# Decomposition of symmetric block matrix

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 positive-semidefinite and $$Y$$ negative semi-definite. I'd like to ask a slightly different question. Suppose that $$X$$ and $$Y$$ are invertible, but not positive-definite nor negative-definite. Does such a decomposition $$MDM^{T}$$ exists? In addition, is it possible to find such a decomposition where $$M$$ itself is invertible?

• you do want $X$ and $Y$ to be symmetric, don't you? Commented Jul 16, 2021 at 12:46
• @CarloBeenakker I do. Didn't mention it. My bad. Commented Jul 16, 2021 at 13:02
• 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 non-positive. So if the product $XY$ is too positive a decomposition is also not available. Commented Jul 16, 2021 at 13:15
• Note that in your linked question, the fact that $X$ and $Y$ have opposite signs avoids such obstructions by design. Commented Jul 16, 2021 at 13:21