that T (v i ) = λ i v i and so Example 1.

Transcription

1 Section 5.1: Defn 1. A linear operator T : V V on a finite-dimensional vector space V is called diagonalizable if there is an ordered basis β for V such that β [T ] β is a diagonal matrix. A square matrix A is called diagonalizable if L A is diagonalizable. Note 1. If A = B [T ] B where T = L A then to find β such that β [I] B B [T ] B B [I] β = β [T ] β is diagonal is the same thing as finding an invertible S such that S 1 AS is diagonal. Defn 2. Let T be a linear operator on V, a non-zero v V is called an eigenvector of T if λ F, such that T (v) = λv. The scalar λ is called the eigenvalue corresponding to the eigenvector v. Let A M n (F ), v F n, v 0 is called an eigenvector of A if v is an eigenvector of L A. That is, Av = λv. Theorem 5.1. A linear operator T on a finite dimensional vector space V is diagonalizable if and only if there is an ordered basis β for V consisting of eigenvectors of T. Furthermore, if T is diagonalizable, β = {v 1, v 2,..., v n } is an ordered basis of eigenvectors of T, and for D = β [T ] β, D is a diagonal matrix and D j,j is the e-val corresponding to v j, 1 j n. Proof. If T is diagonalizable, then β[t ] β = λ λ λ n for basis β = {v 1, v 2,..., v n }, which means that for all i, T (v i ) = λ i v i. So each λ i is an eigenvalue corresponding to the eigenvector v i. Conversely, if there exists an ordered basis β = {v 1, v 2,..., v n } such that each v i is an e-vctr of T then there exist λ 1, λ 2,..., λ n such that T (v i ) = λ i v i and so β[t ] β = λ λ λ n Example 1. [ cos α sin α B[T ] B = sin α cos α whereb = {(1, 0), (0, 1)}, If α = π/2, T has no eigenvectors because the linear transformation is rotation by α. But if α = π, then it does have e-vectors (1, 0) and (0, 1). But this transformation is invertible. (Rotate by the negative angle) So invertible and diagonalizable are not the same. ], 1

2 2 Theorem 5.2. A M n (F ). λ is an e-val of A if and only if det(a λi) = 0. Proof. λ is an e-val of A v 0 such that Av = λv. v 0 such that Av λv = 0. v 0 such that (A λi n )v = 0. det(a λi n ) = 0. Defn 3. For matrix A M n (F ), f(t) = det(a ti n ) is the characteristic polynomial of A. Defn 4. Let T be a linear operator on an n-dimensional vector space V with ordered basis β. We define the characteristic polynomial f(t) of T to be the characteristic polynomial of A = β [T ] β. That is, f(t) = det(a ti n ). Note 2. Similar matrices have the same characteristic polynomials, since if B = S 1 AS, det(b ti n ) = det(s 1 AS ti n ) = det(s 1 AS ts 1 S) = det(s 1 (A ti n )S) = det(s 1 1 ) det(a ti n ) det(s) = det(s) det(a ti n) det(s) = det(a ti n ) So that characteristic polynomials are similarity invariants. If B 1 and B 2 are 2 bases of V then B1 [T ] B1 is similar to B2 [T ] B2 and so we see that the definition of characteristic polynomial of T, does not depend on the basis used in the representation. So, we may say f(t) = det(t ti n ). Theorem 5.3. Let A M n (F ) (1) The characteristic polynomial of A is a polynomial of degree n with leading coefficient ( 1) n. (2) A has at most n distinct eigenvalues. Proof. First, we will prove (1). We need a slightly stronger statement for our induction statement to be of use. If B is a square n n matrix such that for some permutation θ S n, and some subset K [n], (B) i,j = b i,θ(i) t, i K, θ(i) = j (B) i,j = b i,j, i K, j θ(i) (B) i,j = b i,j, i K where for all i, j [n], b i,j is a scalar, then det(b) is a polynomial in t of degree K. Furthermore, if K = n and θ = id, the leading coefficient is ( 1) n. In this case, the entries on the diagonal of B are of the form b i,i t. The proof is by induction. The base case is for is for n = 1. A = [a 1,1 ], B = A ti 1, det(b) = a 1,1 t, which is a polynomial of degree 1 and has leading coefficient ( 1) 1 and if B = A, we have that det(b) = a 1,1 which is a polynomial of degree 0. Assume n > 1 and the theorem is true for n 1 n 1 matrices. Assume B satisfies the hypothesis of the statement. We compute det(b) by expanding on row 1. det(b) = n ( 1) 1+i (B) 1,i B(1 i) i=1

