Given the system $\dot{x} = Ax$, $y = Cx$, it is known that the Gramian, given by
$$ W({t_0},{t_1}) = \int_{t_0}^{t_1} e^{A^T(\tau -t_0)}C^TC e^{A(\tau -t_0)} \,{\rm d}\tau $$
is positive definite for some $t_1 > t_0$ if the system is observable. Assuming I know that the system is observable and hence $W \succ 0$, is it possible to show that $e^{A^T(t_1 -t_0)}C^TC e^{A(t_1 -t_0)} \succ 0$?
I found the following lemma[2]
However, $$e^{A^T(t_1 -t_0)}C^TC e^{A(t_1 -t_0)} \succ 0$$ does not seem to directly follow from this. I would appreciate any help.
Let $A\in\mathbb{R}^{n\times n}$ and $C\in\mathbb{R}^{p\times n}$, $p\le n$. If $p<n$, then the matrix
$$\Psi:=e^{A^T(t_1 -t_0)}C^TC e^{A(t_1 -t_0)}$$
is only positive semidefinite for all $t_1,t_0\in\mathbb{R}$ unless $p=n$ and $C$ is full rank.
First note that
$$u^T\Psi u=v^Tv=||v||_2^2$$
where $v=C e^{A(t_1 -t_0)}u$. Then, we have that the matrix $\Psi$ is positive semidefinite.
If it is positive definite, then $u^T\Psi u=0$ should imply that $u=0$. However, let $N$ be a basis for the null-space of $C$. Then, we have that
$$\Psi e^{-A(t_1 -t_0)}Nv=e^{A^T(t_1 -t_0)}C^TCNv=0$$
for all $v\in\mathbb{R}^n$. As a result, we have that
$$v^TN^Te^{-A^T(t_1 -t_0)}\Psi e^{-A(t_1 -t_0)}Nv=0$$
where $e^{-A(t_1 -t_0)}Nv\ne0$. Therefore, the matrix can only be positive definite if $C$ has no null-space (i.e. $C$ is square and full-rank).
In any way, it is the integral of that which is positive definite as the sum of positive semidefinite matrices can be positive definite. For instance, the sum of the matrices $$M_1=\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}\ \mathrm{and}\ M_2=\begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}$$
is obviously positive definite. This means that one cannot infer the positive definiteness of matrices involved in a sum that is positive definite.