Cfrgtkky

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 $mcdot n times m cdot n$, and I want to show that its characteristic polynomial $chi_T(z) = det(zcdot I_{ncdot m} - T)$ 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 $zcdot I_{nm} - T$ 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 $A = zcdot I_n$, which is invertible over $mathbb{C}(z)$. Using the above notation, one calculates $$D - CA^{-1}B = left[begin{matrix}
z cdot I_n & & & -b_1 - frac 1 z b_0\
-I_n & ddots & &vdots\
& ddots &zcdot I_n&-b_{m-2}\
& &-I_n & z cdot I_n - b_{m-1}
end{matrix}right].
$$
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 $det (z I - T) = z^n det(I_n z^{m-1} - b_{m-1}z^{m-2} - dotsb - b_1 - frac 1 z b_0) = det(I_n z^m - b_{m-1}z^{m-1} - dotsb - b_1 z - b_0)$.