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Ma1bAnSol7

# Ma1bAnSol7 - CALIFORNIA INSTITUTE OF TECHNOLOGY Department...

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CALIFORNIA INSTITUTE OF TECHNOLOGY Department of Mathematics Math 1b; Solutions to Homework Set 7 Due: March 3, 2008 1. Let A = a b c d be a real 2 by 2 matrix. Then A is orthogonal iff AA t = I , or equivalently A t = A - 1 . Further as A is real, Theorem 5.19 says det( A ) = ² = ± 1. Now from Problem 4 on HW1, det( A ) = ad - bc and A - 1 = ² · d - b - c a . Thus A is orthogonal iff a c b d = A t = A - 1 = ² · d - b - c a , which holds iff a = ²d and b = - ²c . That is A is orthogonal iff A = a - ²c c ²a and a 2 + c 2 = 1 since ² = det( A ) = ² ( a 2 + c 2 ). As a and c are real, this last equation is equivalent to a = cos( θ ) and c = sin( θ ) for some angle θ . Therefore (a) is established, since ² = 1 if A is proper. Next A is improper iff ² = - 1 and hence A is improper iff A = cos( θ ) sin( θ ) sin( θ ) - cos( θ ) . Thus we have our description of improper matrices. Further the two matrices in (b) correspond to the cases θ = 0 and 180 degrees, respectively, so they are improper. 2. (a) First char A ( x ) = x 2 - tr ( A ) + det( A ) = x 2 + a 2 = ( x - ia )( x + ia ), so the eigenvalues of A are ± ia , and they are distinct as a 6 = 0. Thus by Theorem 4.10, A is similar to the diagonal matrix D = ia 0 0 - ia

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