Consider the controllability matrix $C= [B, AB, A^2B,...,A^{n-1}B]$ such that $A \in \mathbb{R}^{n\times n}$ and $B \in \mathbb{R}^{n\times m}$.
Now consider the range of $C$, $\mathcal{R}(C)= \{v | v =Mu\}$.
I want to show that $\mathcal{R}(C)$ is $A$-invariant, meaning if $\eta \in \mathcal{R}(C)$, then $A \eta \in \mathcal{R}(C)$.
I suppose $\eta \in \mathcal{R}(C)$, thus there exists $x$ such that
$$[B, AB, A^2B,...,A^{n-1}B]x = \eta$$
Then
$$A \eta = A[B, AB, A^2B,...,A^{n-1}B]x = [AB, A^2B, A^3B,...,A^{n}B]x$$
This is where I get confused. If I recall correctly Cayley-Hamilton theorem states that $$A^n = a_0 I + a_1A + ... + a_{n-1}A^{n-1}=0$$
But I don't see how this gives me any $y$ such that $A\eta =[B, AB, A^2B,...,A^{n-1}B]y$.
Let $p(x)=x^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$ be the characteristic polynomial of $A$. Since $p(A)=0$, we get $$ A^n=-c_{n-1}A^{n-1}-\cdots-c_1A-c_0I. $$ Therefore, if $\eta=[B,AB,\ldots,A^{n-2}B,A^{n-1}B]x$ where $x^T=(x_0^T,\ldots,x_{n-1}^T)$ with each $x_i\in\mathbb R^m$, then \begin{aligned} A\eta &=A[B,AB,\ldots,A^{n-2}B,A^{n-1}B]x\\ &=ABx_0+A^2Bx_1+\cdots+A^{n-1}Bx_{n-2}+A^nBx_{n-1}\\ &=ABx_0+A^2Bx_1+\cdots+A^{n-1}Bx_{n-2}-(c_{n-1}A^{n-1}B+\cdots-c_1AB+c_0B)x_{n-1}\\ &=B(-c_0x_{n-1})+AB(x_0-c_1x_{n-1})+A^2B(x_1-c_2x_{n-1})+\cdots+A^{n-1}B(x_{n-2}-c_{n-1}x_{n-1})\\ &=[B,AB,\ldots,A^{n-2}B,A^{n-1}B]\pmatrix{-c_0x_{n-1}\\ x_0-c_1x_{n-1}\\ x_1-c_2x_{n-1}\\ \vdots\\ x_{n-2}-c_{n-1}x_{n-1}}\\ &=[B,AB,\ldots,A^{n-2}B,A^{n-1}B]y. \end{aligned}