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


$$A = UCU^{-1}$$



Transforming matrix to off-diagonal form
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 $a,b,f$.


