5. Diagonalization. plan given T : V V Does there exist a basis β of V such that [T] β is diagonal if so, how can it be found

Transcription

1 5. Diagonalization plan given T : V V Does there exist a basis β of V such that [T] β is diagonal if so, how can it be found eigenvalues EV, eigenvectors, eigenspaces 5.. Eigenvalues and eigenvectors. T : V W β OB of V, γ OB of W dimv m dimw n recall [T] γ β i-th column is [Tv i] γ β v,...,v m T : V V [T] β [I V ] β β [T] β[i V ] β β Q [T] β Q Q [I V ] β β change-ofcoordinate matrix Definition 5.. T : V V dimv n < linear map T is diagonalizable if OB β of V with [T] β a diagonal matrix. A square matrix A is diagonalizable if L A is. β is called diagonalizing basis. Now if β {v,...,v n } is a diagonalizing basis with [T] β diagλ,...,λ n, then Tv i λ i v i, so T a i v i a i λ i v i and v i 0. Definition 5.2. T : V V linear operator. Assume v 0 and Tv λv. Then we call v eigenvector and λ eigenvalue EV. We say that an eigenvector corresponds to an eigenvalue, and an eigenvalue corresponds to the eigenvector. Theorem 5.3. T : V V is diagonalizable basis of V of eigenvectors of T. Example 5.4. A 3 4 2, v v Av v, Av v 2, 2 0 so if β {v,v 2 }, then [A] β : [L A ] β 0 5

2 2 Example 5.5. T : R 2 R 2 rotation by parallel geomtetrically: no vector goes to a multiple one by 90 rotation thus T has no eigenvalues / eigenvectors not diagonalizable Example 5.6. V C R C -functions on R R. Tf f what are EV of T? f λf f ce λt 0 all λ R are eigenvalues of T f are eigenfunctions for λ 0 the eigenfunctions are the constant functions this cannot happen for operators on f.d. spaces Theorem 5.7. λ EV of A M n n F deta λid n 0 Proof. Av λv v 0 : A λid n v 0 A λid n is not invertible deta λid n 0 Definition 5.8. Let A M n n F. χ A t : deta tid n is called characteristic polynomial of A Example 5.9. A M 2 2 R. 4 t deta tid 2 4 t t2 4 t 3t+ eigenvalues of A are +3,. Definition 5.0. Let T : V V. Let β be OB of V. χ T t det[t] β tid n is called characteristic polynomial of T Theorem 5.. The definition of χ T does not depend on the choice of basis β. Proof. Let β,β be OB of V. Then we know [T] β Q [T] β Q Q [I V ] β β Then [T] β tid n [T tid V ] β Q [T tid V ] β Q Q [T] β tid n Q.

3 3 Thus det[t] β tid n F detq det[t] β tid n detq F detq detq det[t] β tid n }{{} detq QdetId n det[t] β tid n. Example 5.2. V P 2 R T : V V Tf f +x+f β SOB {,x,x 2 } Write [.] β [.] β. 0 A A [T] β t 0 deta tid 3 det 0 2 t t Theorem 5.3. A M n n F t2 t3 t λ EV λ,2,3 χ A t deta tid n is a polynomial in t of degree n with leading coefficient n : [χ A t] n n [χ A t] n n tra T [,0,0] β Tx 2x+ [,2,0] β Tx 2 2xx++x 2... [χ A t] 0 deta. 3x 2 +2x [0,2,3] β Example 5.4. How to find eigenvectors A λ 3 λ 2 calculated before 4 x x x x 2 x x B A λ I eigenvector to λ 3 2x +x 2 0 4x 2x 2 0 x t B 2 A λ 2 I eigenvectortoλ 2 4x +2x 2 0 2x +x 2 0 x t 2 t R\{0} t R\{0} eigenvector of linear operators