3 We see that for each i, by induction, B(1 i) is an n 1 square matrix with K or K 1 entries of the form b i,j t. If there exists an i [n], such that (B) 1,i is of the form b 1,i t then B(1 i) has K 1 entries of the form b r,s t and satisfies the induction hypothesis. det(b(1 i)) is therefore a polynomial of degree K 1 and also for j i, B(1 j) has K 1 entries of the form b r,s t and satisfies the induction hypothesis. det(b(1 i)) is therefore a polynomial of degree K 1. We see in this case that det(b) is a polynomial of degree 1 + K 1 = K. If additionally, K = n and θ = id, then b 1,1 is of the form b 1,1 t and B(1 1) has all of its diagonal entries of the form b i,i t and we see that its leading coefficient is ( 1) n 1 by induction and so the leading coefficient of det(b) is ( 1)( 1) n 1 = ( 1) n. And if for all i [n], (B) 1,i is of the form b 1,i then for all i [n], B(1 i) has K or entries of the form b r,s t and satisfies the induction hypothesis. So we see that in this case, det(b) is a polynomial of degree K. Now (2) follows by the fact from algebra that a polynomial of degree n over a field can have at most n roots. 3 Theorem 5.4. Let T be a linear operator on a vector space V and let λ be an e-val of T. A vector v V is an e-vctr of T corresponding to λ if and only if v 0 and v N(T λi n ). Proof. λ is an e-val of T v 0 such that T (v) = λv. v 0 such that T (v) λv = 0. v 0 such that (T λi n )(v) = 0. v 0 such that v N(T λi n ). Section 5.2. Theorem 5.5. Let T be a linear operator on a vector space V and let λ 1, λ 2,..., λ k be distinct e-vals of T. If v 1, v 2,..., v k are e-vctrs of T such that for all i [k], λ i corresponds to v i, then {v 1, v 2,..., v k } is a linearly independent set. Proof. The proof is by induction on k. Let k = 1. {v 1 } is a linearly independent set. Assume k > 1 and the theorem holds for k 1 distinct e-vals and e-vctrs. Now suppose λ 1, λ 2,..., λ k be distinct e-vals of T and v 1, v 2,..., v k are e-vctrs of T such that for all i [k], λ i corresponds to v i. We wish to show {v 1, v 2,..., v k } is a linearly independent set. Let a 1 v 1 + a 2 v a k v k = 0 (1) for some scalars a 1,..., a k. Applying T λ k I to both sides of the equation, we obtain, i [k 1], (T λ k I)(a 1 v 1 + a 2 v a k v k ) = 0 a 1 (T λ k I)(v 1 ) + a 2 (T λ k I)(v 2 ) + + a k (T λ k I)(v k )) = 0 (2) a i (T λ k I)(v i ) = a i (T (v i ) λ k I(v i )) = a i (λ i v i λ k v i ) = a i (λ i λ k )v i

