we have 1 ftez 27r239ktdt12 11M dt l p l Homework Exercise 1 77 97 11 from page 288 289 These are popular problems fro Quals 13 3 Linear Functionals and Adjoints We start With some preliminary comments 31 Comments and Theorems Let l R or l C and let V be an inner product space over F 1 Given an element 1 E V we can de ne a linear functional f V a lF by fv LU 2 The association Fv f de nes a natural linear map F V a V 3 In fact F is injective Proof Suppose f 0 Then LU 0 for all m E V so v10 0 and hence v 0 4 NOW assume that V has nite dimension Then the natural map F is an isomorphism 5 Again assume V has nite dimension n and assume 61 67 is an orthonormal basis Let f E V Let 1 Z e ei k0 Then f Fv fv That means f LU for all m E V Proof We Will only check f61 611 Which is obvious by or thonormality 6 Assume the same as in Part 5 Then the association Cf Z fe e k0 de nes the inverse G V a V of F 14 7 Remark The map F fails to be isomorphism if V is not nite dimensional Following is an example that shows F is not on to 32 Example Let V be the vector space of polynomial over C For f7 9 E V de ne inner product m D mamas Note fol tjtkdt So7 if fX 21ka and 9X Zkak we have 1 U79 mailm Fix a complex number 2 E C By evaluation at 2 we de ne the functional L V a C as Lf for any f E V We claim that L is not in the image of tha map F V a V In other words7 there is no polynomial g E V such that 1 flt2gt LU my ft9tdt for all f e v To prove this suppose there is such a 9 Write h X72 Givenf 6 V7 we have hf E Vand0 hf2 Lhf 7 So 1 o Lhf hm htft9tdt for all f e v By substituting f X 7 3 we have 1 0 ht 2 t 2 dt 0 l m M l Since h a 0 it follows 9 0 But L a 0 15 31 Adjoint Operatior 33 Theorem Let l R or l C and let V be an nite dimensional inner product space over F Suppose T E LV7 V is a linear operator Then7 there is a unique linear operator TZVgtV such that T11111 11T111 for all 11111 E V De nition This operator T is called the Adjoint of T Proof Fix and element 111 E V Let P V a F be de ned by the diagram V gtT V x 19 1quot 7 l That means N11 T11111 for all 11 E V By Part 2 of thorem 317 there is an unique element 11 such that N11 1511 for all 11 E V That means T11111 1511 for all 11 E V Eqn 7 I Now7 de ne T111 11 It is easy to check that T is linear use Eqn l Uniqueness also follows from Eqn l 34 Theorem Let l R or l C and let V be an nite dimensional inner product space over F Suppose T E LV7 V is a linear operator Let 177 n be an orthomormal basis of V Suppose T E LV7 V be a linear operator of V 1 Write 12739 T j7 and A 042739 Then A is the matrix of T With respect to 177 n 16 2 With respect to 617 76m the matrix of the adjoint T is the conjugate transpose of the matrix A of T Proof To prove Part 17 we need to prove that Te13939 en TM en Tram This follows because7 if T61 A161A262 Anem then Ai T 17 So7 the proof of Part 1 is complete Now7 by Part 17 them matrix of the adjoint T is B T j7 7T j T 7 j Al This completes the proof of Part 2 35 Theorem Projection and Adjoint Let l R or l C and let V be an nite dimensional inner product space over F Let E E LV7 V be an orthogonal projection Then E E Proof For any my 6 V7 we have E7y E7EylyE1l E7Ey because lyEyl L W E7Ey lEl7E17Ey b cause lEl L W Therefore E7y7E1 for My 6 V Hence E E 36 Remarks and Examples let V R and A 6 MAR be a symmet ric matrix Let T E LV7 V be de ned by A Then T T Also note that matrix of T with respect to some other basis may not be symmetric Read Example 17 21 from page 294 296 17 37 Theorem Let l R or l C and let V be an nite dimensional inner product space over F Let T7 U E LV7 V be two linear operator and c E F Then 1 T U T U 2 cT an 3 TU UT 4 my T Proof The proof is direct consequence of the de nition theorem 33 The theorem 37 can be phrased as the map LV7 V a LV7 V that sends T a T is conjugate linear7 anti isomorphism of period two 38 De nition Let l R or l C and let V be an nite dimensional inner product space over F Let T E LV7 V be a linear operator and c E F We say T is selfadjoint or Hermitian if T T Suppose E 61776n is an orhtonorrnal basis of V Let A be the matrix of T