MATH535_HW10

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

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Unformatted text preview: Serge Ballif MATH 535 Homework 10 November 9, 2007 1. Let A ( ) denote the characteristic polynomial of A Mat( F ) . Show that A is diagonalizable over F if and only if (a) A ( ) splits over F into a product of linear factors and (b) whenever ( A- I n ) k v = 0 , then already ( A- I n ) v = 0 . ( ) Suppose that A is diagonal and let 1 ,..., r be the distinct eigenvalues of A with multiplicities m 1 ,...,m r . Then A ( x ) = ( x- 1 ) m 1 ... ( x- r ) m r factors into linear factors. Let E be the eigenspace for the eigenvalue (i.e. E = ker( A- I n )). We can choose a basis B of eigenvectors for A . Let B i = B E i . Since n = r i =1 | B i | r i =1 dim E i r i =1 m i = n , we must have dim E i = m i . Therefore, E i = ker( A- i I n ) = = ker( A- 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 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 F have characteristic zero. Suppose that L : V V satisfies L k = id V for some k > and that L ( ) splits into linear factors over F . 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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MATH535_HW10 - Serge Ballif MATH 535 Homework 10 November...

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