Minimal polynomial $\mu_A(x)$ I have a matrix $A=\begin{bmatrix}1&3&0\\1&-1&0\\0&-2&1\end{bmatrix}$.  The characteristic polynomial, $\chi_A(x)=0=\det(\lambda{I}-A)$ so
$$\det{(\lambda{I}-A)}=(\lambda-1)[(\lambda^2-1)-3]=(\lambda-1)(\lambda-2)(\lambda+2)=\lambda^3-\lambda^2-4\lambda+4$$
So the minimal polynomial, $\mu_A(x)$ must be a product of the linear factors of $\chi_A(x)$, so my candidates are
$$x-1,x-2,x+2,x^2+x-2,x^2-3x+2,x^2-4, \chi_A(x)$$
Calculating, $A^2=\begin{bmatrix}4&0&0\\0&4&0\\-2&0&1\end{bmatrix}$. 
Clearly, none of the linear factors can be $\mu_A(x)$.  Since $A^2\neq 4I$, it can't be $x^2-4.$  That leaves $\chi_A(x)$, or the remaining two quadratics.  But by my calculation, neither work.  That leaves $\chi_A(x)=\mu_A(x)$.  Is there a way to discover this fact without just running through all the polynomial operators?  Or did I make a mistake and is one of the quadratics actually $\mu_A(x)$?

Here's a quicker way:  use the fact that $A$ has $3$ distinct eigenvalues.  Since $\mu_A(x)$ is a polynomial such that $\mu_A(A) = 0$, we must have
$\mu_A(\lambda) = 0$ for any eigenvalue $\lambda$ of $A$; this follows from the fact that $Av = \lambda v$ with $v \ne 0$ implies that $q(A)v = q(\lambda) v$ for any polynomial $q(x)$.  Since there are three distinct eigenvalues, it follows that $\deg \mu_A(x) \ge 3$.  But then we must have $\mu_A(x) = \chi_A(x)$, since $\chi_A(A) = 0$ and $\deg \chi_A(x) = 3$.
Note:  I think it is worth pointing out that this result generalizes to arbitrary dimension $n$ and also holds over any field, at least in the event that the number of distinct eigenvalues of $A$ is $n$.  End of Note.
Hope this helps.  Cheerio,
and as always, 
Fiat Lux!!!
