Let $X,Y\in\mathbb{R}^{n\times n}$ be symmetric matrices. You may assume that $X$ is positive semidefinite and $Y$ negative semidefinite, if needed, but not that they are invertible.
I would like to find a way to factor the $2n\times 2n$ block matrix
$$
\begin{bmatrix}
X & I\\\\ I & Y
\end{bmatrix}
$$
into some form of the kind $MDM^T$, where:
- $D$ should be a "simple" matrix, ideally diagonal or of the form $D=\begin{bmatrix}0 & I\\\\I & 0\end{bmatrix}$, or something similar;
- The factorization should take explicit advantage of the identities being there, without treating them as general matrices and thus depending on too many parameters, so the Cholesky factorization is ruled out.
Is there some nice identity that I am missing?
Best Answer
Assuming that $X$ and $Y$ are invertible (if not, probably a perturbation argument will yield the generalization). Then, the following choice of $M$ works, i.e., $MDM^T$ equals your original matrix with $D$ being the anti-diagonal identity matrix as desired.
(Also, note slightly different notation, I write $-Y$ instead of $Y$ as in the original question, so that both $X$ and $Y$ are positive).
A solution is given by: \begin{equation*} M = \begin{bmatrix} a & b\\\\ c & d \end{bmatrix}, \end{equation*}
where
\begin{equation*} a = X^{1/2},\quad b=\frac{X^{1/2}}{2},\quad c = X^{-1/2} - (X^{-1}+Y)^{1/2},\quad d = \frac{X^{-1/2} + (X^{-1}+Y)^{1/2}}{2}. \end{equation*}
Edit. Typo in $d$ fixed now.
To see that the above matrix provides a solution, simply verify
\begin{equation*} \begin{bmatrix} a & b\\\\ c & d \end{bmatrix}\begin{bmatrix} 0 & I\\\\ I & 0\end{bmatrix}\begin{bmatrix} a^T & c^T\\\\ b^T & d^T\end{bmatrix} = \begin{bmatrix} X & I\\\\ I & -Y\end{bmatrix}, \end{equation*} which boils down to checking the following four equations: \begin{eqnarray*} ad^T + bc^T &=& I\\\\ cb^T + da^T &=& I\\\\ ab^T + ba^T &=& X\\\\ cd^T + dc^T &=& -Y. \end{eqnarray*}