Tv = Tu Tu N. (u 1,...,u M, v 1,...,v N ) T(Sv) = S(Tv) = Sλv = λ(sv).


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1 54 CHAPTER 5. EIGENVALUES AND EIGENVECTORS 5.6 Solutions 5.1 Suppose T L(V). Prove that if U 1,...,U M are subspaces of V invariant under T, then U U M is invariant under T. Suppose v = u u N U U M, with u n U n. Then Tv = Tu Tu N. Since each U n is invariant under T, Tu n = U n, so Tv U U M. 5.2 Suppose T L(V). Prove that the intersection of any collection of subspaces of V invariant under T is invariant under T. Suppose we have a set of subspaces {U r }, with each U r invariant under T. Let v r U r. Then Tv U r for each r, and so r U r is invariant under T. 5.3 Prove or give a counterexample: if U is a subspace of V that is invariant under every operator on V, then U = {0} or U = V. We ll prove the contrapositive: if U is a subspace of V and U {0} and U V, then there is an operator T on V such that U is not invariant under T. Let (u 1,...,u M ) be a basis for U, which we extend to a basis (u 1,...,u M, v 1,...,v N ) of V. The assumption U {0} and U V means that M 1 and N 1. Define a linear map T by Tu 1 = v 1, Tu n = 0, n > 1. Since v 1 / U, the subspace U is not invariant under the operator T. 5.4 Suppose S, T L(V) are such that ST = TS. Prove that null(t λi) is invariant under S for every λ F. Suppose that T v = λv. Then Sv satisfies T(Sv) = S(Tv) = Sλv = λ(sv). Thus if v is an eigenvector of T with eigenvalue λ, so is Sv. 5.5 Define T L(F 2 ) by T(w, z) = (z, w).
2 5.6. SOLUTIONS 55 Find all eigenvalues and eigenvectors of T. Suppose (w, z) (0, 0) and T(w, z) = (z, w) = λ(w, z). Then Of course this leads to z = λw, w = λz. w = λz = λ 2 w, z = λw = λ 2 z. Since w 0 or z 0, we see that λ 2 = 1, so λ = ±1. A basis of eigenvectors is (w 1, z 1 ) = (1, 1), (w 2, z 2 ) = ( 1, 1), and they have eigenvalues 1 and 1 respectively. 5.6 Define T L(F 3 ) by T(z 1, z 2, z 3 ) = (2z 2, 0, 5z 3 ). Find all eigenvalues and eigenvectors of T. Suppose (z 1, z 2, z 3 ) (0, 0, 0) and T(z 1, z 2, z 3 ) = (2z 2, 0, 5z 3 ) = λ(z 1, z 2, z 3 ). If λ = 0 then z 2 = z 3 = 0, and one checks easily that v 1 = (1, 0, 0) is an eigenvector with eigenvalue 0. If λ 0 then z 2 = 0, 2z 2 = λz 1 = 0, 5z 3 = λz 3, so z 1 = 0 and λ = 5. The eigenvector for λ = 5 is v 2 = (0, 0, 1). These are the only eigenvalues, and each eigenspace is one dimensional. 5.7 Suppose N is a positive integer and T L(F N ) is defined by T(x 1,...,x N ) = (x x N,...,x x N ). Find all eigenvalues and eigenvectors of T.
