Adv Alegbra HW Solutions 61

Adv Alegbra HW Solutions 61 - 61 If AB = B A, then cw + dy...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
61 If AB = BA , then c w + dy = ay + cz ; that is, c (w z ) = y ( a d ) . There are two cases to consider. If a = d , then de f ne B = · 01 11 ¸ . The 2,1 entry of is d = a , while the 2,1 entry of is a + c .As c ±= 0, we have ±= .If a ±= d ,de f ne B = · 10 ¸ . Now the 2,1 entry of is c + d , while the 2,1 entry of is c + a . Since a ±= d , these entries are different, and so ±= in this case as well. Therefore, A is not in the center of GL ( 2 , R ) . A similar argument holds if b ±= 0, (This result is generalized to n × n matrices in Corollary 4.86. The proof using linear transformations is much shorter and much simpler.) 2.85 Let ζ = e 2 π i / n be a primitive n th root of unity, and de f ne A = h ζ 0 0 ζ 1 i and B = £ ± . (i) Prove that A has order n and that B has order 2. Solution. It is clear that A n = I = B 2 .I f1 k < n , then A k ±= I , for the 1,1 entry of A k is ζ k ±= 1. (ii) Prove that BAB = A 1 . Solution. One multiplies the matrices. (iii) Prove that the matrices of the form A i and i
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

Ask a homework question - tutors are online