hw2sol

# hw2sol - HOMEWORK 2 SOLUTIONS 1(a Recall that a symmetric...

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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 ij -entry is ca ij , and it’s ji -entry 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 ij -entry of their sum is a ij + b ij and the ji -entry 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 ji -entries 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. If A = ( a ij ), B = ( b ij ) have zero trace, then tr( A + B ) = ( a 11 + b 11 ) + ( a 22 + b 22 ) = ( a 11 + a 22 ) + ( b 11 + b 22 ) = 0 + 0 = 0, hence T is closed under addition. 0

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