3 56 CHAPTER 5. EIGENVALUES AND EIGENVECTORS First, any vector of the form v 1 = (α,..., α), α F, is an eigenvector with eigenvalue N. If v 2 is any vector v 2 = (x 1,...,x N ), x n = 0, then v 2 is an eigenvector with eigenvalue 0. Here are N independent eigenvectors: and n v 1 = (1, 1,..., 1), v n = (1, 0,..., 0) E n, n 2), where E n denotes the nth standard basis vector T L(V) is invertible and λ F\{0}. Prove that λ is an eigenvalue of T if and only if 1/λ is an eigenvalue of T 1. Suppose v 0 and Tv = λv. Then v = T 1 Tv = λt 1 v, or T 1 v = 1 λ v, and the other direction is similar Suppose S, T L(V). Prove that ST and T S have the same eigenvalues. Suppose v 0 and ST v = λv. Multiply by T to get TS(Tv) = λtv. Thus if Tv 0 then λ is also an eigenvalue of TS, with nonzero eigenvector Tv. On the other hand, if Tv = 0, then λ = 0 is an eigenvalue of ST. But if T is not invertible, then rangets ranget is not equal to V, so TS has a nontrivial null space, hence 0 is an eigenvalue of TS Suppose T L(V) is such that every vector in V is an eigenvector of T. Prove that T is a scalar multiple of the identity operator.
4 5.6. SOLUTIONS 57 Pick a basis (v 1,...,v N ) for V. By assumption, Tv n = λ n v n. Pick any two distince indices, m, n. We also have Write this as Since v m and v n are independent, T(v m + v n ) = λ(v m + v n ) = λ m v m + λ n v n. 0 = (λ λ m )v m + (λ λ n )v n. λ = λ m = λ n, and all the λ n are equal Suppose S, T L(V) and S is invertible. Prove that if p P(F) is a polynomial, then p(sts 1 ) = Sp(T)S 1. First let s show that for positive integers n, (STS 1 ) n = ST n S 1. We may do this by induction, with nothing to show if n = 1. Assume it s true for n = k, and consider Then (STS 1 ) k+1 = (STS 1 ) k (STS 1 ) = ST k S 1 STS 1 = ST k+1 S 1. Now suppose p(z) = a n z N + + a 1 z + a 0. p(sts 1 ) = a n (STS 1 ) n = a n ST n S 1 = S( a n T n )S 1 = Sp(T)S Suppose F = C, T L(V), p P(C), and α C. Prove that α is an eigenvalue of p(t) if and only if α = p(λ) for some eigenvalue λ of T. Suppose first that v 0 is an eigenvector of T with eigenvalue λ. That is Tv = λv.
5 58 CHAPTER 5. EIGENVALUES AND EIGENVECTORS Then for positive integers n, and so T n v = T n 1 λv = = λ n v, p(t)v = p(λ)v. That is α = p(λ) is an eigenvalue of p(t) if λ is an eigenvalue of T. Suppose now that α is an eigenvalue of p(t), so there is a v 0 with or p(t)v = αv, (p(t) αi)v = 0. Since F = C, we may factor the polynomial p(t) αi into linear factors 0 = (p(t) αi)v = n (T λ n I)v. At least one of the factors is not invertible, so at least one of the λ n, say λ 1, is an eigenvalue of T. Let w 0 be a eigenvector for T with eigenvalue λ 1. Then 0 = (T λ N I) (T λ 1 I)w = (p(t) αi)w, so w is an eigenvector for p(t) with eigenvalue α. But by the first part of the argument, p(t)w = p(λ 1 )w = αw, and α = p(λ 1 ) Show that the result in the previous exercise does not hold if C is replaced with R. Take T : R 2 R 2 given by T(x, y) = ( y, x). We ve seen previously that T has no real eigenvalues. On the other hand, T 2 (x, y) = ( x, y) = 1 (x, y) Suppose V is a complex vector space and T L(V). Prove that T has an invariant subspace of dimension j for each j = 1,...,dim V.
6 5.6. SOLUTIONS 59 Let (v 1,...,v N ) be a basis with respect to which T has an upper triangular matrix. Then by Proposition 5.12 T : span(v 1,...,v j ) span(v 1,...,v j ) Give an example of an operator whose matrix with respect to some basis contains only 0 s on the diagonal, but the operator is invertible. ( ) 0 1 T = Give an example of an operator whose matrix with respect to some basis contains only nonzero numbers on the diagonal, but the operator is not invertible. We can use the idea of problem 5.7, taking ( ) 1 1 T =. 1 1 If v = [1, 1], then Tv = 0.
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