Finding minimal polynomial of big blocks diagonal matrix Consider the following matrix:  
\begin{bmatrix}
-3 & 1 & -1 &  &  &  &  & \\ 
-7 & 5 & -1 &  & 0 &  &  & 0\\ 
-6 & 6 & -2 &  &  &  &  & \\ 
 &  &  &  4 & 0 & 1 & & \\ 
 & 0 &  &  0 & 1 & 0 & & 0\\ 
 &  &  &  0 & 0 & 4 & & \\ 
 &  &  &  &  &  & -1 & -1\\ 
 & 0 &  &  & 0 &  &  1 & -3
\end{bmatrix}
I need to find its minimal polynomial.
I found that the characteristic polynomial is: $P(\lambda)=(\lambda+2)^4(\lambda-4)^3(1-\lambda)$ using the fact that for a block matrix:
$$A=\begin{bmatrix} {A}_{1} & 0 & \cdots & 0 \\ 0 & {A}_{2} & \cdots &  0 \\\vdots & \vdots & \ddots & \vdots \\0 & 0 & \cdots & {A}_{n} \end{bmatrix} $$
It holds that: $\det(A)=\prod_{i=1}^{n}\det(A_i)$
Is there an easy trick to find the minimal polynomial as well?

$p(A)=0$ is equivalent to $p(A_1)=\dots=p(A_n)=0$ 
So it is the minimum common multiple of the blocks minimum polynomials.
