The matrix
A= \begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}
has 2 as a double eigenvalue. Its characteristic polynomial is \lambda^2- 4\lambda + 4= (\lambda- 2)^2. Of course, the matrix itself satisfies A^2- 4A+ 4I= 0. But, here, it also satisfies A- 2I= 0. That is its "minimal" polynomial.
On the other hand,
B= \begin{bmatrix}2 & 1 \\ 0 & 2\end{bmatrix}
also has 2 as a double eigenvalue and, of course, satisfies its charateristic equation, (2- A)^2= 0, but does not satisfy A- 2= 0 so its minimal polynomial is the same as its characteristic polynomial.
One can show that all eigenvalues must be "represented", as factors, in the minimal polynomial so, if all eigenvalues are distinct, all the linear factors must be there- the minimal polynomial is the same as the characteristic polynomial. But if a matrix has a multiple eigenvalue with more than one independent corresponding eignvector, then we can remove some of those factors and get a minimal polynomial of degree lower than the characteristic polynomial.
I also recommend you look at the website Simon Bridge links to.
