Limit of a defined positive symmetry

Question

Let $A:\mathbb{R}\rightarrow S_n^{++}(\mathbb{R})$ continuous with a continuous derivative.

We suppose that $\forall t\in\mathbb{R},\mathrm{tr}(A'(t))\le-\mathrm{tr}(A(t))$. Show that $\displaystyle\lim_{t\rightarrow\infty}A(t)=0$

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Notations : $S_n^{++}(\mathbb{R})$ : Symmetric linear function $f$ such that $\forall x, \left<f(x)\mid x\right> > 0$

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I am looking for at least a good enough hint.

Answer 1

Hint: it is sufficient to show that $\DeclareMathOperator{\tr}{tr} \lim_{t \to \infty} \tr(A(t)) = 0$, noting that $\tr(A(t)) \geq 0$ by positivity.

We note that $$ f(t) = e^t(\tr(A(t))) $$ is non-increasing. It follows that $\tr(A(t)) \leq \tr(A(0))e^{-t}$.