Given $$ \begin{align*} \dot{x} &= Ax + Bu \\ y &= Cx \end{align*} $$ where $A \in \mathbb{R}^{n \times n}$, $B \in \mathbb{R}^{n \times m}$, $C \in \mathbb{R}^{p \times n}$.
How to prove the title?
I think I should use the matrix:
$$ \begin{bmatrix}C \\ CA \\ \vdots \\ CA^{n-1} \end{bmatrix} $$
Let
$$ \mathcal{O} = \begin{bmatrix}C \\ CA \\ \vdots \\ CA^{n-1} \end{bmatrix} $$
We need to show that if $v \in \text{Ker} \mathcal{O}$, then $Av \in \text{Ker} \mathcal{O}$. If $v \in \text{Ker} \mathcal{O}$, then
$$ \begin{bmatrix}C \\ CA \\ \vdots \\ CA^{n-1} \end{bmatrix} v = \begin{bmatrix}Cv \\ CAv \\ \vdots \\ CA^{n-1}v \end{bmatrix} = 0 $$
Now, let us look at $\mathcal{O} A v$, which is
$$ \begin{bmatrix}CA \\ CA^2 \\ \vdots \\ CA^{n} \end{bmatrix} v = \begin{bmatrix}CAv \\ CA^2v \\ \vdots \\ CA^{n}v \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ \vdots \\ CA^{n}v \end{bmatrix} $$
The last term is also $0$ because of the Cayley-Hamilton theorem.