So I had a question regarding proving the Cayley-Hamilton theorem using discrete states i.e. $x(k+1)=Ax(k)+bu(k)$ & $y(k)=c^Tx(k)$ where $x(k),b,c \in R^n$. The question stated that for an integer n and real numbers $a_0, a_1,..a_n$ we define an input sequence $u(k)$ s.t $u(0)=a_0, u(-1)=a_1,..., u(-n)=a_n$ and another condition to solve the problem is that $x(-n)=0$
And in order to shorten the question, we define the characteristic polynomial of $a(z)$ as $a(z)=a_nz^n+a_{n-1}z^{n-1}+...+a_1z_1+a_0$.
The first section was to prove that $x(1)=a(A)b$ which was pretty straight-forward and the second section was to find an expression for $y(1)$ using the first question which came as $y(1)=c^Ta(A)b$.
The section I had an issue with was proving that $y(1)=0$
For this section we defined the Z-transform of the sequences in question i.e. $y(k), x(k)$ and $u(k)$.
Without showing much of the derivation as I don't feel it relevant to the question, (I can add it below if deemed necessary) I obtained, $U(z)=a(z), (zI-A)X(Z)=U(z)$ $Y(z)=c^T(zI-A)^{-1}bU(z)=c^T(zI-A)^{-1}ba(z)$. For A inveritble, I had no problem in showing that $y(1)=0$ because I plugged in $z=0$ into the Z transform and I obtained $Y(0)=-c^TA^-1ba_0=-c^TA^-1b\cdot det(A)$ which is non-zero and because of that we know that we don't have a pole at zero and that means the ROC contains $|z|=0$ and thus the system is anti-causal. As a result, $y(k)=0$ for $k>0$ and thus $y(1)=0$.
This isn't enough though as the Cayley-Hamilton theorem holds for all square matrices so I necessarily have to show it for A that's non-invertible.
I'm not sure how to go about this as I'm not too well versed in proving such items mathematically or rigorously but a push in what approach to adopt to solve it for the non-invertible A case would really help.
update: So I was informed that the logic of considering the value of $Y(0)$ and checking if $Y(0)$ is finite doesn't hold with the following counter-example $\sum_{i=0}^\infty az^{-1}=F(z):=z/(z-a)$ for $z=0, F(0)=0$ but the ROC is $|z|>a$ which doesn't contain $|z|=0$. I did manage to expand $(zI-A)^-1$ into a power series but there may be more information needed to complete the proof.
The solution is actually much simpler. As usual, when a problem is approached from the right angle, the solution becomes natural.
Consider the input
$$ u:\{-n,\ldots\}\mapsto\{a_n,\ldots,a_1,a_0,0,\ldots\} $$ and the initial condition $x(-n)=0$. We know already that $x(1)=a(A)b$ and, hence, that $y(1)=c^Ta(A)b$. To see that $y(1)=0$, note first that the $z$-transform of the sequence $u$ is given by
$$ U(z)=\sum_{i=0}^na_iz^i=a(z)=\det(zI-A). $$
Therefore, we have that
$$ Y(z)=H(z)U(z)=N(z) $$ where $H(z):=c^T(zI-A)^{-1}b=:N(z)/a(z)$, where $N(z)$ is the numerator of the transfer function associated with $(A,b,c^T)$. For completeness, the the numerator is given by
$$ N(z):=\begin{bmatrix} zI-A & b\\c^T & 0 \end{bmatrix}. $$
Now, since there is no direct feedthrough (i.e. the output $y$ does not depend directly on the input $u$), the degree of $N(z)$ is at most $n-1$ and, we have that
$$ Y(z)=\alpha_{n-1}z^{n-1}+\ldots+\alpha_{1}z+\alpha_{0}. $$ This can be rewritten as $$ Y(z)=y(-n+1)z^{n-1}+\ldots+y(-1)z+y(0)+y(1)z^{-1}+y(2)z^{-2}+\ldots, $$ which implies that $y(k)=0$ for all $k\ge1$.
Now, it is enough to observe that this holds regardless of the value for $b$ and $c$, which means that $c^Ta(A)b=0$ for all $b,c\in\mathbb{R}^n$. This implies that we necessarily have that $a(A)=0$.