This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: HOMEWORK 2  SOLUTIONS 1. (a) Recall that a symmetric matrix ( a ij ) 1 ≤ i,j ≤ n is one such that for all 1 ≤ i,j ≤ n , we have a ij = a ji . First, we show that S n × n ( C ) is a subspace. • First, it’s clear that the zero matrix is symmetric. • If A = ( a ij ) is symmetric, then cA = c ( a ij ) = ( ca ij ). It’s ijentry is ca ij , and it’s jientry is ca ji . Since we assumed A = ( a ij ) was symmetric, we know that a ij = a ji and so ca ij = ca ji so cA = c ( a ij ) is symmetric. • Suppose A = ( a ij ) and B = ( b ij ) are symmetric. Then the ijentry of their sum is a ij + b ij and the jientry of the sum is a ji + b ji and these two things are clearly equal by our assumption that A and B were symmetric. Let E ij denote the matrix with 1’s in the ij and jientries and 0 elsewhere. Then { E ij : 1 ≤ i ≤ j ≤ n } is a basis for S n × n ( C ) of size 1 + 2 + ··· + n = n ( n +1) 2 , hence dim( S n × n ( C )) = n ( n +1) 2 . (b) Let’s denote this set by T . First we show T is a subspace: • Clearly, the zero matrix has zero trace. • If A = a 11 a 12 a 21 a 22 has zero trace, then tr( cA ) = ca 11 + ca 22 = c ( a 11 + a 22 ) = c · 0 = 0, so T is closed under scalar multiplication....
View
Full
Document
This note was uploaded on 03/28/2012 for the course MATH 110 taught by Professor Gurevitch during the Fall '08 term at Berkeley.
 Fall '08
 GUREVITCH
 Math

Click to edit the document details