Exercise: Why is the following "proof" false?
$\text{ch}_A(xI-A)=\det(xI-A)$
Substitude $A$ for $x$, and obtain $\det(A-A)=0=\text{ch}_A(A)$.
Solution: Explanation
We cannot substitude $A$ for $x$ because $A$ is a matrix and $x$ represents a scalar.
$xI$ represents the matrix $\begin{pmatrix} x\\ &x\\ &&\ddots\\ &&&x\\ \end{pmatrix}$, and we clearly cannot substitute the matrix $A$ into the entries.