Sol10 - SOLUTION TO ASSIGNMENT 10 - MATH 251, WINTER 2007...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 non-singular matrix then also A is a diagonalizable....
View Full Document

This note was uploaded on 08/26/2008 for the course MATH 251 taught by Professor Goren during the Spring '08 term at McGill.

Page1 / 5

Sol10 - SOLUTION TO ASSIGNMENT 10 - MATH 251, WINTER 2007...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online