4 4 and a k (T λ k I)(v k ) = a k (T (v k ) λ k I(v k )) = a k (λ k v k λ k v k ) = a k (λ k λ k )v k So (2) becomes a 1 (λ 1 λ k )v 1 + a 2 (λ 2 λ k )v a k 1 (λ k 1 λ k )v k 1 = 0 By induction, {v 1, v 2,..., v k 1 } is a linearly independent set and so for all i [k 1], a i (λ i λ k ) = 0. But, since λ i λ k 0, it must be that a i = 0. Now looking back at equation (1), we have that a k v k = 0. But since v k is not the zero vector, it must be that a k = 0 as well. Therefore, {v 1, v 2,..., v k } is a linearly independent set. Cor 1. Let T be a linear operator on an n-dimensional vector space V. If T has n distinct e-vals, then T is diagonlizable. Proof. Suppose λ 1, λ 2,..., λ n are distinct e-vals of T with corresponding e-vctrs v 1, v 2,..., v n. By Theorem 5.5, {v 1, v 2,..., v n } is a linearly independent set. By Theorem 5.1, T is diagonalizable. Defn 5. A polynomial f(t) in P (F ) splits over F if there are scalars c, a 1,..., a n such that f(t) = c(t a 1 )(t a 2 ) (t a n ). Theorem 5.6. The characteristic polynomial of any diagonalizable linear operator splits. Proof. Let T be a diagonalizable linear operator on V. Suppose β is a basis of V such that D = β [T ] β is diagonal. λ D = 0 λ λ n f(t) is the characteristic polynomial so f(t) = det(d ti) = λ 1 t λ 2 t λ n t = (λ 1 t)(λ 2 t) (λ n t). Defn 6. Let λ be an e-val of a linear operator (or matrix) with characteristic polynomial f(t). The algebraic multiplicity (or just multiplicity) of λ is the largest positive integer k for which (t λ) k is a factor of f(t). Defn 7. Let T be a linear operator on a vector space V and let λ be an e-val of T. Define E λ = {x V T (x = λx} = N(T λi). The set E λ is called the eigenspace of T corresponding to λ. The eigenspace of a matrix A M n (F ) is the e-space of L A. Fact 1. E λ is a subspace. Proof. Let a F, x, y E λ. Then T (ax + y) = at (x) + T (y) = aλx + λy = λ(ax + y).

5 Theorem 5.7. Let T be a linear operator on a finite dimensional vector space V, and let λ be an e-val of T having multiplicity m. Then 1 dim(e λ ) m. Proof. Let {v 1, v 2,..., v p } be a basis of E λ. Extend it to a basis β = {v 1, v 2,..., v p, v p+1,..., v n } of V. Let A = [T ] β, then ( ) λip B A = 0 C since i p, T (v i ) = λv i. So, A ti = ( ) (λ t)ip B 0 C ti n p Expanding on the 1st column, we see that det(a ti n ) = (λ t) p det(c ti n p ) = (λ t) p q(t). So the multiplicity of λ is greater than or equal to p and the dim(e λ ) = p. 5 Lemma 1. Let T be a linear operator on a vector space V and let λ 1, λ 2,..., λ k be distinct e-vals of T. For each i = 1, 2,..., k, let v i E λi. If v 1 +v 2 + v k = 0, then i [k], v i = 0. Proof. Renumbering if necessary, suppose for 1 i p, v i 0 and for p + 1 i k, v i = 0. Then, v 1 + v 2 +, v p = 0. But this contradicts Theorem 5.5. Thus, i [k], v i = 0. Theorem 5.8. Let T be a linear operator on a vector space V and let λ 1, λ 2,..., λ k be distinct e-vals of T. For each i = 1, 2,..., k, let S i E λi, be a finite linearly independent set. Then S = S 1 S 2 S k is a linearly independent subset of V. Proof. For all i suppose S i = {v i,1, v i,2,..., v i,ni }. Then S = {v i,j : 1 i k, 1 j n i }. Suppose there exists {a i,j } such that k n i a i,j v i,j = 0. i=1 j=1 For each i, let w i = n i j=1 a i,jv i,j. Then w i E λi, for all i and w 1 + w w k = 0. So, by the lemma, w i = 0, i. But each S i is linearly independent, so for all j, a i,j = 0. Theorem 5.9. Let T be a linear operator on a finite dimensional vector space V such that the characteristic polynomial of T splits. Let λ 1, λ 2,..., λ k be distinct e-vals of T. Then (1) T is diagonalizable if and only if the multiplicity of λ i is equal to dim(e λi ), i. (2) If T is diagonalizable and β i is an ordered basis for E λi, for each i, then β = β 1 β 2 β k is an ordered basis for V consisting of eigenvectors of T. Proof. For all i [k], let m i be the multiplicity of λ i, d i = dim(e λi ), and dim(v ) = n. We will show (2) first. Suppose T is diagonalizable. let β i be a basis for E λi, i [k]. We know β = {β 1 β 2 β k } is a linearly independent set by Theorem 5.8. By Theorem 5.1, there is a basis γ of V such that γ consists of eigenvectors of T. Let x V. x Span(γ), so x = a 1 v 1 + a 2 v a n v n where v 1, v 2,..., v n are eigenvectors of T.

