Cayley–Hamilton And Invertible Matrix In my lecture notes, it was mentioned that if the Cayley–Hamilton polynomial has a free element then it is invertible. Namely, $P_A(x) = a_n x^n + \dots + a_1 x + a_0$ there $a_0 \neq 0$. Why is it correct?
 A: Yes, it is correct, because the constant term of the characteristic polynomial of a matrix $A$ is $\det A$. Thefore, it is not $0$ if and only if the matrix is invertible.
A: The characteristic polynomial vanishes at the matrix $A$. This is Cayley–Hamilton theorem.
Therefore $0= P_A(A)=a_n A^n+...+a_1 A+a_0 I$
Which can be writen as$$
I = A \cdot \left[\left( \frac{-1}{a_0} \right) (a_1 + \cdots + a_n A^{n-1}) \right]$$
Proving that $A$ is invertible.
A: If your question is:

if the characteristic polynomial does not have zero as a root, then $T$ is invertible

we can show the contrapositive: if $T$ is not invertible, then it has $0$ as an eigenvalue, and hence $x$ divides the characteristic polynomial.
In particular, we have that $\mathrm{det}(T)=0$, which means that 
$\det(T-Ix)$ is zero when $x=0$, so $x \mid P_T(x)$.

A concrete example: $$\begin{pmatrix}1&0\\0&0 \end{pmatrix}$$
we can take $$\mathrm{det}\begin{pmatrix}1-x&0\\0&-x \end{pmatrix}=(x^2-x).$$
A: If $a_0\ne 0$, then Cayley-Hamilton $p(A)=0$ gives an explicit inverse for $A$:
\begin{align}
           I &=  -\frac{1}{a_0}(a_n A^{n-1}+a_{n-1}A^{n-2}+\cdots+a_1)A \\
             &= A\left(-\frac{1}{a_0}(a_n A^{n-1}+a_{n-1}A^{n-2}+\cdots+a_1)\right)\end{align}
