MATH535_HW10

# MATH535_HW10 - Serge Ballif MATH 535 Homework 10 November 9...

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Serge Ballif MATH 535 Homework 10 November 9, 2007 1. Let χ A ( λ ) denote the characteristic polynomial of A Mat( ) . Show that A is diagonalizable over if and only if (a) χ A ( λ ) splits over into a product of linear factors and (b) whenever ( A - λ 0 I n ) k v = 0 , then already ( A - λ 0 I n ) v = 0 . ( ) Suppose that A is diagonal and let λ 1 , . . . , λ r be the distinct eigenvalues of A with multiplicities m 1 , . . . , m r . Then χ A ( x ) = ( x - λ 1 ) m 1 . . . ( x - λ r ) m r factors into linear factors. Let E λ be the eigenspace for the eigenvalue λ (i.e. E λ = ker( A - λI n )). We can choose a basis B of eigenvectors for A . Let B i = B E λ i . Since n = r i =1 | B i | ≤ r i =1 dim E λ i r i =1 m i = n , we must have dim E λ i = m i . Therefore, E λ i = ker( A - λ i I n ) = · · · = ker( A - λ i I n ) k = · · · . ( ) Suppose that (a) and (b) hold. Then m i = d i for all 1 i r . Choose an ordered basis B i for E λ i . Then the set B = S r i =1 B i is a basis for V consisting of eigenvalues. Hence, A is diagonalizable. 2. Let have characteristic zero. Suppose that L : V V satisfies L k = id V for some k > 0 and that χ L ( λ ) splits into linear factors over . Show that V has a basis of eigenvectors for L . Hint: Use the Jordan Normal Form. Let A be the matrix for L in Jordan Normal Form. If there is an elementary Jordan block B of size greater than 1, then B k is not the identity matrix. This would contradict the assumption that L k = id V . Thus, all blocks are of size 1, i.e., A is diagonal. Thus, the columns of A give us a basis of eigenvectors for L .

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