Full rank of the matrix $Ce^{At}B$

Let $A \in \mathbb R^{n \times n}$. Fix $m <n$ and let $B \in \mathbb R^{n \times m}$, $C \in \mathbb R^{(m-1) \times n}$ be two matrices with full rank.

I want to find algebraic conditions (which I suspect to be related to observability of $(A,C)$ and/or controllability of $(A,B)$) on the matrices $A,B,C$, such that the kernel

\begin{align*} \ker Ce^{At}B \end{align*}

is one-dimensional for infinitely many times $t\ge 0$. Or, said differently (due to dimensionality reasons), I want to find conditions such that the matrix

\begin{align*} Ce^{At}B \end{align*}

has full (row) rank.

The conjecture is false. Let $n \geq 2m$ and let $A=0$, then $e^{At}$ is the identity matrix. Now let $B \in \mathbb{R}^{n \times m}$ be a full rank matrix such that the first $n-m$ lines are zero. Analogously, let $C \in \mathbb{R}^{(m-1) \times n}$ be a full rank matrix such tat the last $n+1-m$ columns are zero. Then $CB$ is the zero matrix. The same argument works if $A$ is a diagonal matrix.
• Would the conjecture be true though ,if I assume that the rank of $\begin{pmatrix} CB \\ CAB \\ \vdots \\ CA^{n-1}B \end{pmatrix}$ is $n$ ? – samsa44 Apr 30 '14 at 12:33