With respect to E Then T is self adjoint if and only if A 14 18 4 Unitary Operators Let me draw your attention that the expression 77isomorphism77 means dif ferent things in di erent context like group isomorphism vector space iso morphism module isomorphism In this section we talk about isomorphisms of inner product spaces 41 De nition and Facts Let l R or l C and let V W be two in ner product spaces over F A linear map T V a W is said to preserve inner product if TxTy Ly for all my 6 V We say T is an an isomorphism of inner product spaces if T preserves inner product and is one to one and onto 42 Facts Let l R or l C and let V W be two inner product spaces over F Let T V a W be a linear map 1 If T preserves inner product then HTHHH for all xEV 2 If T preserves inner product then T is injective ie one to one 3 If T is an isomorphism of inner product spaces then T 1 is also an isomorphism of inner product spaces Proof Obvious 43 Theorem Let l R or l C and let V W be two nite dimensional inner product spaces over lF with dimV dimW 72 Let T V a W be a linear map The the following are equivalent 1 T preserves inner product 2 T is an isomorphism of inner product spaces 3 If 61 en is an orthonormal basis of V than T 1 Ten an orthonormal basis of W 19 4 There is an orthonormal basis 61 en ofV such that T 1 Ten also an orthonormal basis of W Proof 1 i 2 Since T preserves inner product T is injective Also since dimV dimW it follows Hence T is also onto So Part 2 is established 2 i 3 Suppose 61 en is an orthonormal basis of V Since also pre serves preserves inner product T 1 Ten is an orthonormal set Since dimW n and since orthonormal set are independent T 1 Ten is a basis of W So Part 3 is established 3 i 4 Since V has an orthonormal basis Part 4 follows from Part 3 4 i 1 Let 61 en be an orthonormal basis ofV such that T 1 Ten also an orthonormal basis of W Since TeiTe j for all 2 it follows TxTy Ly for all my 6 V So Part 1 is established 44 Corollary Let l R or l C and let VW be two nite dimen sional inner product spaces over F Then V and W are isomorphic as inner product spaces if and only is dimV dimW Proof If V and W are isomorphic then clearly dimV dimW Conversely if dimV dimW n then we can nd an orthonormal basis 61 en of V an an orthonormal basis E1 En of W The association de nes and isomorphism of V to W 45 Example Let l R or l C and let V be a nite dimensional inner product spaces over F with dimV 72 Fix a basis E 61 en of V Consider the linear map f V a ll given by f11 1 amen 11 an With respect to the usual inner product on F this map is an isomorphism of of ineer product spaces if and only if E 61 en is an orthonormal basis 20 Proof Note that f sends E to the standard basis So7 by theorem 437 E is orthonormal f is an isomorphism Homework Read Example 23 and 25 from page 301 302 Question This refers to Example 25 Let V be the inner product space of all continuous Rivalued functions on 017 With inner product M 01ftgtdt Let T V a V be any linear operator What can we say about When T perserves inner product 46 Theorem Let l R or l C and let V7W be two inner product spaces over F Let T V a W be any linear tansformation Then T perserves inner product if and only if HTvH M for an zev Proof If perserves inner product then clearly the condition holds Now suppose the condition above holds and my 6 V Then HTyH2 llyll2 Since HTMH HM and HTyH illH7 it follows that T7Ty Ty7T 711 11 Hence R lT7T1l R l7yl lf l R then the proof is complete Also7 since Similar arguments With m 7 y7 Will give Imw7z Rew7iz for all w72EV or W the proof is complete 21 47 De nition and Facts Let l R or l C and let V be an inner product space over F An isomorphism T V V of inner product spaces is said to be an unitary operator on V 1 Let M 72 MV denote the set of all unitary operators on V 2 The identity I E 1V 3 If U1 U2 6 WV then U1U2 6 WV 4 If U 6 WV then U 1 6 WV 5 So7 MV is a group under composition It is a subgroup of the group of linear isomorphisms of V Notationally MV Q GLV Q LV7 V 6 If V is nitel dimensional then a linear operator T E LV7 V is unitary if and only