1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$ \begin{pmatrix}A & -B\\ B & A \end{pmatrix}^{-1}=\begin{pmatrix}E & -F\\ F & E \end{pmatrix}\qquad \text{if }(A+iB)^{-1}=E+iF. $$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I know that formula, but it does not suffice for my purposes: M might be invertible even when A or the Schur complement are not invertible. This is the case also for the analogous standard formula that expresses the inverse of M in terms of the Schur complement of B. However, I guess that M can be invertible even if neither A nor B are invertible. I had hoped that in the given special situation, i.e. A symmetric and B skew-symmetric, there is a simple formula of the sort "M is invertible iff XXX holds, and then the inverse is given by YYY". $\endgroup$ – B K Nov 20 '16 at 13:45
  • $\begingroup$ 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. $\endgroup$ – B K Nov 20 '16 at 17:06
  • $\begingroup$ You are correct, I forgot that $A$ and $B$ need to commute. $\endgroup$ – K. Miller Nov 20 '16 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.