HWsolutions13

# Part b suppose rank u m rank u m 1 this means that

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Part (b): Suppose rank ( U m ) = rank ( U m +1 ). This means that dim( U m ( V )) = dim( U m +1 ( V )). However, U m +1 ( V ) U m ( V ). Since these are both finite-dimensional vector spaces, of the same dimension, this means U m ( V ) = U m +1 ( V ). Applying U to both sides we get U m +1 ( V ) = U m +2 ( V ). Therefore U m ( V ) = U m +2 ( V ). Repeating this process, we have U m ( V ) = U m + n ( V ) for any n 0. If k m then k = m + n for some n 0, and so U m ( V ) = U k ( V ). Taking dimensions on both sides yields rank ( U m ) = rank ( U k ). Part (c): Suppose rank ( U m ) = rank ( U m +1 ). Let k m . By part (b), we have rank ( U m ) = rank ( U k ). Let n = dim( V ). Then, applying the rank-nullity theorem, nullity ( U m ) = n - rank ( U m ) = n - rank ( U k ) = nullity ( U k ). However, by part (a), N ( U m ) N ( U k ). Since these two finite-dimensional spaces have equal dimensions, they must be equal. Hence N ( U m ) = N ( U k ). Part (d): Let T be a linear operator on V , and λ an eigenvalue of T . Suppose that rank (( T - λI ) m ) = rank (( T - λI ) m +1 ). Then, by part (c), N (( T - λI ) m ) = N (( T - λI ) k ) for any k m . By Theorem 7.2.b, we know that K λ = N (( T - λI ) p ) where p is the multiplicity of λ . But, just from the definition of K λ , it is clear that we can replace p by any larger number and this equality still holds (see problem 1.h above for the argument). In particular, putting k = max { p, m } we have K λ = N (( T - λI ) k ) = N (( T - λI ) m ).

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6 HOMEWORK 13 SOLUTIONS Part (e): Let T be a linear operator on V whose characteristic polynomial splits, and let λ 1 , . . . , λ k be the eigenvalues of T . First suppose that rank ( T - λI ) = rank (( T - λI ) 2 ) for 1 i k . Then by part (d), K λ i = N ( T - λ i I ) = E λ i for every i . So, by the Corollary on page 488, T is diagonalizable. On the other hand, suppose that T is diagonalizable. Then, again by the Corollary on page 488, K λ i = E λ i = N ( T - λ i I ) for each i . But then K λ i = N (( T - λ i I ) 2 ) (since we can raise the exponent, using the proof in problem 1.h). Therefore N ( T - λ i I ) = N (( T - λ i I ) 2 ) for every i . Using the rank-nullity theorem, we easily have rank ( T - λ i I ) = rank (( T - λ i I ) 2 ). Part (f): Let T be a diagonalizable linear operator on V , and let W be a T -invariant subspace. Let λ be an eigenvalue for T W . Then λ is an eigenvalue for T . Since T is diagonalizable, part (e) implies rank ( T - λI ) = rank (( T - λI ) 2 ). From part (c), N ( T - λI ) = N (( T - λI ) 2 ). Therefore, when restricted to W , these two maps again have the same null-space. In other words, N ( T W - λI W ) = N (( T W - λI W ) 2 ) inside W (where we know we can bring W inside since W is both T and I -invariant). But then, by another easy application of the rank-nullity theorem, rank ( T W - λI W ) = rank (( T W - λI W ) 2 ). Since this holds for all eigenvalues, λ , of W , then by part (d) T W is diagonalizable. Problem ( § 7.2 # 1) . Label the following statements as true or false. Assume that the characteristic polynomial of the matrix or linear operator splits. Proof. (a) The Jordan canonical form of a diagonal matrix is the matrix itself. True. In fact, any Jordan matrix is the Jordan canonical form of itself. For a proof, let J be a Jordan matrix. Let β be the standard basis, then [ L J ] β = J , hence J is the Jordan canonical form for J .
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