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# 4.3 - Math 431 Assignment 7 Solutions Section 1.3 28...

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Math 431 - Assignment 7 Solutions Section 1.3 28 . Clearly, the n × n zero matrix O is in W 1 . Let A, B be in W 1 and let c be in F . Then, A + B and cA are in W 1 because ( A + B ) t = A t + B t = - A - B = - ( A + B ) , and ( cA ) t = cA t = c ( - A ) = - ( cA ) . Therefore, W 1 is a subspace of M n × n ( F ). Now assume that char ( F ) 6 = 2. If A is in W 1 W 2 , then A = A t = - A, and so 2 A = O . Since char ( F ) 6 = 2, we conclude that A = O , i.e. W 1 W 2 = { O } . To complete the proof of M n × n ( F ) = W 1 W 2 , we need only show that every matrix M in M n × n ( F ) can be written as a sum of symmetric matrix A and a skew-symmetric B . That is, M = A + B, with A t = A and B t = - B . Assuming that we have such a sum, taking the transpose of both parts gives us M t = A t + B t = A - B. Adding and subtracting the above two equations gives us M + M t = 2 A and M - M t = 2 B, respectively. Therefore, our candidates are A = M + M t 2 and B = M - M t 2 . Finally, it is easy to verify that A t = A and B t = - B to complete the proof. 30 . Assume that V = W 1 W 2 . Then each v in V can be written as the sum x 1 + y 2 with x 1 W 1 and y 2 W 2 , because V = W 1 + W 2 . Assume that v = x 1 + x 2 = y 1 + y 2 , for some x i , y i W i (i=1,2). Then x 1 - y 1 = y 2 - x 2 ,

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and so x 1 - y 1 and y 2 - x 2 are in W 1 W 2 = { 0 } . Therefore, x 1 = y 1 and x 2 = y 2 , that is, each v in V can be written uniquely as the sum x + y with x W 1 and y
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