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This is a question in elementary linear algebra, though I hope it's not so trivial to be closed.

Real symmetric matrices, complex hermitian matrices, unitary matrices, and complex matrices with distinct eigenvalues are diagonalizable, i.e. conjugate to a diagonal matrix.

I'd just like to see an example of a complex symmetric $n\times n$ matrix that is not diagonalizable.

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  • $\begingroup$ It's nontrivial but surely falls under "homework" rubric $\endgroup$ – Victor Protsak May 5 '10 at 21:12
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    $\begingroup$ It's homework level difficulty but, if you look at this user's past history, it probably isn't homework. $\endgroup$ – David E Speyer May 5 '10 at 21:15
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    $\begingroup$ It appears to be well-known (or in my case, I was told in a seminar) that every complex square matrix is similar to a complex symmetric one, but not necessarily unitarily so. (See the comments on page 1 of arxiv.org/abs/0907.2728v2 ) In particular, any non-trivial nilpotent would do. $\endgroup$ – Yemon Choi May 5 '10 at 21:36
  • $\begingroup$ @David: I know, for I gave it to him (mathoverflow.net/questions/23478/…)! :) $\endgroup$ – Victor Protsak May 5 '10 at 23:30
  • $\begingroup$ @Victor: excuse me, but I don't understand how can Gram-Schmidt and isotropic vectors be related to this problem (ok, "exercise" ;) ), as I'm looking for diagonalizing a (symmetric) matrix $A$ as an operator ($BAB^{-1}$), not as a quadratic form ($B^tAB$). A priori, $A$ could be non-orthogonally diagonalizable. Right? $\endgroup$ – Qfwfq May 6 '10 at 9:55
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$$\begin{pmatrix} 1 & i \\ i & -1 \end{pmatrix}.$$

How did I find this? Non-diagonalizable means that there is some Jordan block of size greater than $1$. I decided to hunt for something with Jordan form $\left( \begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix} \right)$. So I want trace and determinant to be zero, but the matrix not to be zero. The diagonal entries made sure the trace vanished, and then the off diagonal entries were forced.

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  • $\begingroup$ @David In what sense is the above matrix symmetric? (Definitely not in the sense of it being being equal to its transpose). Further one knows that $3\times 3$ symmetric traceless real matrices support an irreducible representation of $SO(3)$. Is there any analogous statement known for complex symmetric matrices or for the double cover $SU(2)$ or $SL(2,\mathbb{C})$ ? $\endgroup$ – Anirbit Dec 16 '10 at 10:16
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    $\begingroup$ In the sense that it is equal to its transpose. There are two off diagonal entries, and they are equal. $\endgroup$ – David E Speyer Dec 16 '10 at 13:12
  • $\begingroup$ @David. Sorry for the first question. It was stupid of me. So any comments about the representation theory aspect of complex symmetric matrices as in comparison to real case which I was mentioning? $\endgroup$ – Anirbit Dec 16 '10 at 18:11
  • $\begingroup$ How about this quiz - generalizing autonne takagi factorization: mathoverflow.net/q/302741/27004 $\endgroup$ – wonderich Jun 14 '18 at 22:24
  • $\begingroup$ @DavidESpeyer Why do we look for trace and determinant to be 0? $\endgroup$ – rannoudanames May 11 at 14:39

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