if T preserves inner product 48 Theorem Let l R or l C and let V be an inner product space over F Let U E LV7 V be a linear operator Then U is unitary if and only if the adjoint U of U exists and UU UU 1 Proof Suppose U is unitary Then U has an inverse U l So7 for my 6 V we have U7y U7UU 1y 7U 1y So7 U exists and U U l Conversely7 assume the adjoint U exists and UU UU I We need to prove that U preserves inner product For my 6 V we have U7Uy 7UUy 711 So7 the proof is complete Homework Read example 277 page 304 49 Theorem Let l R or l C and A be an n X 72 matrix Let T F a F be the linear operator de ned bt TX AX With usual inner product on an7 we have T is unitary if and only if AA I 22 Proof Suppose AA I Then AA AA I Therefore7 Conversely7 suppose T is unitary Then7 yAAx ya for all my 6 lF With appropriate choice of L y we can show that AA I So7 the proof is complete 410 De nition An n X 727 matrix A 6 MAR is called an orthogonal matrix if AtA In The subset 0n Q MAR of all orthognal matrices from a subgroup of GLAR An n X 727 matrix B 6 MAC is called an unitary matrix if BB In The subset 72 Q MAC of all unitary matrices from asubgroup of GLAC 411 Theorem Let l R or l C and let V be a nite dimensional inner product space over F Let U E LV7 V be a linear operator Then U is unitary if and only if the matrix of U in with respect to some or every orthonormal basis is unitary Proof By theorem 48 7 U is unitary if and only if U exists and UU UUU 1 Suppose E 61776n is an orthonormal basis of V and A be the matrix on U with respect to E Assume U is unitary Since UU UU I7 we have AA AA I So A is unitary Conversely7 assume A is unitary Then AA AA I Write A 17 Therefore U 7 Uej n n 71 Z ak mzak k 2akiek7akjek 20419in 52 767 k1 k1 k1 191 So7 the proof is complete 412 Exercise 1 A matrix A 6 MAR is orthogonal if and only if A 1 At 2 A matrix B 6 MAC is unitary if and only if B 1 B 413 Theorem Suppose A E GLAC be an invertible matrix Then there is a lower triangular matrix M E GLAC such that MA 6 UAC and diagonal entries of M are positive F urhter7 such an M is unique 23 Proof Write vi A 2 vn Where vi 6 C are the rows of A Use Gram Schmidt orthogonalization the orern 28 and de ne k we 23411 67 6k 1971 H vie 271vk7 j j ll Note 617 6 is an orthogonalnorrnal basis of C Also k 6k E ckjvj with CM 6 C and 073 a 0 j1 So7 we have 6 011 0 0 0 v 1 021 022 0 0 1 62 02 031 032 033 0 e 0 1 TL TL Cnl CnZ on 39 39 39 Crm Since 61 62 6n unitary7 the existance is established For uniqueness7 assume UMB and U1NB Where U7U E 72 and M7N are lower triangular With diagonal entries positive Then MN I UUjf lL is unitary Note that N 1 is also a lower triangular matrix With positive diagonal entries Therefore7 24 1 MN 1 A 61172 2 A is a diagonal matrix7 3 Diagonal entries of A are positive Therefore A MN l I and M N So7 the proof is complete Homework Read example 287 page 307 414 Theorem Let l R or l C and let V be a nite dimensional inner product space over F Suppose E 61 7671 and 5 617quot 76 are two orthonormal basesof V Let 177en 177 nP for some matrix P Then P is unitary Proof We have 61 E1 61en Pt 62 6177enP en En Sorry7 this approach is naive and fails because such products of matri ces of vectors and scalars matrix products and inner products are not associative In any case7 the correct proof is left as an exercise 415 Exercise Consider V C 7 With the usual inner product Think of the elements of V as row vectors Let v17v277vn E V and let vi A U2 5 1 Prove that v115 71 forms a basis if and only if A invertible 2 Prove that v115 1 forms an orthonormal basis if and only if A an unitary matrix ie AA I 3 We can make similar statements about vectors in R and orthogonal matrices 25 5 Normal Operators Let V be an inner product space and T E LV V Main objective of this section is to nd necessary and suf cient conditions for T so that there is an orthonormal basis 61 67 of V such that each 6239 is an eigen vector of T 51 Theorem Let F R or F C and let V be a nite dimensional inner product space over F Let T E LV V be a linear operator Suppose E 61 67 is an orthonormal basis