I came across this question and got really interested about it. There, the OP asks whether is possible to decompose a $2n \times 2n$ block matrix: $$ \begin{pmatrix} X & I \\ I & Y \end{pmatrix} $$ into $MDM^{T}$, where: $$D = \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix} $$ and $I$ is the identity matrix. The matrices $X$ and $Y$ where assumed to be noninvertible, but $X$ could be assumed to be positivesemidefinite and $Y$ negative semidefinite. I'd like to ask a slightly different question. Suppose that $X$ and $Y$ are invertible, but not positivedefinite nor negativedefinite. Does such a decomposition $MDM^{T}$ exists? In addition, is it possible to find such a decomposition where $M$ itself is invertible?
$\begingroup$
$\endgroup$
4

$\begingroup$ you do want $X$ and $Y$ to be symmetric, don't you? $\endgroup$– Carlo BeenakkerCommented Jul 16, 2021 at 12:46

$\begingroup$ @CarloBeenakker I do. Didn't mention it. My bad. $\endgroup$– InMathweTrustCommented Jul 16, 2021 at 13:02

2$\begingroup$ Let $n = 2$ and $X = Y = \mathrm{diag}(1,1)$, then both $X$ and $Y$ are invertible, but the matrix $[[X, I]; [I, Y]]$ is not, and so there does not exist a decomposition with invertible $M$. More generally, the determinant of your matrix is equal to $\det(XY  I)$, and the determinant of $MDM^T$ is nonpositive. So if the product $XY$ is too positive a decomposition is also not available. $\endgroup$– Willie WongCommented Jul 16, 2021 at 13:15

2$\begingroup$ Note that in your linked question, the fact that $X$ and $Y$ have opposite signs avoids such obstructions by design. $\endgroup$– Willie WongCommented Jul 16, 2021 at 13:21
Add a comment
