linear algebra - How to compute the characteristic polynomial of a companion matrix to a polynomial with matrix-valued coefficients.

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.

Answer

We first note that any block matrix $ M =\left[\begin{matrix}
A & B \\C & D
\end{matrix}\right],$
where $A$ and $D$ are square matrices and $A$ is invertible, can be factorised in the form
$$M =

\left[\begin{matrix}
A & B \\
C & D
\end{matrix}\right]
=
\left[\begin{matrix}
A & 0 \\
C & 1
\end{matrix}\right]
\left[\begin{matrix}

1 & A^{-1}B \\
0 & D - CA^{-1}B
\end{matrix}\right],
$$
so that $\det M = \det A \cdot \det(D - CA^{-1}B)$.

If we now calculate formally in the function field $\mathbb{C}(z)$, the upper-left block of $z \cdot I_{nm} - T$ is just $A = z\cdot I_n$, which is invertible over $\mathbb{C}(z)$. Using the above notation, one calculates $$D - CA^{-1}B = \left[\begin{matrix}
z \cdot I_n & & & -b_1 - \frac 1 z b_0\\
-I_n & \ddots & &\vdots\\
& \ddots &z\cdot I_n&-b_{m-2}\\

& &-I_n & z \cdot I_n - b_{m-1}
\end{matrix}\right].
$$
Now inductively we know that $\det(D - CA^{-1}B) = \det(I_n z^{m-1} - b_{m-1}z^{m-2} - \dotsb - b_1 - \frac 1 z b_0)$. Thus $\det (z I - T) = z^n \det(I_n z^{m-1} - b_{m-1}z^{m-2} - \dotsb - b_1 - \frac 1 z b_0) = \det(I_n z^m - b_{m-1}z^{m-1} - \dotsb - b_1 z - b_0)$.