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A $n$ by $n$ hermitian matrix $A$ is such that

a) $A^{*}=A$ where $*$ represents complex conjugate transpose.

Now what does it mean for $B$ to be complex symmetric? Correct me if I am wrong on this but I think that

1) $B^{T}=B$ where $T$ represents transpose.

Both $A$ and $B$ are normal matrices so they both have an orthogonal set of eigenvectors that form a basis. Say for matrix $A$, these eigenvectors form a unitary matrix $U$ such that $UAU^{*}=D$ where $D$ is a diagonal matrix with eigenvalues along the diagonal.

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    $\begingroup$ Did you mean complex symmetric in the title? $\endgroup$ – Marc van Leeuwen Nov 18 '14 at 5:14
  • $\begingroup$ yes sorry for the typo $\endgroup$ – Clannad Nov 23 '14 at 12:33
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Let

$$B = \begin{bmatrix} 1 & x \\ x & 0 \end{bmatrix}, \quad x \in \mathbb{C}.$$

Obviously, $B = B^T$. However,

$$BB^* - B^*B = \begin{bmatrix} 0 & \overline{x} - x \\ x - \overline{x} & 0 \end{bmatrix} = \begin{bmatrix} 0 & -2 \operatorname{Im} x \\ 2 \operatorname{Im} x & 0 \end{bmatrix} \ne 0,$$

for $x \not\in \mathbb{R}$. In other words, complex symmetric matrices need not be normal.

An appropriate decomposition for them is the Takagi's decomposition,

$$B = V D V^T,$$

where $D$ is a real nonnegative diagonal matrix, and $V$ is unitary. Note, however, that this is not a similarity relation, like the eigenvalue decomposition, because $V^T$ is not the inverse of $V$ (unless $V$ happens to be real, of course).

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