Inverse of certain symmetric 2x2 block matrices

Question

Let $M$ be a complex $2n\times 2n$-matrix of the form $$ M=\begin{pmatrix}A&B\\ -B &A\end{pmatrix}, $$ where $A$ is a symmetric $n\times n$-matrix and $B$ a skew-symmetric $n\times n$-matrix. In particular, $M$ is symmetric.

I would like to know the precise conditions on $A$ and $B$ such that $M$ is invertible, and then a formula for $M^{-1}$ in terms of $A$ and $B$ which is as easy as possible. In particular, the formula should take into account that A and B are symmetric and skew-symmetric, respectively.

The literature on general block matrix inversion formulas is so overwhelming that I can find only results which are way too general for my purpose (e.g. the inversion formulas given in this Wikipedia article).

Thank you very much in advance for any helpful comments and answers.

Answer 1

Suppose that $A$ and the Schur complement $S = A + BA^{-1}B$ are invertible. Then the inverse of $M$ is given by

$$ M^{-1} = \begin{bmatrix} A^{-1}(I - S^{-1}BA^{-1}) & -A^{-1}BS^{-1}\\ S^{-1}BA^{-1} & S^{-1} \end{bmatrix}, $$ which is just the usual block inversion formula.

Answer 2

In the meantime, I found out a satisfying answer myself, at least in the case where $A$ and $B$ are real, which suffice for my purposes. In the book Matrix Theory by Fuzhen Zhang (2nd edition, p. 48), it says that if $A$ and $B$ are real $n\times n$-matrices, then $$ det\begin{pmatrix}A & -B\\ B & A \end{pmatrix}=|det(A+iB)|^2, $$ which implies that the matrix on the left hand side is invertible iff $A+iB$ is, and the book also says that $$ \begin{pmatrix}A & -B\\ B & A \end{pmatrix}^{-1}=\begin{pmatrix}E & -F\\ F & E \end{pmatrix}\qquad \text{if }(A+iB)^{-1}=E+iF. $$