MTH50 Spring 07 HW Assignment 7 {From [FIS0]}: Sec 44 #4a h 6; Sec 5 #ad ac 4ae 4 7 The due date for this assignment is 04/05/7 Sec 44 #4a h Evaluate the erminant of the following matrices by any legitimate method a A 0 4 5 6 ; h B 0 5 4 9 9 0 7 8 9 4 9 4 5 Solution The simplest way is to convert the matrices to upper triangular form such as using the forward pass in Gaussian elimination and use properties -4 of the erminant a A 0 R +R R A 4 5 6 7 8 9 0 6 0 0 0 4R+R R 7R +R R A and so by properties and 4 of the erminant we have b B 0 R +R R B R+R4 R4 A 0 A A 0 0 5 4 9 9 0 4 9 4 5 0 4 0 0 5 5 0 0 5 5 0 6 0 6 5R +R R ; 9R +R R B 4R +R 4 R 4 R+R4 R4 B and so by properties and 4 of the erminant we have B 0 B B B 5 0 00 Sec 44 #6 Prove that if M M n n F can be written in the form A B M 0 C where A and C are square matrices then M A C 0 4 0 0 5 5 0 0 0 0 0 4 0 4 7 77 0

Proof If C exists then A BC I 0 M 0 C A B A B 0 C so that A BC M I 0 0 C and by property of erminants we have I 0 C 0 0 C and hence by exercise 5 of this section which is proved using erminant expanions along rows and columns for instance we have A BC A C 0 C and hence A BC M I 0 0 C A C If C isnt invertible then by property 7 of erminants we have C 0 and there exists invertible matrices E F of same dimensions as C of product of elementary matrices such that Ir 0 ECF 0 0 and hence I 0 A B I 0 M 0 E 0 E 0 C 0 F I 0 A BF I 0 0 E 0 ECF 0 F I 0 A BF I 0 0 E 0 ECF 0 F I 0 0 F implying that A BF 0 ECF 0

since its bottom row of that matrix is all zeros and hence I 0 A BF I 0 M 0 E 0 ECF 0 F 0 A C This completes the proof Sec 5 #ad For each of the following linear operators T on a vector space V and ordered bases compute [T ] and ermine whether is a basis consisting of eigenvectors of T a V R T a b 0a 6b 7a 0b { and d V P R T a + bx + cx 4a + b c 7a + b + 7c x + 7a + b + 5c x { x x + x x + x } Solution a We have T 0 7 0 λ for any real number λ Therefore is not a basis of eigenvectors of T d We have T x x 4 + 4x 4x λ x x for any real number λ Therefore is not a basis of eigenvectors of T Sec 5 #ac For each of the following matrices A M n n F i Determine all of the eigenvalues of A; ii For each eigenvalue λ of A find the set of eigenvectors corresponding to λ; iii If possible find a basis for F n consisting of eigenvectors of A; iv If successful in finding such a basis ermine an invertible matrix Q and a diagonal matrix D such that Q AQ D a A for F R; i c A for F C i Solution a i First which has roots A λi λ λ λ λ 4 } λ λ 6 λ ± 9 4 4 ± 5 4 so that A has eigenvalues λ λ 4

ii Second λ N A λ I N λ { } N span λ N A λ I N λ N { x R x + y 0 : y x y 0 } } span { so that eigenpairs of A are λ v ; λ 4 v iii Third a basis for R consisting of eigenvectors of A is { } {v v } iv Fourth b i First Q λ 0 0 AQ D 0 λ 0 4 Q v v A λi i λ i λ λ λ λ + i λ i λ so that A has eigenvalues λ λ 4

ii Second i λ N A λ I N i λ i + N i { } x C i + x + y 0 : y x + i y 0 { } span i + i λ N A λ I N i λ i N i { } x R i x + y 0 : y x + i y 0 { } span i so that eigenpairs of A are λ v i + ; λ v i iii Third a basis for R consisting of eigenvectors of A is { } {v v } i + i iv Fourth Q λ 0 0 AQ D 0 λ 0 Q v v i + i Sec 5 #4ae For each linear operator T on V find the eigenvalues of T and an ordered basis for V such that [T ] is a diagonal matrix a V R and T a b a + b 0a + 9b e V P R and T f x xf x + f x + f Solution a Let α {e e } be the standard ordered basis for R Then [T ] α 0 9 5

