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I have two square matrices $Y$ and $Z$ size $n$, and matrix $M = Z^{-1}YZ$ eigenvalue is the same as Matrix $Y$'s eigenvalue. I have been able to prove that the eigenvalues are the same, and thus the characteristic polynomial of $Z^{-1}YZ$ $=$ $Y$ as the $|Y| = |Z^{-1}YZ|$ because the determinants are commutative and the determinant of an inverse matrix is $1/|Matrix|$. However, the eigenvectors will be different, am stuck here.

To put it more clearly:

What are the eigenvectors of matrices $Y$ and $Z^{-1}YZ$, they are both square matrices $n$ and the eigenvalues of $Y$ are the same as $Z^{-1}YZ$?

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  • $\begingroup$ Can you state your question in the form of a question? $\endgroup$ – John Hughes Feb 17 '15 at 17:52
  • $\begingroup$ What are the eigenvectors of matrices $Y$ and $Z^{-1}YZ$, they are both square matrices $n$ and the eigenvalues of $Y$ are the same as $Z^{-1}YZ$? $\endgroup$ – liujm Feb 17 '15 at 18:10
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Hint: Choose a vector $w$ such that applying $Z$ to it gives an eigenvector $v$ of $Y$. Then $Z^{-1}YZ(w) = Z^{-1}Yv = Z^{-1}(\lambda v) = \lambda(Z^{-1}v) = \lambda w$. Can you take it from here?

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  • $\begingroup$ I don't see the connection between $\lambda(Z^{-1} v)$ equaling $\lambda w$. Should one assume that $\lambda v$ of $Av = \lambda v$ equals $\lambda w$? I thought that b/c the eigenvectors are different, $\lambda v$ and $\lambda w$ would be different. $\endgroup$ – liujm Feb 17 '15 at 18:23
  • $\begingroup$ Since $v = Zw$, it follows that $Z^{-1}v = w$. $\endgroup$ – rogerl Feb 17 '15 at 19:43
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If $v$ is an eigenvector for Y, try putting the vector $Z^{-1}v$ into $Z^{-1}YZ$, see what you get.

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