of V and each 6239 is an eigen vector of T 1 So we have Tei 026239 for some scalars c E F 2 So the matrix of T With respect to E is the diagonal matrix A diagonaKcl Cg 0 3 f F R then the matrix of the adjoint operator T is A A diagonalcl02 0 Therefore in the real case a suf cient condition is that T is self adjoint 4 f F C the matrix of the adjoint operator T is A diagonaMc h Q Therefore TT TT So in complex case a necessary condition is that T commutes With the adjoint T Compare with theorem 48 52 De nition Let F R or F C and let V be a nite dimensional inner product space over F Let T E LV V be a linear operator We say T is a normal operator if TT TT Therefore self adjoint operators are normal 26 53 Theorem Let l R or l C and let V be a nite dimensional inner product space over F Let T E LV V be a linear operator If T is self adjoint then 1 Eigen values of T are real 2 Eigen vectors associated to distinct eigen values are orthogonal Proof Suppose c E l be an eigen valuea and e E V be the corresponding eigen vector Then Te cc and cee 066 Tee eTe eTe 666 Eee So 0 E and c is real Now suppose Te cc and T d6 where c a d scalars and 66 E V be nonzero Then cw cw Te amen we we 8a Since d a c we have 66 0 So the proof is complete 54 Theorem Let l R or l C and let V be a nite dimensional inner product space over F Let T E LVV be a self adjoint operator Assume dimV gt 0 Then T has a non zero eigen vector Proof Let E 61 67 be an orthonormal basis of V Let A be the matrix of T Since T is self adjoint A A Now we deal with real and complex case seperately Real Case Obviously A 6 MAR and A A means A is symmetric In any case consider the map UC 7C where UXAX for XEC Since A A by theorem 53 U has only real eigen values So detXI 7 A 0 has ONLY real solution Since C is algebraically closed we can pick a real solution c E R so that detcI 7 A 0 Therefore CI 7 AX 0 has a real non zero solution 1 Wt E R Write e 161 767 then 6 a 0 and Te 06 Note we went upto C to get a proof in the real case Complex Case Proof is same only easier Here we know detXI7 A 0 has a solution c E C and the rest of the proof is identical 27 55 Exercise Let V be a nite dimensional inner product space over C Let T E LV V be a self adjoint operator Let QX be the characteristic polynomial of T Then QX E RX Proof We can repeat some of the steps of theorem 54 Let E 61 en be an orthonormal basis of V Let A be the matrix of T Since T is self adjoint A 14 The QX detXI 7 A Then QX X 7 clX 7 02 X 7 on where c E C By arguments in theorem 54 c E lR for 239 1 n 56 Example 29 page 313 Let V be the vector space of continuous C7valued functions on the interval 01 As usual for fg E V de ne inner product 1 my ftytdt 0 Let T V 7 V be the operator de ned by Tft tft 1 Then T is self adjoint This is true because tf79f7ty for all 139 EV 2 T has no non zero eigen vector Proof Suppose f E V and Tf cf for some 0 E C Then tft cft for all t 6 01 Since f is continuous f 0 3 This example shows that theorem 54 fails if V is in nite dimensional 57 Theorem Let l R or l C and let V be a nite dimensional inner product space over F Let T E LV V be a linear operator Suppose W is a T7invariant subspace of V Then the orthogonal complement WL of W is invariant under T Proof Let m E W and y E WL Since T E W we have Ty Ty 0 807 Ty E Wt 28 58 Theorem Let l R or l C and let V be a nite dimensional inner product space over F Let T E LV7 V be a self adjoint linear operator Then V has an orthonormal basis E 617 7an such that each 6239 is an eigen vector of T Proof We assume that dimV n gt 0 By theorem 547 T has an eigen vector 1 Let 1 61 H v H39 lf dimV 17 we are done Now we will use induction and assume that the theorem to true for inner product spaces of dimension less than dimV Write W F61 and V1 WL Since W is invariant under T7 by theorem 577 WL is invariant under T T Therefore7 V1 has a orthonormal basis lt227 7an such that 627 6 are eigen vectors of TM hence of T Therefore 617627 7an is an orthonormal basis of V and each 6239 is eigen vectors of of T So7 the proof is complete 59 Theorem 1 Let A 6 MAC be aHermitian self