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Hey! I’m quite stuck on the question below. Thanks!

Let $v$ be an eigenvector of the $n \times n$ matrix $M$ corresponding to the eigenvalue $\lambda$. Set $Q = [v,a_2,a_3,\ldots,a_n]$ where ${v,a_2,a_3,\ldots,a_n}$ is a set of $n$ linearly independent vectors. Show that the first column of $Q^{-1}MQ$ is $(\lambda,0,0,\ldots,0)$

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1 Answer 1

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Hint: consider $M$ as the matrix representation of a linear function (say $f$) in a particular basis. Then, $Q^{-1}MQ$ corresponds to the matrix representation of this function in another basis, namely $(\vec v, \vec a_2, \ldots, \vec a_n)$. I think from there you can conclude quite easily. Good luck.

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I tried your method but Im still not getting it. I do understand the question - Q^-1MP will give you a diagnoal matrix hence the lamda,0,0...0 vector. Im just not still sure how to show this in general... –  user4645 May 22 '11 at 13:11

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