How to prove that the numerator of the transfer function has a degree of m. Consider we have the following state equation: (n-dimensional) 
$ \dot x = Ax + bu $    , $ y = cx $ 
how can I show that $\hat g(s)$ [transfer function] has m zeros if and only if 
$ cA^i b = 0 \space\space\space for\space\space i=0,1,2,..., n-m-2$
and 
$\quad cA^i b \neq 0 \quad i = n - m - 1 , ... , n $.
I don't have a single clue how should I prove this, I just know that $\hat g(s) = c(sI-A)^{-1}b $. 
 A: To prove the forward path : differentiating the output consecutively $n$ times it holds true that
$$\dot{y}=c\dot{x}=cAx+cbu=cAx\\
\ddot{y}=cA\dot{x}=cA(Ax+bu)=cA^2x+cAbu=cA^2x\\ \vdots 
\\ y^{(n-m-1)}=cA^{n-m-1}x+cA^{n-m-2}bu=cA^{n-m-1}x\\
y^{(n-m)}=cA^{n-m}x+(cA^{n-m-1}b)u\\
y^{(n-m+1)}=cA^{n-m+1}x+(cA^{n-m}b)u+(cA^{n-m-1}b)\dot{u}\\
\vdots\\
y^{(n)}=cA^{n}x+(cA^{n-1}b)u+(cA^{n-2}b)\dot{u}+\cdots+(cA^{n-m-1}b)u^{(m)}$$
Let $\chi(s)$ the characteristic polynomial of $A$ i.e.
$$\chi(s):=\det(sI-A)=s^n+a_1s^{n-1}+\cdots+a_{n-1}s+a_n$$
with the property (Cayley-Hamilton theorem) that
$$A^n+a_1A^{n-1}+\cdots+a_{n-1}A+a_nI=0$$
Then, multiplying each of the equations describing $y^{(n-i)}$ by $a_i$ and taking their sum we result in
$$y^{(n)}+a_1y^{(n-1)}+\cdots+a_{n-1}\dot{y}+a_ny=c\left[A^n+a_1A^{n-1}+\cdots+a_{n-1}A+a_nI\right]x+\\
+c(a_mA^{n-m-1}+a_{m-1}A^{n-m}+\cdots+a_1A^{n-2}+A^{n-1})bu+c(a_{m-1}A^{n-m-1}+\cdots+a_1A^{n-3}+A^{n-2})b\dot{u}+\cdots+c(a_1A^{n-m-1}+A^{n-m})bu^{(m-1)}+(cA^{n-m-1}b)u^{(m)}$$
or equivalently
$$y^{(n)}+a_1y^{(n-1)}+\cdots+a_{n-1}\dot{y}+a_ny=
b_0u+b_1\dot{u}+\cdots+b_{m-1}u^{(m-1)}+b_mu^{(m)}\qquad\qquad (1)$$
with
$$b_m:=cA^{n-m-1}b\\
b_{m-1}:=c(a_1A^{n-m-1}+A^{n-m})b\\
\vdots\\
b_1:=c(a_{m-1}A^{n-m-1}+\cdots+a_1A^{n-3}+A^{n-2})b\\
b_0:=c(a_mA^{n-m-1}+a_{m-1}A^{n-m}+\cdots+a_1A^{n-2}+A^{n-1})b$$
and $b_m\neq 0$. Considering the Laplace transform of both sides of (1) (with zero initial values) we obtain
$$\chi(s)Y(s)=(b_ms^m+b_{m-1}s^{m-1}+\cdots+b_1s+b_0)U(s)$$
and therefore the transfer function is given by
$$\frac{Y(s)}{U(s)}=\frac{b_ms^m+b_{m-1}s^{m-1}+\cdots+b_1s+b_0}{\chi(s)}$$
Thus, the numerator of the transfer function is a polynomial of order $m$ and therefore there are exactly $m$ zeros.
We have proved the if part. The reverse can also be proven with a similar reasoning and a contradiction argument i.e. if we assume that $cA^{i}b\neq 0$ for some $i=0,1,2,\cdots,n-m-2$ then repeating the same process a polynomial of order higher than $m$ will appear in the numerator of the transfer function.
