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Problem. Let $t\in\mathbb{R}_{+}\mapsto M(t)$ such that

  1. $M(t)$ is symmetric positive semidefinite matrix, and
  2. $\forall\: t \geq^{\cdot}\! 0$, $\operatorname{Tr}\left(M'(t)\right) \leq -\operatorname{Tr}\left(M(t)\right)$.

Prove that $$\lim_{t\to+\infty} M(t) = 0$$

All I could prove is the easy case where $M$ is a scalar, but I could not do more...

Egor Ivanov
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Sabrebar
  • 147

1 Answers1

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We invoke the following lemma (which is the special case of the fact that the operator norm of a matrix is equal to its largest singular value):

Lemma. Let $A$ be a $d \times d$ symmetric real positive semidefinite matrix. Then its operator norm $\|A\|$ is equal to the largest eigenvalue of $A$.

Proof. Let $\lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_d \geq 0$ denote the eigenvalues of $A$, arranged in decreasing order. By the spectral theorem, there exists an orthonormal basis $u_1, \ldots, u_d$ of $\mathbb{R}^d$ such that $Au_i = \lambda_i u_i$ for each $i = 1, \ldots, d$. Then for any vector $v \in \mathbb{R}^d$,

\begin{align*} \| A v \| &= \left\| A \sum_{i=1}^{d} \langle u_i, v \rangle u_i \right\| = \left\| \sum_{i=1}^{d} \lambda_i \langle u_i, v \rangle u_i \right\| \\ &= \left( \sum_{i=1}^{d} \lambda_i^2 \langle u_i, v \rangle^2 \right)^{1/2} \leq \left( \lambda_1^2 \sum_{i=1}^{d} \langle u_i, v \rangle^2 \right)^{1/2} = \lambda_1 \|v\|, \end{align*}

proving the desired claim. $\square$

Next, note that

$$ \frac{\mathrm{d}}{\mathrm{d}t} e^t \operatorname{Tr}(M(t)) = e^t [\operatorname{Tr}(M(t)) + \operatorname{Tr}(M'(t))] \leq 0. $$

So, $ t \mapsto e^t \operatorname{Tr}(M(t))$ is non-increasing and we get $e^t \operatorname{Tr}(M(t)) \leq \operatorname{Tr}(M(0))$, or equivalently,

$$ \operatorname{Tr}(M(t)) \leq e^{-t} \operatorname{Tr}(M(0)). \tag{1} $$

Finally, invoke the lemma and positive semidefiniteness of $M(t)$ to get

$$ \| M(t) \| \leq \operatorname{Tr}(M(t)). \tag{2} $$

Combining $\text{(1)}$ and $\text{(2)}$, we conclude the desired claim.

Sangchul Lee
  • 167,468
  • Thanks a lot! The last part with the lemma seems natural to me. But how did you think about multiplying by $e^t$? – Sabrebar Jan 16 '24 at 18:55
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    @Sabrebar, As someone who learned basic ODE stuffs, I guess I am trained to associate an expression of the form $f(t)+f'(t)$ with the integrating factor $e^t$. – Sangchul Lee Jan 17 '24 at 11:09