6 6 Each v i Span(β j ), for some j [k]. So it can be expressed as a linear combination of vectors in β j. Thus x Span(β) and we have span(()β) = V. Now we show (1). ( :) We know d i m i, i. But since by (2), β is a basis, we have n = d 1 + d d k m 1 + m m k = n Thus, by the squeeze principle we have d 1 + d d k = m 1 + m m k and m 1 d 1 + m 2 d m k d k 0 But i [k], m i d i 0 and so m i = d i. ( :) Suppose i, m i = d i. We know m 1 + m m k = n since T splits and by Theorem 5.3 f(t) has degree n. Thus, we know d 1 + d d k = n and if i, β i is an ordered basis for E λi, by Theorem 5.8 β = β 1 β 2 β k is linearly independent. And, since β = n, by Corollary 2, (b) to Theorem 1.10, β is a basis of V. Then by Theorem 5.1, T is diagonalizable. Note 3. Test for diagonalization. A linear operator T on a vector space V of dimension n is diagonalizable if and only if both of the following hold. (1) The characteristic polynomial splits. (2) For each λ, eigenvalue of T, the multiplicity of λ equals the dimension of E λ. Notice that E λ = {x (T λ I)(x) = 0} = N(T λi) and n = nullity(t λi) + rank(t λi). So, dim(e λ ) = nullity(t λi) = n rank(t λi). Proof. Assume T is diagonalizable. By Theorem 5.6, the characteristic polynomial of T splits. Now by Theorem 5.9, for each λ, eigenvalue of T, the multiplicity of λ equals the dimension of E λ. Now assume (1) and (2) hold. Then by Theorem 5.9 again, T is diagonalizable. Defn 8. Let W 1, W 2,..., W k be subspaces of a vector space V. The sum is: k W i = {v 1 + v v k : v i W i, i [k]} i=1 Fact 2. The sum is a subspace. Defn 9. Let W 1, W 2,..., W k be subspaces of a vector space V. We call V the direct sum of W 1, W 2,..., W k, written V = W 1 W 2 W k If V is the sum of W 1, W 2,..., W k and j [k], W j i j W i = {0}. Example 2. V = R 4 W 1 = {(a, b, 0, 0) a, b R} W 2 = {(0, 0, c, 0) c R} W 3 = {(0, 0, 0, d) d R} Theorem Let W 1, W 2,..., W k be subspaces of a finite-dimensional vector space V. Tfae

