Linear Algebra

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Unformatted text preview: s invertible (and thus by part (a), a0 = 0), we can multiply this entire equation by −a−1 A−1 to give 0 the desired result. 5.4.20 If there exists a g such that U = g (T ), then it is clear that U T = T U . On the other hand, suppose that U T = T U . Suppose that V is the T cyclic subspace of itself generated by v . By exercise 5.4.13 (using U (v ) in place of w), we can find a polynomial g so that g (T )(v ) = U (v ). We then claim that U = g (T ). Notice first that U (T k (v )) = T k (U (v )) = T k (g (T )(v )) = g (T )(T k (...
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This note was uploaded on 04/02/2013 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.

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