adjoint matrix Then there is unitary matrix P such that P lAP A is a diagonal matrix 2 Let A 6 MAR be a symmetric matrix Then there is orthogonal real matrix P such that P lAP A is a diagonal matrix Proof To prove Part 17 consider the operator T C a C where TX AX for X E C Since A is Hermitian7 so is T By theorem 587 there is an orthonormal basis E lt217 7an of C such that each ei is an eigen vector of T So7 we have T 17T 27 7T n 617627 7enA where A diagonaucl7 7on is a diagonal matrix and Tei 626239 Sup pose 17 76 is the standard basis of C 7 and 177en 177 nP for some matrix P E Then P is unitary We also have T 177T n 177 nA 29 Combining all these7 we have A PAPA So7 the proof of Part 1 is complete The proof of Part 2 is similar 30 51 Regarding Normal Operators 510 Theorem Let l R or l C and let V be a nite dimensional inner product space over F Let T E LV V be a normal operator 1 Then T T for all a E V 2 Suppose v E V Then 1 is an eigen vector of T with eigen value c if and only if v is an eigen vector of T with eigen value E In other words Tv cv gt Tv Ev Proof We have TT TT and T H2 T7T 7TT 7TT T7T 11 TW H2 So Part 1 is established To prove Part 2 for a c E l write U T 7 c So U T 75 Since TT TT we have UU UU Therefore by 1 H Tid v 1111 T EIXU H Therefore the proof of Part 2 is complete 511 De nition A matrix A 6 MAC is said to be normal if AA AA 512 Theorem Let l R or l C and let V be a nite dimensional inner product space over F Let T E LVV be a linear operator on V Let E 61 en be an orthonormal basis and let A the matrix of T with respect to E Assume A is upper triangular Then T is normal if and only if A is diagonal Proof Since E is orthonormal matrix of T is A Assume A is diagonal Then AA AA Therefore TT TT and T is normal Conversely assume T is normal So TT TT and hence AA AA First we will assume n 2 and prove A is diagonal Write AltaM a12gtand so AltE 0 an 112 122 31 We have AAltla11l2lltl12l2 112 gt 1221 12 l 122 l2 and Ami lt lam l2 a nau gt 1 12 111 l 112 l2 l 122 l2 Since AA AA we have l 111 2 llt112 l2llt111l2 Hence 112 0 and A is diagonal For n gt 27 we nish the proof by similar computations To see this7 write 111 112 113 39 39 39 am 0 122 123 39 39 39 a2n A 0 0 133 39 39 39 13m 0 0 0 am Comparing the diagonal entries of of the equation AA AA we get7 n 2 l We l2l 122 l2 k2 for 239 17 772 So we have 1 0 for all k gt 239 Therefore7 A is a diagonal matrix 513 Theorem Let V be a nite dimensional inner product space over C Let T E LV7 V be a linear operator on V Then there is an orthonormal basis E 617 7an such that the matrix of T with respect to E is upper triangular Proof We will use induction on dirnV Note that theorem is true when dirn V 1 Since C is algebraically closed7 the adjoint T has an eigen vector v a 0 Write e v v The e is an eigen vector of T and T416 06 for some 0 E C 32 Let WCe and VFWL Since W is T7invariant7 by theorem 577 V1 is Tiinvariant Let T1 TM be the restriction of T Then T1 6 LV17V1 By induction7 there is an orthonormal basis E 61776n1 of V1 such that the matrix of T1 With respect to E is upper triangular Write en 6 then the matrix of T With respet to E0 6177671717671 is upper triangular 514 Let A 6 MAC be any matrix Then there is an unitary matrix U E Q MAC7 such that U lAU is upper triangular Proof Proof is an immediate application of theorem 513 to the map C a C de ned by the matrix A 515 Theorem Let V be a nite dimensional inner product space over C Let T E LV7 V be a normal operator on V Then there is an orthonormal basis E 617 7an such that each 6239 is an eigen vector of T Proof By theorem 5137 V has an orthonormal basis E lt217 7an such that that matrix A of T With respect to E is upper triangular Since T is normal7 by therem 5127 A is diagonal Therefore ei is an eigen vector of T The following is the matrix version of this theorem 515 516 Let A 6 MAC be a NORMAL matrix Then there is an unitary matrix U E 72 Q MAC7 such that U lAU is diagonal Proof Proof is an immediate application of theorem 515 to the map C a C de ned by the matrix A 33
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