HWsolutions13 - HOMEWORK 13 SOLUTIONS Problem (7.1 # 1)....

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HOMEWORK 13 SOLUTIONS Problem ( § 7.1 # 1) . Label the following statements as true or false. Proof. (a) Eigenvectors of a linear operator T are also generalized eigenvectors of T . True. By Theorem 7.1.a, K λ E λ . (b) It is possible for a generalized eigenvector of a linear operator T to correspond to a scalar that is not an eigenvalue of T . False. Let x be a generalized eigenvector, for the scalar λ . Let p be the smallest number so that ( T - λI ) p ( x ) = 0. Setting y = ( T - λI ) p - 1 ( x ) 6 = 0 we have ( T - λI )( y ) = ( T - λI ) p ( x ) = 0, so y is an eigenvector with eigenvalue λ . Another way to see this is that if λ is not an eigenvalue, we can think of it as an eigenvalue but with multiplicity 0. Then by Theorem 7.4, dim( K λ ) = 0. (c) Any linear operator on a finite-dimensional vector space has a Jordan canonical form. False. This is true if and only if the characteristic polynomial splits. For a specific counterexample, consider A = ± 0 1 - 1 0 ² , over R . This matrix has no eigenvalues, and so no eigenvectors. But every matrix made up of Jordan blocks has at least one eigenvector. Therefore, A does not have a Jordan canonical form over R . (d) A cycle of generalized eigenvectors is linearly independent. True. This is just the Corollary on page 490. (e) There is exactly one cycle of generalized eigenvectors corresponding to each eigen- value of a linear operator on a finite-dimensional vector space. False. Let T be the identity operator on R 2 . Then T has two cycles corre- sponding to its single eigenvalue λ = 1. The zero operator will also work to give a counter-example. (f) Let T be a linear operator on a finite-dimensional vector space whose character- istic polynomial splits, and let λ 1 2 ,...,λ k be the distinct eigenvalues of T . If, for each i , β i is a basis for K λ i then β 1 β 2 ∪ ··· ∪ β k is a Jordan canonical basis for T . False. We need each β i to actually be a basis consisting of cycles. To give a concrete counter-example, consider the operator T = L A on R 2 , where A = ± 0 1 0 0 ² . The only eigenvalue is λ = 0, and K λ = R 2 . Let β = { (1 , 1) t , (0 , 1) t } . Then β is a basis for K 0 . However, [ T ] β = ± - 1 - 1 1 1 ² is not a Jordan canonical form. Therefore, β is not a Jordan canonical basis for T . (g) For any Jordan block J , the operator L J has Jordan canonical form J . True. Let β be the standard basis. Then [ L J ] β = J is a Jordan block, and hence the Jordan canonical form for L J . 1
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2 HOMEWORK 13 SOLUTIONS (h) Let T be a linear operator on an n -dimensional vector space whose characteristic polynomial splits. Then, for any eigenvalue λ of T , K λ = N (( T - λI ) n ).
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HWsolutions13 - HOMEWORK 13 SOLUTIONS Problem (7.1 # 1)....

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