Inequality in Lyapunov matrix equation

Question

Let $Q \in M_n(\mathbb R)$ be positive definite. Let $X, Y \in M_n(\mathbb R)$ satisfy following matrix equation/inequality \begin{align*} A^T X A + Q - X = 0 \\ A^T Y A + Q - Y \succ 0, \end{align*} where $\rho(A) < 1$. Condition on $A$ guarantees $X$ exists and unique. Is it possible to conclude $Y \succ X$?

Answer 1

Assuming that $X$ and $Y$ are assumed to be symmetric and $\succ 0$ just means positive definite, the answer to your question is no. In fact, the exact opposite holds: $X\succ Y$. To see this, let $Z := Y-X$. Then your assumptions imply $A^TZA\succ Z$.

First, we observe that $(A^n)^TZ(A^n)\succ Z$ holds for all $n\in\Bbb N$. This can be proved by induction. Now, choose $n$ such that $\|A^n\| < 1$ (which is possible due to the well known formula $\rho(A) = \lim_{n\to\infty}\|A^n\|^{1/n}$) and set $B := A^n$. Then we have $B^TZB\succ Z$ and $\|B\|<1$.

Let us first prove that $Z\preceq 0$. For this, let $x\in\Bbb R^n$, $\|x\|=1$, such that $x^TZx = \sup\{y^TZy : \|y\|=1\}$. We have to show that $x^TZx\le 0$. If $Bx=0$, then $x^TZx\le x^TB^TZBx = 0$, and we are done. So, assume that $Bx\neq 0$. Then, setting $y := Bx/\|Bx\|$, we have $\|y\|=1$ and get $$ x^TZx\le x^TB^TZBx = \|Bx\|^2y^TZy. $$ Now, if $x^TZx > 0$, then also $y^TZy > 0$ and $$ x^TZx\le\|Bx\|^2y^TZy\le\|B\|^2y^TZy < y^TZy, $$ which is a contradiction to the choice of $x$. Hence, $x^TZx\le 0$, which shows that $Z\preceq 0$.

Remark: Note that, up to this point, we have only used $B^TZB\succeq Z$.

Finally, assume that $x^TZx = 0$. Then, as $B^TZB-Z$ is positive definite and $x^T(B^TZB-Z)x = x^TB^TZBx\le 0$, we conclude that $x=0$. Hence, indeed, $Z\prec 0$.