7 7 (1) V = W 1 W 2 W k (2) V = k i=1 W i and i, v i W i if v 1 + v 2 + v k = 0 then v i = 0, i (3) Each vector v V can be written uniquely as v = v 1 + v 2 + v k, where i [k], v i W i. (4) If i [k], γ i is an ordered basis for W i then γ 1 γ 2 γ k is an ordered basis for V. (5) For each i [k] there is an ordered basis γ i for W i such that γ 1 γ 2 γ k is an ordered basis for V. Proof. (1) (2): Suppose for i [k], v i W i v 1 + v v k = 0. Let 1 i k. Then v i = j i v j. But since by (1), V = W 1 W 2 W k, we know that v i = j i v j = 0. (2) (3): By (2), each vector v V can be written as v = v 1 + v 2 + v k, where i [k], v i W i. To show uniqueness, suppose that v = v 1 +v 2 + v k and v = w 1 +w 2 + w k, where i [k], v i, w i W i. Then we have v 1 + v 2 + v k = w 1 + w 2 + w k v 1 w 1 + v 2 w v k w k = 0. or each i [k], v i w i W i, so by (2), v i w i = 0 and we have that v i = w i. (3) (4): Let i [k] and γ i = {w i,1, w i,2,..., w i,ni } be an ordered basis for W i. By (3), we know that γ 1 γ 2 γ k spans V. To show linear independence, we suppose for some {a i,j : 1 i k, 1 j n i }, k n i a i,j w i,j = 0 Notice that for each i [k], n i j=1 a i,jw i,j W i. We also have i=1 j=1 k 0 = 0 i=1 So by uniqueness, we have that n i j=1 a i,jw i,j = 0. Now since γ i is linearly independent, it must be that j [n i ], a i,j = 0. (4) (5): We know that for each i [k], W i has a finite basis, γ i. Thus γ 1 γ 2 γ k is an ordered basis for V by (4). (5) (1): Let i [k] and γ i = {w i,1, w i,2,..., w i,ni } be an ordered basis for W i. Since γ 1 γ 2 γ k spans V, we have that V = k i=1 W i. Let j [k] and consider v W j k i=1,i j W i. Since v W j, we have that But also, v = a j,1 w j,1 + a j,2 w j,2 + + a j,nj w j,nj k v = i=1,i j x i

8 8 for some vectors x i W i, which are linear combinations of the vectors in γ i. We have that k a j,1 w j,1 + a j,2 w j,2 + + a j,nj w j,nj x i = 0 i=1,i j and hence all of the coefficients of vectors in γ 1 γ 2 γ k are zero. This implies that v = 0. Theorem A linear operator T on a finite-dimensional vector space V is diagonalizable if and only if V is the direct sum of the eigenspaces of T. Proof. Let λ 1, λ 2,..., λ k be the distinct eigenvalues of T. ( ) Let T be diagonalizable. Then i [k], let γ i be an ordered basis of E λi. By Theorem 5.9, γ 1 γ 2 γ k is an ordered basis for V. By Theorem 5.10, V is the direct sum of E λi s. ( ) If V = k i=1 E λ i. Choose a basis γ i for each E λi. By Theorem 5.10, γ 1 γ 2 γ k is an ordered basis for V. Since there is a basis for V of E-vcts of T, T is diagonalizable, by Theorem 5.1. Section Skip. Section Invariant subspaces and the Cayley-Hamilton Theorem. Defn 1. Let T be a linear operator on a vector space V. A subspace W of V is called a T -invariant subspace of V if (W ) W. Defn 2. If T is a linear operator on V and W is a T -invariant subspace of V, then the restriction T W of T to W is a mapping from W to W and it follows that T W is a linear operator on W. Lemma 2. Exercise 21 from Section 4.3. If M M n (F ) can be expressed as ( ) A B M =, O C where A M r (F ), C M s (F ), s + r = n, and O is the all s r matrix of all zeros. Proof. The proof is by induction on r. If r = 1, we form det M be expanding on column 1. Then det M = a 1,1 M(1 1) = det A det C. Now assume for all such matrices M where A is r 1 r 1. Again, we expand on column 1 of M. det M = a 1,1 det M(1 1) a 2,1 det M(2 1) + ( 1) r+1 det M(r 1) For each i, M(i 1) has the form ( A(i 1) B O where B is a submatrix of B. By induction, det M(i 1) = det A(i 1) det C. So, we have: det M = a 1,1 det A(1 1) det C a 2,1 det A(2 1) det C + ( 1) r+1 det A(r 1) det C = det C(a 1,1 det A(1 1) a 2,1 det A(2 1) + ( 1) r+1 det A(r 1)) = det C det A C ),

