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Let $G = GL(n, \mathbb{C})$ and $M = \{Q \in GL(n, \mathbb{C}) | Q^t = Q\}$. Let $P\in M$ be non-diagonalizable. Question: For any $A \in G$, can we say that $APA^t$ is also non-diagonalizable? Or there's no definite conclusion?

p.s. For simplicity, take $P=\left[\begin{array}{cc} 2i & 1 \\ 1 & 0\end{array}\right]$.

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    Are you sure you mean $APA^t$ and not $APA^{-1}$? In general, $APA^t$ does not conserve eigenvalues, hence the new eigenvalues will be distinct ("almost surely", say) so it will diagonalizable.2012-04-24
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    Yes, I'm certain it's $APA^t$...2012-04-24

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