# 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.

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 \begin{align*}\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).\end{align*}