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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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    Yes, I'm certain it's $APA^t$...2012-04-24

2 Answers 2

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$\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}2i&1\\1&0\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}^\top=\begin{pmatrix}(1+i)-\sqrt{1+2i}&(1+i)+\sqrt{1+2i}\\1&1\end{pmatrix}\begin{pmatrix}(1+i)-\sqrt{1+2i}&0\\0&(1+i)+\sqrt{1+2i}\end{pmatrix}\begin{pmatrix}(1+i)-\sqrt{1+2i}&(1+i)+\sqrt{1+2i}\\1&1\end{pmatrix}^{-1}$

and

$\begin{pmatrix}1&-1\\1&1\end{pmatrix}\begin{pmatrix}2i&1\\1&0\end{pmatrix}\begin{pmatrix}1&-1\\1&1\end{pmatrix}^\top=\begin{pmatrix}i&-i/2\\1&0\end{pmatrix}\begin{pmatrix}2i&1\\0&2i\end{pmatrix}\begin{pmatrix}i&-i/2\\1&0\end{pmatrix}^{-1}$


If $\mathbf A$ is real orthogonal, then the diagonalizability of your original matrix is preserved; all bets are off, otherwise.

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    hmm, ok thanks...2012-04-24
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In fact the symmetric part of $P$, i.e., $P_S= \tfrac12 (P + P^t)$ can alway be diagonalized by a transformation matrix $A$. In fact it is possible to obtain $ A P_S A^t = \begin{pmatrix} 1 \\ & \ddots\\ & & -1\\ & & &\ddots \\ & & & &0\\ & & & & &\ddots \end{pmatrix} =\Sigma;$ with $\Sigma$ the signature of the symmetric part of $P$ having $n_+$-times $1$ and $n_-$-times $-1$ and $n-n_+-n_-$-times 0 on the diagonal.

On the other hand, the antisymmetric part $P_A = \tfrac12 (P- P^t)$ remains antisymmetric under the congruence relation and thus cannot be diagonalized. We thus have $A P A^t = \Sigma + W$ with $W=-W^t$.

The question whether $A P A^t$ can be diagonalized can now be restated: does a $U\in GL(n, \mathbb{C})$ exist with $U\Sigma U^{-1} = \Sigma$ and $U W U^{-1}$ diagonal which is equivalent to the question if the eigenvectors of generalized eigenvalue problem $ \det(\Sigma- \lambda W) =0$ span the whole $\mathbb{C}^n$.

For two cases the answer is obviously yes. If $n_+ =n$ or $n_-=n$, we have the spectral theorem of skew-symmetric matrices which tells us that in fact $APA^t$ can be diagonalized. Thus if $P_S$ is positive or negative definite then the answer is that there exist a congruent matrix to $P$ which can be diagonalized. For $n_+ = n_- =0$ ($P$ is an antisymmetric matrix) then $\Sigma$ remains obviously invariant and the answer is again yes.

For the other cases we should know whether there exist an $U\in SU(n_+,n_-)$ which diagonalizes a skew-symmetric matrix...