so that the characteristic polynomial of T is λ T λi [T ] α λi 0 9 λ λ 9 λ + 0 λ 7λ + which has roots λ 7 ± 49 4 so that T and [T ] α have eigenvalues λ λ 4 7 ± 4 To compute the eigenvectors of [T ] α we have λ N [T ] α λ I N 0 9 λ 5 N 0 6 { } span 5 λ N [T ] α λ I N 0 9 λ 6 N 0 5 { } span Thus the eigenpairs of T are λ v e + 5 e ; 5 λ 4 v e + e 5 and the ordered basis {v v } for V makes [T ] diagonal e Let α { x x } be the standard ordered basis for P R Then [T ] α 9 4 0 0 6

so that the characteristic polynomial of T is T λi [T ] α λi λ λ λ λ [ λ λ ] λ 9 λ 4 0 0 λ λ [ λ 4λ ] λ λ 4 λ so that T and [T ] α have eigenvalues λ 0 λ λ 4 To compute the eigenvectors of [T ] α we have λ 9 N [T ] α λ I N λ 4 0 0 λ N 9 4 0 0 span 0 λ 9 N [T ] α λ I N λ 4 0 0 λ 9 N 4 0 0 0 4 span 4 λ 9 N [T ] α λ I N λ 4 0 0 λ 9 N 4 0 0 span 0 7

Thus the eigenpairs of T are and the ordered basis λ 0 v x; λ v 4 4 x + x ; λ 4 v + x {v v v } for V makes [T ] diagonal Sec 5 # Let T be a linear operator on a finite-dimensional vector space V over a field F let be an ordered basis for V and let A [T ] In reference to Figure 5 prove the following: a If v V and φ v is an eigenvector of A corresponding to the eigenvalue λ then v is an eigenvector of T corresponding to λ b If λ is an eigenvalue of A and hence of T then a vector y F n is an eigenvector of A corresponding to λ if and only if φ y is an eigenvector of T corresponding to λ Proof Recall the coordinate map φ : V F n is an isomorphism satisfying φ λv [λv] λ[v] λφ v for all λ F and all for all v V ; [T ] φ v [T ] [v] [T v] for all v V And hence φ : F n V is an isomorphism a If v V and φ v is an eigenvector of A corresponding to the eigenvalue λ then φ v 0 and φ λv λφ v Aφ v [T ] φ v φ T v implies from this and the fact that φ : V F n is an isomorphism that v 0 and T v λv This proves v is an eigenvector of T corresponding to λ and therefore a is true b Suppose λ is an eigenvalue of A and hence of T If y F n is an eigenvector of A corresponding to λ then y 0 so that v φ y 0 and since Ay λy then implies φ T v [T ] φ v Ay λy φ λv v φ y 0 T v λv implying φ y is an eigenvector of T corresponding to λ Conversely v y is an eigenvector of T corresponding to λ then φ v 0 T v λv 8

implying y φ v 0 and Ay [T ] φ v φ T v φ λv λφ v λy y F n is an eigenvector of A corresponding to λ This proves b Sec 5 #4 For any square matrix A prove that A and A t have the same characteristic polynomial and hence the same eigenvalues Proof Recall by Sec 44 property 8 of erminants we have for any square matrix B Therefore B B t A λi A λi t A t λi t A t λi which means A and A t have the same characteristic polynomial This completes the proof Sec 5 #7 Let T be the linear operator on M n n R defined by T A A t a Show that ± are the only eigenvalues of T b Describe the eigenvectors corresponding to each eigenvalue of T c Find an ordered basis for M R such that [T ] is a diagonal matrix c Find an ordered basis for for M n n R such that [T ] is a diagonal matrix for n > Proof Let E ij i j n be the standard basis vectors of M n n R ie E ij has all zero entries except a in the ith row jth column entry A basis of orthonormal in the Frobenius inner product on M n n R eigenvectors for T corresponding to the eigenvalue are the vectors : E ii i n and E ij + E ji i < j n which gives the dimension of the corresponding eigenspace which is just the set of all real symmetric n n matrices n n + A basis of orthonormal in the Frobenius inner product on M n n R eigenvectors for T corresponding to the eigenvalue - are the vectors : E ij E ji i < j n which gives the dimension of the corresponding eigenspace which is just the set of all real skew-symmetric n n matrices n n Then is a basis an orthonormal basis in fact in the Frobenius inner product on M n n R of M n n R such that giving any order and giving any order [T ] is a diagonal matrix in fact it is the block matrix [T ] I nn+ 0 0 I nn 9