<s>[INST] Minimalpolynomial of a Matrix without calculating the characterstic polynomial Is there a way calculating the minimal polynomial of a matrix without using the characteristic polynomial?
For example following matrix $$M = \left(
\begin{array}{cccc}
 1 & 0 & -2 & 0 \\
 -2 & -1 & 4 & 2 \\
 0 & 0 & -1 & 0 \\
 0 & 0 & 2 & 1 \\
\end{array}
\right).$$
Calculating the characteristic polynomial I get $\chi_M = (x-1)^2(x+1)^2$ with eigenvalues $-1$ and $1$ with algebraic multiplicity 2. Using the characteristic polynomial I have to check all possibilities for the minimal polynomial $\mu_M$ which are $$(x-1), (x-1)^2, (x+1), (x+1)^2, (x-1)(x+1), (x-1)^2(x+1), (x-1)(x+1)^2, (x-1)^2(x+1)^2.$$
Is there a way to see from the form of the matrix, what the minimal polynomial could be? Since $M$ could be partitioned as follows:
$$M = \left(
\begin{array}{cccc}
 1 & 0 & -2 & 0 \\
 -2 & -1 & 4 & 2 \\
 0 & 0 & -1 & 0 \\
 0 & 0 & 2 & 1 \\
\end{array}
\right)= \left(
\begin{array}{cc}
 A & B \\
 0 & C \\
\end{array}
\right),$$ with $A =\left(
\begin{array}{cc}
 1 & 0 \\
 -2 & -1 \\
\end{array}
\right), B = \left(
\begin{array}{cc}
 -2 & 0 \\
 4 & 2 \\
\end{array}
\right)$ and $C = \left(
\begin{array}{cc}
 -1 & 0 \\
 2 & 1 \\
\end{array}
\right)$.
 [/INST] In fact, if we take 
$$
A = \pmatrix{1&-2\\0&-1}, \quad
B = \pmatrix{1&0\\-2&-1}
$$
Then we have
$$
M = \pmatrix{a_{11}B & a_{12}B\\a_{21}B & a_{22}B} = A \otimes B
$$
Where $\otimes$ denotes the Kronecker product. Since both $A$ and $B$ have eigenvalues $\{-1,1\}$, $A \otimes B$ will have the spectrum $\{-1,-1,1,1\}$.
Note also that the Kronecker product of diagonalizable matrices is diagonalizable, so $M$ must have the minimal polynomial $q_M(x) = (x-1)(x+1)$.  
In general, the exponent of a linear term in the minimal polynomial of $A \otimes B$ cannot exceed any of the exponents in the minimal polynomials of $A$ or $B$.
In fact, given the Jordan forms of $A$ and $B$, we can deduce the Jordan form of $A \otimes B$.

Another approach: the characteristic polynomial of a matrix of the form
$$
M = \pmatrix{A&B\\0&C}
$$
with $A,C$ square will be $\chi_M(x) = \chi_A(x)\chi_C(x)$.
</s>