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 $\mu_A(x)$?

Hint: In this case the characteristic polynomial consists only of distinct linear factors, AND EVERY linear factor of the characteristic polynomial must be a factor of the minimum polynomial (ah, gave it away)
