Transforming matrix to off-diagonal form I wonder if one can write the following matrix in the form $A = \begin{pmatrix} 0 & B \\ B^* & 0 \end{pmatrix}.$
The matrix I have is of the form
$$ C = \begin{pmatrix} 0 & a & b & 0 & 0 & 0 \\
\bar a & 0 & 0 &b & 0& 0\\
\bar b & 0 & 0 & a & f & 0 \\
0 & \bar b & \bar a & 0 & 0 &f \\
0 & 0 & \bar f & 0 & 0 & a\\
0 & 0 & 0 & \bar f & \bar a & 0
 \end{pmatrix}.$$
The reason I believe it should be possible is that the eigenvalues of $A$ are symmetric with respect to zero $\pm \vert a \vert, \pm \sqrt{ \vert a \vert^2+ \vert b \vert^2 + \vert f \vert^2 \pm \vert a \vert^2( \vert b \vert^2 + \vert f \vert^2)}$ where in the latter case all sign combinations are allowed.
Hence, I wonder if there exists $U$ such that
$$A = UCU^{-1}$$
 A: The general recipe to accomplish the block off-diagonalization is as follows. The matrix $C$ has eigenvalues $\pm\lambda_1,\pm\lambda_2,\ldots \pm\lambda_3$. Define $\Lambda=\text{diag}\,(\lambda_1,\lambda_2,\lambda_3)$, and decompose
$$C=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast,$$
with $U$ the unitary matrix of eigenvectors of $C$;
Then the matrix product
$$\Omega^\ast U^\ast CU\Omega =\begin{pmatrix}0&\Lambda\\  \Lambda&0\end{pmatrix},$$
with $\Omega=2^{-1/2}\begin{pmatrix}1&1\\ -1 &1\end{pmatrix}$, has the desired form.
The explicit form of $U$ is complicated for arbitrary complex numbers $a,b,f$.
A: The $C$ is a particular block matrix, $C\in \mathbb{M}_3(\mathbb{M}_2(\mathbb{C}))$. For $V$ unitary let $V\begin{pmatrix}0&a\\\bar a&0\end{pmatrix}V^*=\begin{pmatrix}s&0\\0&-s\end{pmatrix}$, $P$ the perfect shuffle matrix (permutation), $D=\text{diag}(1,1,1,1,-1,1)$ and $U=\frac{1}{\sqrt{2}}\begin{pmatrix}I&I\\-I&I\end{pmatrix}\in \mathbb{M}_2(\mathbb{M}_3(\mathbb{C}))$.
Applying $DP\begin{pmatrix}V&0&0\\0&V&0\\0&0&V\end{pmatrix}C \begin{pmatrix}V&0&0\\0&V&0\\0&0&V\end{pmatrix}^*P^*D$ gives $\begin{pmatrix}s&b&0&0&0&0\\\bar b&s&f&0&0&0\\0&\bar f&s&0&0&0\\0&0&0&-s&-b&0\\0&0&0&-\bar b&-s&-f\\0&0&0&0&-\bar f&-s\end{pmatrix}=G$, $UGU^*$ has the required form.
