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I have following block matrices:

$$M_1 = \left(\begin{array}{cc}A & B\\B' & D\end{array}\right)$$ and

$$M_2 = \left(\begin{array}{cc}A & -B\\-B' & D\end{array}\right)$$ I want to show that $\mathrm{eig}(M_1) = \mathrm{eig}(M_2)$. How can I prove that?

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    [Similar matrices](http://en.wikipedia.org/wiki/Similar_matrix) have the same eigenvalues. Find a non-singular matrix $P$ such that $P M_1 = M_2 P,$ or $M_1 = P^{-1} M_2 P.$ This operation is called [conjugation](http://en.wikipedia.org/wiki/Conjugacy_class). Jyrki's answer provides such a $P.$2012-07-10

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