4 4 V T L A V T : V V β OB of V A [T] β φ β F n A [T] β F n φ β φ β [.] β Lemma5.5. v is an eigenvector of T with EV λ [v] β is eigenvector of A with EV λ. diag commutes Proof. A[v] β Aφ β v φ β Tv φ β λv λφ β v λ[v] β. since φ β is isomorphism, v 0 [v] β φ β v 0 similar so, to find eigenvectors of T, we can work in any OB β. Write [.] β φ β. Thus β a.. a n n a i v i for β {v,...,v n }. i Example 5.6. V P 2 R Tf f +x+f β {,x,x 2 } 0 A λ,2,3 calculated before Let λ B A λ Id 0 2 kerb t EVec of T for EV λ is t 0 0 β t R. check: f t Tf f +x+f t+x+t t f 0 Let λ 2 2 B 2 A λ 2 Id kerb 2 t EVec of T for EV λ 2 2 is t 0 β t+tx t R. check: f t+tx Tf f +x+f t+tx+x+t+tx t+tx+x+t 2t+tx 2f

5 2 0 λ 3 3 B kerb EVec f t+2x+x 2 check: r.r.e.f. t t R Tf Tt+2x+x 2 t+2x+x 2 +t+x2x+2 t+x 2 +tx+2x+2 3t+x 2 3f Diagonalizability. - test whether operator can be diagonalized - eigenbasis to find T : V V β with [T] β diagonal, find β Theorem 5.7. T : V V λ,...,λ k distinct eigenvalues with eigenvectors v i. Then {v,...,v k } linearly independent. Proof. Induction over k. When k, {v } linearly independent v 0. Now induction step Let {v,...,v k } linearly independent. k 0 a i v i 2 i k 0 T λ k I0 T λ k I a i v i i k a i λ i λ k v i Now by induction assumption, we have a i λ i λ k 0 but λ i λ k by assumption a i 0 i,...,k i k i i,...,k 2 0 a k v k k 0. v k 0 all a i 0 i,...,k v i i,...,k linear independent. Corollary 5.8. If T : V V dimv n If T has n distinct EV, then T diagonalizes. Proof. {v,...,v n } eigenvectors to λ i are linearly independent eigenbasis. Remark 5.9. Converse is not true: Id has only one EV, but diagonalizable.

6 6 F[t] Definition5.20. A polynomial ft PF splits over F if c,a,...,a n F, c 0 with ft ct a t a 2... t a n not necessarily distinct The algebraic multiplicity of a i in f is µ ai f : #{j : a j a i }. Note that a i are the roots of f fa i 0 and using factorization, one can see every polynomial splits every non-const. polynomial has a root Definition 5.2. F is algebraically closed if every polynomial in PF splits in F. Example ft t 2 + PR does not split in R R is not algebraically closed Theorem Fundamental Theorem of Algebra C is algebraically closed. Theorem The characteristic polynomial of any diagonalizable operator on a f.d. VS splits. Proof. T : V V χ T t χ [T]β t β OB of V so choose eigenbasis. Then [T] β is diagonal diagλ,...,λ n n so χ [T]β t n t λ i χ T splits. Example i A χ A t χ Id t t 2 splits, λ only EV 0 0 If A is diagonalizable, then [A] β [L A ] β A Id. 0 So χ A splits, but A does not diagonalize. Definition5.26. T : V V linearoperator E λ kert λid λ EV is called eigenspace of T for EV λ Theorem dime λ µ λ χ T t algebraic mult. of λ in χ T t. not always equal: Example A χ A t t 2 λ has algebraic multiplicity µ λ 2 0 dime λ? E dim. 0 E λ R 2 dim 2. If dime λ 2 E λ R 2 Eλ L A λid so if E λ R 2, then Eλ R L A L A Id Id. So dime λ < 2 µ λ. 2 R 2

