# Inverse of certain symmetric 2x2 block matrices

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).

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 $$\begin{pmatrix}A & -B\\ B & A \end{pmatrix}^{-1}=\begin{pmatrix}E & -F\\ F & E \end{pmatrix}\qquad \text{if }(A+iB)^{-1}=E+iF.$$
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.
• Do you have a source for the relation $det(M)=det(A^2+B^2)$? I don't see directly why it holds if $A$ and $B$ do not commute. The inverse of $M$ appears in some formulas of my research, and I am just curious whether it can be expressed more explicitly. – B K Nov 20 '16 at 17:06
• You are correct, I forgot that $A$ and $B$ need to commute. – K. Miller Nov 20 '16 at 17:18