Diagonalizing a symmetric block matrix

Question

Let us consider the matrix

$$ A = \begin{pmatrix} a & c+ib \\ c-ib& a \end{pmatrix},$$ then this matrix has eigenvalues $a\pm \sqrt{c^2+b^2}.$

Now, let us consider a block matrix

$$ A = \begin{pmatrix} A_1 & A_2 \\ A_2& A_1 \end{pmatrix},$$

then this block matrix can be block-diagonalized to

$$ \begin{pmatrix} A_1-A_2 & 0 \\ 0& A_1+A_2 \end{pmatrix},$$ by conjugating it with $1/\sqrt{2}\begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}.$

I would like to know: Is there any similar construction that allows me to explicitly block-diagonalize

$$ A = \begin{pmatrix} A_1 & A_2 \\ A_2^*& A_1 \end{pmatrix},$$ where $A_1$ is a real-symmetric matrix?

This is a block-matrix version of the first scalar matrix that I have written down, but unlike that one, I do not see right away how to block-diagonalize $A$.

Answer 1

Say that the blocks are $n\times n$ (hence $A$ is $2n\times2n$). Your question amounts to finding explicit $n$-dimensional subspaces that are stable under $M$. Generically, such spaces are of the form $$E_M=\{(x,Mx);x\in{\mathbb C}^n\},$$ for suitable matrices $M$.

An elementary calculation shows that these matrices are the solutions of a Ricatti equation $$MA_2M+MA_1-A_1M-A_2^*=0_n.$$ This cannot be solved explicitly in general, but there are existence results and approximation procedures.

Mind that if $M$ is a solution, then so is $N=-M^{-*}$, and the spaces $E_M,E_N$ are transversal to each other, because $M^*M$ is positive semi-definite.