Is a complex symmetric matrix with positive definite real part diagonalizable? 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.
 A: The answer is no. Based on your example,
$$
\pmatrix{3&i\\i&1}
$$
will not be diagonalizable, but has a positive definite real part.
A: 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
$$
\left(
\begin{array}{cc}
1 &\sqrt 3 \\
-1 & \sqrt 3
\end{array}
\right)
\left(
\begin{array}{cc}
3 & i \\
i & 1
\end{array}
\right)
\left(
\begin{array}{cc}
1 & -1 \\
\sqrt 3 & \sqrt 3
\end{array}
\right) =
\left(
\begin{array}{cc}
6 + 2i\sqrt 3 & 0 \\
0 & 6 - 2i\sqrt 3
\end{array}
\right)
$$
A: 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.
