Consider we have a polynomial $p = z^m + b_{m-1}z^{m-1} + \dotsb + b_0$ with matrix coefficients $b_i \in M_n(\mathbb{C})$. Then we might consider the companion matrix $$T = \left[ \begin{matrix} 0_n & 0_n &\dots & b_0 \\ I_n & 0_n &\dotsb & b_1 \\ & \ddots && \vdots \\ &&I_n & b_{m-1} \end{matrix} \right], $$ where $I_n$ is the identity matrix, and $0_n$ is the zero matrix in $M_n(\mathbb{C})$.
$T$ is a block matrix, so we might consider it as a complex-valued matrix of dimension $m\cdot n \times m \cdot n$, and I want to show that its characteristic polynomial $\chi_T(z) = \det(z\cdot I_{n\cdot m} - T)$ equals $\det(p(z))$, where we consider $p(z)$ as a matrix with polynomial-valued entries.
If we naively compute the characteristic polynomial of $T$ over the ring of matrices we simply obtain $p$, but I'm not shure if I can even apply the Laplace expansion theorem over non-commutative rings. And even then, computing block-wise determinants does and then the determinant of the resulting matrix does in general not yield the determinant of the matrix one started with.
Applying Laplace expansion directly to $z\cdot I_{nm} - T$ yields rather ugly mixed terms that I don't know how to handle.
I also found this nice answer on how to show that the characteristic polynomial of the companion matrix equals the polynomial you started with, but I don't see if this generalizes: The claim that $\chi_T = \mu_T$ ($\mu_T$ is the minimal polynomial) is wrong in my situation. For example if $b_i = 0$ for all $i$, then $T^m = 0$, so in general the minimal polynomial has lower degree.
