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Let $M \in \mathbb{C}^{n \times n}$ be a complex-symmetric $n \times n$ matrix. That is, $M$ is equal to its own transpose (without conjugation). If the real part of $M$ is positive-definite, then is $M$ necessarily diagonalizable?

Of course, if we drop the positive-definiteness requirement, then there are examples of non-diagonalizable complex-symmetric matrices. For example, $$\begin{bmatrix} 1 & i \\ i & -1 \end{bmatrix}$$ is one such example. However, its real part $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ is clearly not positive-definite.

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    $\begingroup$ Adding a positive real multiple of the identity will take care of the positive definite condition, but will not affect diagonalisability. See the answer by Om(nom)${}^3$. $\endgroup$ May 10 '16 at 5:45
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The answer is no. Based on your example, $$ \pmatrix{3&i\\i&1} $$ will not be diagonalizable, but has a positive definite real part.

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  • $\begingroup$ Can't believe this didn't occur to me earlier, thanks! $\endgroup$ May 10 '16 at 20:35
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With symmetric matrices I assume people are talking about $P^T AP$ rather than $Q^{-1} A Q.$ It does appear that the book Horn and Johnson gives some precedence to using the word "diagonalizable" for $Q^{-1} A Q.$ On page 215, exercise 15 is exactly the example given in the original question.

From Horn and Johnson, on symmetric complex matrices. Corollary 4.4.4, if complex matrix $A=A^T$ there exists a unitary $U$ and a real nonnegative diagonal $\Sigma$ such that $$A = U \Sigma U^T.$$ This is called Takagi's factorization.

I also did the example from the earlier answer by the method I had described, $R^T MR,$ as

$$ \left( \begin{array}{cc} 1 &\sqrt 3 \\ -1 & \sqrt 3 \end{array} \right) \left( \begin{array}{cc} 3 & i \\ i & 1 \end{array} \right) \left( \begin{array}{cc} 1 & -1 \\ \sqrt 3 & \sqrt 3 \end{array} \right) = \left( \begin{array}{cc} 6 + 2i\sqrt 3 & 0 \\ 0 & 6 - 2i\sqrt 3 \end{array} \right) $$

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  • $\begingroup$ You're right, I should have specified that by "diagonalizable" I mean "similar to a diagonal matrix," i.e., $A = PDP^{-1}$. But this Takagi factorization seems very useful, thanks for sharing! $\endgroup$ May 10 '16 at 20:34
  • $\begingroup$ One reason to prefer not using the term "diagonalizable" for the existence of a complex invertible matrix $P$ such that $P^TAP$ is diagonal, is that thus defined is would be equivalent to "symmetric" (in other words the condition is always satisfied in the case you suggest). One can even restrict the diagonal matrix to have diagonal entries $1$ or $0$ only. This is a consequence of the classification of symmetric blilinear forms over$~\Bbb C$. $\endgroup$ May 11 '16 at 7:59
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Write this as $M = A + iB,$ where both $A,B$ are real and $A$ is positive definite. It follows that $A$ is invertible. From Horn and Johnson, Theorem 4.5.15, page 228 in the first edition, define $C = A^{-1} B.$ If there is a real invertible matrix $R$ with $R^{-1}CR$ diagonal, there is then a nonsingular real matrix $S$ with $SAS^T$ and $SBS^T$ diagonal. It follows that $SMS^T$ is diagonal complex.

Not sure what happens when $A^{-1}B$ has nontrivial Jordan form.

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    $\begingroup$ Not sure whether this is a yes or no answer. $\endgroup$ May 10 '16 at 7:22
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    $\begingroup$ @MarcvanLeeuwen people do not typically specify whether they want some $P^{-1}AP$ or some $Q^T AQ$ to be diagonal. When $A$ is real symmetric, the second one is the reasonable choice. More in a few minutes, tea ready. $\endgroup$
    – Will Jagy
    May 10 '16 at 18:27

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