Off-diagonalize a matrix

Question

Consider a self-adjoint matrix $M$ that has block form

$$M = \begin{pmatrix} M_{11} & M_{12} \\ M_{12}^* & M_{11} \end{pmatrix}.$$

I am wondering if there exists any criterion to decide if this matrix can be transformed by some invertible matrix $T$

such that $$TMT^{-1} = \begin{pmatrix}0 & C \\ C^* & 0 \end{pmatrix}$$ for some suitable matrix $C?$

Notice that one restriction that $\begin{pmatrix}0 & C \\ C^* & 0 \end{pmatrix}$ already puts is that the spectrum of $M$ has to be symmetric with respect to zero as conjugation by $$\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}$$ shows.

As a first step, one might ask when we can achieve a form

$$TMT^{-1} = \begin{pmatrix}0 & C \\ D & 0 \end{pmatrix}$$

where $C$ and $D$ are arbitrary matrices?

Answer 1

This is a so-called chiral symmetry. The restriction on the symmetry of the spectrum of $M$ is the only restriction you need, you can then bring $M$ to the desired off-diagonal form by a unitary transformation: $$M=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast\Rightarrow \Omega^\ast U^\ast MU\Omega =\begin{pmatrix}0&\Lambda\\ \Lambda&0\end{pmatrix},$$ for $\Omega=2^{-1/2}\begin{pmatrix}1&1\\ -1 &1\end{pmatrix}$.

Here $U$ is the unitary matrix of eigenvectors of $M$; the eigenvalues are contained in the diagonal matrix $\Lambda$.