I am asked to find the minimal polynomial of the matrix
\begin{bmatrix}
4&-2&2\\
6&-3&4\\
3&-2&3\end{bmatrix}
I've calculated the characteristic polynomial is $\Delta(x)=(x-2)(x-1)^2$.
As we know that the minimal polynomial $m(x)$ must divide $\Delta(x)$, so the possibilities of being $m(x)$ are $(x-2),(x-1),(x-1)^2,(x-2)(x-1),(x-2)(x-1)^2$.
Among these, by Cayley-Hamilton theorem, for which least degree factor $\Delta(A)=O$ is satisfied will be the minimal polynomial.
But for some reason, the first three above mentioned factors were ignored in my book. What can be the possible reason for this?
Because every root of the characteristic polynomial is always a root of the minimal polynomial. In the case, for instance, of $x-1$, $2$ is not a root, but it is a root of that characteristic polynomial.