9 Theorem Let T be a linear operator on a finite-dimensional vector space V, and let W be a T -invariant subspace of V. Then the caracteristic polynomial of T W divides the characteristic polynomial of T. Proof. Choose an ordered basis γ = {v 1, v 2,..., v k } for W, and extend it to an ordered basis β = {v 1, v 2,..., v k, v k+1,..., v n } for V. Let A = [T ] β and B 1 = [T W ] γ. Observe that A can be written in the form ( ) B1 B A = 2. O B 3 Let f(t) be the characteristic polynomial of T and g(t) the characteristic polynomial of T W. Then ( ) B1 ti f(t) = det(a ti n ) = det k B 2 = g(t) det(b O B 3 ti 3 ti n k ) n k by the Lemma. Thus g(t) divides f(t). The following was presented by N.Vankayalapati. He provided a handout. Defn 3. Let T be a linear operator on a vector space V and let x be a nonzero vector in V. The subspace W = span({x, T (x), T 2 (x),...}) is called the T -cyclic subspace of V generated by x. Theorem Let T be a linear operator on a finite-dimensional vector space V, and let W denote the T -cyclic subspace of V generated by a nonzero vector v V. Let k = dim(w ). Then (a) {v, T (v), T 2 (v),..., a k 1 T k 1 (v)t k (v)} is a basis for W. (b) If a 0 v + a 1 T (v) + + T k 1 (v) = 0, then the characteristic polynomial of T W f(t) = ( 1) k (a 0 + a 1 t + + a k 1 t k 1 + t k ). The following was presented by J. Stockford. He provided a handout. Theorem (Cayley-Hamilton). Let T be a linear operator on a finite-dimensional vector space V, and let f(t) be the characteristic polynomial of T. Then f(t ) = T 0, the zero transformation. The following was presented by Q. Ding. He provided a handout. Cor 1. (Cayley-Hamilton Theorem for Matrices). Let A be an n n matrix, and let f(t) be the characteristic polynomial of A. Then f(a) = O, the zero matrix. We did not cover the following theorems. Theorem Let T be a linear operator on a finite-dimensional vector space V, and suppose that V = W 1 W 2 W k, where W i is a T -invariant subspace of V for each i (1 i k). Suppose that f i (t) is the characteristic polynomial of T Wi (1 i k). Then f 1 (t) f 2 (t) f k (t) is the characteristic polynomial of T. 9 is

10 10 Defn 4. Let B 1 M m (F ), and let B 2 M n (F ). We define the direct sum of B 1 and B 2, denoted B 1 B 2 as the (m + n) (m + n) matrix A such that (B 1 ) i,j for 1 i, j m A i,j = (B 2 ) (i m),(j m) for m + 1 i, j n + m 0 otherwise If B 1, B 2,..., B k are square matrices with entries from F, then we define the direct sum of B 1, B 2..., B k recursively by B 1 B 2 B k = (B 1 B 2 B k 1 ) B k Of A = B 1 B 2 B k, then we often write A = B 1 O O O B 2 O... O O B k Theorem Let T be a linear operator on a finite-dimensional vector space V, an d let W 1, W 2,..., W k be T invariant subspaces of V such that V = W 1 W 2 W k. For each i, let β i be an ordered basis for W i, and let β = β 1 β 2 β k. Let A = [T ] β and B i = [T Wi ] βi for each i. Then A = B 1 B 2 B k.