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!

  • $\begingroup$ @JochenGlueck: stability is needed to ensure convergence of the matrix series. Otherwise, the inequality does not make much sense... $\endgroup$
    – Ludwig
    Feb 14, 2018 at 19:19
  • $\begingroup$ @ Ludwig Argh... You're right, of course. Sorry for the spam, I'll delete my comment. $\endgroup$ Feb 14, 2018 at 19:22

1 Answer 1


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.


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