Schur complement for negative semidefinite

Question

I have: $$X^{-1}(1-\alpha^2) - A^T(I+AXA^T)^{-1}A\preceq0,$$ where $X\succ0$, $\alpha$ is scalar, and $A$ is arbitrary matrix. Can I find an equivalent condition for the inequlity using the Schur complement?

Answer 1

Let $A$ have shape $m \times n$ (so that $X$ is $n \times n$). The matrix that you are looking at is the Schur complement $M/(I + AXA^T)$ where $$ M = \pmatrix{I + AXA^T&A\\A^T&(1-\alpha^2)X^{-1}}. $$ Because $I + AXA^T$ is positive definite, the fact that this Schur complement is negative definite is equivalent to $M$ having at exactly $m$ (strictly) positive eigenvalues, or if you prefer exactly $n$ negative or zero eigenvalues.

If we consider the Schur complement relative to $(1 - \alpha^2)X^{-1}$, we get $$ I + AXA^T - A[(1 - \alpha^2)X^{-1}]^{-1}A^T = \\ I + AXA^T - \frac 1{1 - \alpha^2} AXA^T = \\ I - \frac{\alpha^2}{1 - \alpha^2}X. $$ In the case where $|\alpha| < 1$, $(1 - \alpha^2)X^{-1} \succ 0$ which means that your condition can only hold if $n \leq m$; given this assumption, the condition is equivalent to $I - \frac{\alpha^2}{1 - \alpha^2}X$ having exactly $m-n$ positive eigenvalues.

In the case where $|\alpha| > 1$, $(1 - \alpha^2)X^{-1} \prec 0$, which means that your condition is equivalent to $I - \frac{\alpha^2}{1 - \alpha^2}X \succ 0$.

Note that the eigenvalues of $I - \frac{\alpha^2}{1 - \alpha^2}X$ are of the form $1 - \frac{\alpha^2}{1 - \alpha^2}\lambda$ for each eigenvalue $\lambda$ of $X$.