linalg_friedberg_ch5 - Partial Solutions for Linear Algebra...

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Partial Solutions for Linear Algebra by Friedberg et al. Chapter 5 John K. Nguyen December 7, 2011 5.2.11. Let A be an n × n matrix that is similar to an upper triangular matrix and has the distinct eigen- values λ 1 , λ 2 , ..., λ k with corresponding multiplicities m 1 , m 2 , ..., m k . Prove the following statements. (a) tra ( A ) = k X i =1 m i λ i . (b) det ( A ) = ( λ 1 ) m 1 ( λ 2 ) m 2 ... ( λ k ) m k . Proof of (a). Since A be an n × n matrix that is similar to an upper triangular matrix, say B, that has the distinct eigenvalues λ 1 , λ 2 , ..., λ k with corresponding multiplicities m 1 , m 2 , ..., m k . By definition of similar, there exists an invertible matrix Q such that A = Q - 1 BQ . Then, in consideration of Exercise 2.3.13 , tr ( A ) = tr ( Q - 1 BQ ) = tr ( Q - 1 QB ) = tr ( B ) = k X i =1 m i λ i . Proof of (b). Since A be an n × n matrix that is similar to an upper triangular matrix, say B, that has the distinct eigenvalues λ 1 , λ 2 , ..., λ k with corresponding multiplicities m 1 , m 2 , ..., m k . By definition of similar, there exists an invertible matrix Q such that A = Q - 1 BQ . Recall Theorem 4.7 which states that for any A, B M 2 × 2 ( F ), det ( AB ) = det ( A ) det ( B ). In consideration of this theorem and the corollary on page 223, we have that det ( A ) = det ( Q - 1 BQ ) = det ( Q - 1 ) det ( B ) det ( Q ) = 1 det ( Q ) det ( B ) det ( Q ) = det ( B ) = ( λ 1 ) m 1 ( λ 2 ) m 2 ... ( λ k ) m k as required. 5.2.12. Let T be an invertible linear operator on a finite-dimensional vector space V . (a) Recall that for any eigenvalue λ of T , λ - 1 is an eigenvalue of T - 1 . Prove that the eigenspace of T corresponding to λ is the same as the eigenspace of T - 1 corresponding to λ - 1 . (b) Prove that if T is diagonalizable, then T - 1 is diagonalizable. Proof of (a). Pick v E λ . Then, by definition, T ( v ) = λv . Taking the inverse of both sides, we get T - 1 T ( v ) = T - 1 ( λv ) which means v = λT - 1 ( v ). Then, by definition, v E λ - 1 so we have that the eigenspace of T corresponding to

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