hw2sol - HOMEWORK 2 - SOLUTIONS 1. (a) Recall that a...

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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 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....
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This note was uploaded on 03/28/2012 for the course MATH 110 taught by Professor Gurevitch during the Fall '08 term at Berkeley.

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hw2sol - HOMEWORK 2 - SOLUTIONS 1. (a) Recall that a...

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