3
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

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*}

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .