Adv Alegbra HW Solutions 64

Adv Alegbra HW Solutions 64 - 64 Solution It is easy to see...

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64 Solution. It is easy to see that ϕ is a homomorphism: ϕ( MN ) = QMNQ 1 = QMQ 1 QNQ 1 = M )ϕ( N ). Next, if M = · ab cd ¸ , then a + c = 1 and b + d = 1. It follows that im ϕ A , for matrix multiplication gives · 10 11 ¸· ¸ = · a bb 01 ¸ (the bottom row of the last matrix is ( a + c ) ( b + d ) b + d , and this is 0 1 because M is stochastic. Finally, we must show that ϕ is surjective. This is obvious from the last calculation. If A = · ¸ A , then A = M ) , where M = · a + 1 a b 1 b ¸ . 2.91 Prove that the symmetry group 6(π n ) , where π n is a regular polygon with n vertices, is isomorphic to a subgroup of S n . Solution. Every isometry ϕ n ) permutes the n vertices X
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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