# Off-diagonalize a matrix

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?

• Another common term for this form (block) anti-diagonal. – Igor Khavkine Jul 1 '20 at 10:06
• Looks like you're working in quantum mechanics? – Nike Dattani Jul 2 '20 at 0:33

## 1 Answer

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 $$\Lambda$$.