Suppose we have a positive definite symmetric matrix $\mathbf V(0) \in \mathbb S^{n}_{++}$, which changes with time according to the following equation,
$\dot{\mathbf V}(t) = \mathbf A \mathbf V(t) + \mathbf V(t) \mathbf A^\text{T}$,
where $\mathbf A \in \mathbb R^{n\times n}$ is an arbitrary constant matrix.
I would like be able to claim that $\mathbf V(t) \succ \mathbf 0$ for all $t \geq 0$. Is this true?
At least one solution of the above differential equation is \begin{equation} \mathbf V(t) = e^{\mathbf A t} \mathbf V(0) e^{\mathbf A^\text{T} t}. \end{equation} Since $e^{\mathbf A^\text{T} t}= (e^{\mathbf A t})^\text{T}$, this can be written as \begin{equation} \mathbf V(t) = e^{\mathbf A t} \mathbf V(0) (e^{\mathbf A t})^\text{T}. \end{equation} The matrix $e^{\mathbf A t}$ is nonsingular. Thus, $\mathbf V(t) \succ 0$.