# Inverse statement of the linear ODE question

I am working on the following problem, which is the inverse statement of the cited question:

Let a system $$x' = A(t)x$$ and suppose there are values positives $$k, \beta$$ such that a positive fundamental matrix $$X(t)$$ satisfies $$\|X(t)\| \leq k$$, $$t \geq 0$$ and $$X^{-1}(t)$$ is bounded in $$[0,\infty)$$. Show that: $$\liminf_{t \rightarrow \infty} \int^t_\beta \operatorname{tr}(A(s))\,ds > - \infty.$$

My attempt:

Note that for some constants $$M>0$$ so that $$\|X^{-1}\|=\frac{\|adj(X(t))\|}{|det(X(t))|}\le M$$ Then $$|det(X(t))|\ge\frac{\|adj(X(t))\|}{M}$$

If I apply Liouville's Formula: $$det(X(t))=det(X(0))e^{\int_{0}^t tr(A(s))ds}$$ Then $$|det(X(0))|e^{\int_{0}^t tr(A(s))ds}\ge\frac{\|adj(X(t))\|}{M}$$

But I am stuck on there. Because we try to show that $$e^{\int_{0}^t tr(A(s))ds}$$ is bounded away from constant. I cannot take $$t\to \infty$$ since $$\|adj(X(t))\|$$ dependent on $$t$$. How to deal with this term? we just know that upper bound but not lower bound.

I am unsure what exactly a positive fundamental matrix means, but I guess it is just a fundamental matrix considered for $$t \ge 0$$, is that correct? I.e., we have $$X(0) = I$$ and $$\dot{X}(t) = A(t)X(t)$$, $$t \ge 0$$. I also guess that we can take $$\beta = 0$$ since $$\beta$$ is not actually defined.
Coefficients of $$X(t)$$ are continuous in $$t$$, and so the determinant of $$X(t)$$ is a continuous function of time. At $$t=0$$, we have $$X(0)=I$$ and $$\det X(0)=1$$. Saying that the inverse exists and is bounded implies that the determinant of $$X$$ does not cross zero and remains strictly positive. So, for $$d := \det X$$, there exists $$d_0>0$$ such that $$d(t)\ge d_0$$ for all $$t\ge 0$$; note that $$d(0)=1$$ and $$d$$ is continuous and bounded.
Next, by Jacobi's formula, we have $$\dot{d}(t) = d(t)\operatorname{tr}\left(X^{-1}(t)\dot{X}(t)\right) =d(t)\operatorname{tr}\left(X^{-1}(t)A(t)X(t)\right) = d(t)\operatorname{tr}\left(A(t)\right).$$
So (we can also come here with Liouville's Formula), $$d(t) = e^{\int_0^\infty\operatorname{tr}\left(A(s)\right)ds}d(0)$$ and since $$d(t)\ge d_0>0$$, we conclude $$\int_0^\infty\operatorname{tr}\left(A(s)\right)ds \ge \ln d_0 > -\infty$$.