7 7 Theorem Assume T : V V λ,...,λ k distinct EV β i basis of E λi EV λ i of T. Then β β 2 β k 3 is linearly independent. Proof. Similar to Theorem 5.7. Theorem T : V V diagonalizable λ i EV of T, i,...,k dime λi µ λi χ T and χ T splits or k i dime λ i n Then 3 is an eigenbasis. Proof. Theorem Theorem Test for diagonalization - determine characteristic polynomial of T find zeros eigenvalues λ i + multiplicities µ λi - for each distinct eigenvalue λ i, solve T λ i Ix 0 determine m i dime λi n rkt λ i I - if for all i, m i µ λi χ, then T diagonalizable, else not Example 5.3. f f 2 f 3 A f f 2 f 3 linear differential equation system an application of diagonalization f If A diagonalizes, then Q : Q AQ D D diagλ i 3 Q f D Q f Q ft c i e λ it c i R 5.3 skip solution ft Q i c i e λ it 3 i Invariant subspaces and Cayley-Hamilton theorem. f f 2 f 3 f Definition T : V V W V is T-invariant subspace if Example TW W, i.e., Tw W w W. T arbitrary {0}, V, kert, ImT, E λ for any eigenvalue λ of T. Example T : R 3 R 3 Ta,b,c a+b,b+c,0 W { x,y,0 : x,y R } T-invariant only finitely Definition T : V V x V Σ T x : span { x,tx,t 2 x,... } many are linearly independent T-cyclic subspace of V generated by x f f 2 f 3 f i : R R Exercise:aW Σ T xist invariant,b if x W and W is T-invariant, then W W W is the smallest T-invariant subspace x

8 8 Example T : R 3 R 3 Ta,b,c b+c,a+c,3c x,0,0 e Te 0,,0 e 2 T 2 e Te 2,0,0 e [ T 3 e e 2 T 4 e e ] W span {e,e 2 } { x,y,0 : x,y R } R[z] R[z] Example T : PR PR Tf f x z 2 Σ T x span{z 2,2z,2} P 2 R PR Theorem T : V V W invariant subspace. Then χ χ T Proof. γ OB of W β γ OB of V B B 2 [T] β χ T t 0 B 3 B tid B 2 0 B 3 tid detb t Id detb 3 t Id }{{}}{{} χ t F[t] Theorem Let T : V V W Σ T v, k dimw. Then a {v, Tv,...,T k v} is a basis of W b If a 0 v+a Tv+ +a k T k v+t k v 0, then χ t k a 0 +a t+ +a k t k +t k. Proof. a Let j be largest positive integer such that β {v,tv,...,t j v} is linearly independent. T j v spanβ spanβ is T-invariant Σ Tv spanβ Σ T v exercise β is basis of Σ T v j k a Now b. Work in OB β 0 0 a 0 [T W ] β 0 a a k def of β def of k β j dimσ T v k χ t k.... Example continue example 5.36 T : R 3 R 3 Ta,b,c b+c,a+c,3c

9 9 check using determinant β {e,e 2 } 0 [T W ] β 0 W Σ T e Te e 2 T 2 e e k 2 T 2 e +0 Te + e 0 Th 5.39 χ 2 +0 t+ t 2 t 2 +. ii χ t t t2 + Definition 5.4. If P m a i t i PF and T : V V, A M n n F, then define i0 PT m a i T i, with T 0 Id, T T, and T n } T T {{} i0 n times PA m a i A i, with A 0 I n, A A, etc. matrix mult i0 Theorem Let P PF and T : V V, n dimv <. a If PT 0, then for each eigenvalue λ of T we have Pλ 0. b If for each eigenvalue λ of T we have Pλ 0, and T is diagonalizable, then PT 0. proof is exercise Example A projection T satisfies T 2 T. Thus PT 0 for Pt t 2 t. Thus all possible eigenvalues of a projection are λ 0 and λ. Example A reflection T satisfies T 2 Id. Thus PT 0 for Pt t 2. Thus all possible eigenvalues of a reflection are λ and λ. Remark It is not claimed and not true in general that all roots of P occur as EV of T! Theorem Cayley-Hamilton theorem A linear operator T : V V satisfies its characteristic equation χ T T 0.

10 0 Proof. Let v 0. We prove χ T Tv 0. Let W Σ T v k dimw Th 5.39a a i with a 0 v+a Tv+...+a k T k v+t k v 0 Th 5.39b χ k a 0 +a t+ +a k t k +t k χ Tv k a 0 Id+a T + +a k T k +T k v 0. Now χ χ T by theorem 5.38, so χ T Tv 0. Example T : R 2 R 2 β e,e 2 [T] β Ta,b a+2b, 2a+b A χ A t t 2 2 t t2 +4 t 2 2t χ A A A 2 2A+5Id