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Unformatted text preview: Serge Ballif MATH 535 Homework 10 November 9, 2007 1. Let χ A ( λ ) denote the characteristic polynomial of A ∈ Mat( F ) . Show that A is diagonalizable over F if and only if (a) χ A ( λ ) splits over F into a product of linear factors and (b) whenever ( A λ I n ) k v = 0 , then already ( A λ 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 F have characteristic zero. Suppose that L : V → V satisfies L k = id V for some k > and that χ L ( λ ) 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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 Fall '08
 BROWNAWELL,WOODRO
 Math, Linear Algebra, Algebra, Factors, minimal polynomial, Serge Ballif, elementary divisors, minimal polynomial x2

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