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Unformatted text preview: SOLUTION TO ASSIGNMENT 10  MATH 251, WINTER 2007 1. Let A be a matrix in block form: A = A 1 A 2 . . . A k . Prove that A = A 1 A 2 A r , and m A = lcm { m A 1 ,m A 2 , ,m A r } . You may use the formula A b = A b 1 A b 2 . . . A b k for every positive integer b . Proof. We may argue by induction on the number of blocks. If we have one block the assertion is obviously true. Assume the result for r blocks and consider a matrix with r + 1 blocks. A = A 1 A 2 . . . A r +1 . We think of this matrix as consisting of two blocks A = B 1 B 2 , where B 2 = A r +1 and B 1 is a matrix in blocks A 1 ,...,A r . Let n be the size of A , n i the size of B i . Now, A = det( tI n A ) = det( tI n 1 B 1 I n 2 B 2 ) = det( tI n 1 B 1 )det( tI n 2 B 2 ) = B 1 B 2 . Since B 1 consists of r blocks we may use the induction hypothesis and get that B 1 = A 1 A 2 A r and the result concerning the characteristic polynomial follows. Let m i be the minimal polynomial of B i . Then m A ( A ) = m A ( B 1 ) m A ( B 2 ) = 0 . 1 2 SOLUTION TO ASSIGNMENT 10  MATH 251, WINTER 2007 This implies that m A ( B i ) = 0 and hence that m i  m A . It follows that lcm ( m 1 ,m 2 )  m A . On the other hand let p = lcm ( m 1 ,m 2 ) = p 1 m 1 = p 2 m 2 . Then 0 = p ( A ) = ( p 1 m 1 )( B 1 ) ( p 2 m 2 )( B 2 ) = m 1 ( B 1 ) m 2 ( B 2 ) p 1 ( B 1 ) p 2 ( B 2 ) = 0 0 0 0 p 1 ( B 1 ) p 2 ( B 2 ) = 0 . Therefore m A  p and hence m A = lcm ( m 1 ,m 1 ). We now apply the induction hypothesis and get m a = lcm ( m 1 ,m 2 ) = lcm ( m 1 ,m A r +1 ) = lcm ( lcm ( m A 1 ,...,m A r ) ,m A r +1 ) = lcm ( m A 1 ,...,m A r +1 ) . / 2. Let A be an n n matrix over an algebraically closed field of characteristic different from 2, such that A 2 is diagonalizable. Prove that if A is a nonsingular matrix then also A is a diagonalizable....
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This note was uploaded on 08/26/2008 for the course MATH 251 taught by Professor Goren during the Spring '08 term at McGill.
 Spring '08
 Goren
 Math

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