Part d for an upper triangular matrix the diagonal

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Part (d): For an upper triangular matrix, the diagonal entries are the eigenvalues.
Problem ( § 7.2 # 14) . Prove the converse of Exercise 13(d): If T is a linear operator on an n -dimensional vector space V and ( - 1) n t n is the characteristic polynomial of T , then T is nilpotent.
8 HOMEWORK 13 SOLUTIONS Problem ( § 7.2 # 17) . Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial splits, and let λ 1 , λ 2 , . . . , λ k be the distinct eigenvalues of T . Let S : V V be the mapping defined by S ( x ) = λ 1 v 1 + λ 2 v 2 + · · · + λ k v k , where, for each i , v i is the unique vector in K λ i such that x = v 1 + v 2 + · · · + v k . (This unique representation is guaranteed by Theorem 7.3 (p. 486) and Exercise 8 of Section 7.1.) (a) Prove that S is a diagonalizable linear operator on V . (b) Let U = T - S . Prove that U is nilpotent and commutes with S , that is SU = US . Proof. Part (a): Let β i be a basis for K λ i (where K λ i is in terms of T as above). Then put β = β 1 ∪ · · · ∪ β k . This is a basis for V , by Theorem 7.4, since the characteristic polynomial of T splits. But clearly, every vector in K λ i is an eigenvector of S (with eigenvalue λ i ) and hence [ S ] β is a diagonal matrix. In particular, S is a diagonalizable linear operator. Part (b): Let U = T - S . Let β i (for each i ) be as in part (a), except also assume that it is a Jordan canonical basis for K λ i , and let β = β 1 ∪ · · · ∪ β k . Then [ T ] β is the Jordan canonical form for T , while [ S ] β is a diagonal matrix. Further, the diagonal entries on both of these matrices agree, since they are just the eigenvalues of T , with the correct multiplicities. Thus [ U ] β = [ T - S ] β = [ T ] β - [ S ] β is an upper-triangular matrix, with zeros down the diagonal. In particular, from our work above, we know that the characteristic polynomial of U is ( - 1) n t n , and hence U is nilpotent by Exercise 14 of this section. To show commutativity we need a lemma: Lemma 1. Let A = λI n be the n × n identity matrix. Let B be the n × n matrix with ones down the superdiagonal and zeros elsewhere. (In other words, a Jordan block for the eigenvalue 0.) Then A and B commute. Proof. One easily calculates that AB = λIB = λB and BA = BλI = λB . This gives us the lemma. To finish the problem, notice that [ U ] β is just the matrix formed from [ T ] β by removing the diagonal, and that [ S ] β consists of [ T ] β with everything above the diagonal removed. So, breaking [ U ] β into Jordan blocks, we see that the eigenvalues along the diagonal in [ S ] β match these blocks. Hence if we apply the lemma to each of these blocks, we see that [ U ] β and [ S ] β commute on each block, and hence commute. This implies U and S commute.