Let $X\in\mathbb{R}^{n\times n}$ be a positive semi-definite matrix and $A\in\mathbb{R}^{n\times n}$ be a stable matrix, i.e. a matrix whose eigenvalues are strictly inside the left-half complex plane. Consider two positive reals $T\_1,T\_2>0$ such that $T\_1\le T\_2$.



> 
> **My question.** Does the following inequality hold true
>  $$
> \sum\_{k\ge 0}e^{A T\_1 k} X e^{A^\top T\_1 k}\ge \sum\_{k\ge 0}e^{A T\_2 k} X e^{A^\top T\_2 k},
> $$
>  where $e^\cdot$ denotes the matrix exponential?
> 
> 
> 


I'm able to prove the above inequality in the special case $T\_2= m T\_1$ where $m$ is an integer, however I have no clue about how to prove (or disprove) it in the general case. Thanks in advance for your help!



A matrix monotonicity question
The answer is no, in general. Here is a counterexample:


Let
\begin{align\*}
 X =
 \begin{pmatrix}
 1 & 0 \\
 0 & 0
 \end{pmatrix},
 \quad \text{and} \quad
 A =
 \begin{pmatrix}
 -1 & -1 \\
 1 & -1
 \end{pmatrix}
 = 
 \begin{pmatrix}
 0 & -1 \\
 1 & 0
 \end{pmatrix}
 - I,
\end{align\*}
where $I \in \mathbb{R}^{2 \times 2}$ denotes the identity matrix. Note that the eigenvalues of $A$ are $-1\pm i$, so $A$ is stable. For every $t \in [0,\infty)$ and every $k \in \mathbb{N}\_0$ we have
\begin{align\*}
 e^{tkA} =
 e^{-kt}
 \begin{pmatrix}
 \cos(kt) & -\sin(kt) \\
 \sin(kt) & \cos(kt)
 \end{pmatrix}.
\end{align\*}
Choose $y = (0,1) \in \mathbb{R}^2$. Then we have
\begin{align\*}
 \langle y, e^{tkA} X e^{tkA^T}y\rangle = \langle y, e^{tkA}X(e^{tkA})^T y\rangle = e^{-2kt}\sin^2(kt),
\end{align\*}
for all $t \in [0,\infty)$ and $k \in \mathbb{N}\_0$, so
\begin{align\*}
 \langle y, \sum\_{k\ge 0} e^{tkA} X e^{tkA^T} y\rangle = \sum\_{k \ge 0} e^{-2kt} \sin^2(kt) =: \varphi(t).
\end{align\*}
for all $t \in [0,\infty)$. Now, set $t\_1 := \pi$ and $t\_2 := \frac{3}{2}\pi$. Then $\varphi(t\_1) - \varphi(t\_2) = -\varphi(t\_2) < 0$, so the matrix
\begin{align\*}
 \sum\_{k\ge 0} e^{t\_1kA} X e^{t\_1kA^T} - \sum\_{k\ge 0} e^{t\_2kA} X e^{t\_2kA^T}
\end{align\*}
is not positive semi-definite.


