Let $A\in\mathbb{R}^{n\times n}$. Assume that $m$ is a divisor of $n$ and let $k=\frac n m$. Consider the characteristic polynomial of $A$, namely, $$ p_A(s) = \det(sI_n -A). $$
My question. Is it possible to rewrite $p_A(s)$ as $$\tag{1}\label{eq:1} p_A(s) = \det(s^k I_m + s^{k-1} L_{k-1}+\dots + sL_1 + L_0), $$ for suitable matrices $L_{i}\in\mathbb{R}^{m\times m}$? If so, how to compute the matrices $L_i$ in \eqref{eq:1}?
A few remarks:
- If $m=1$, then matrices $L_i$ in \eqref{eq:1} are scalars and they can be computed explicitly using a well-known expression.
- If $m=n$, $L_0=-A$, and \eqref{eq:1} coincides exactly with the standard definition of characteristic polynomial.
- I suspect that \eqref{eq:1}, if true, must be a well-known result. If this is indeed the case, pointers to textbooks/papers are very much appreciated.
There certainly are matrices $L_i$ that satisfy your equation. Let $$p_A(s)=s^n+a_{n-1}s^{n-1}+\ldots+a_1s+a_0$$ and define $$L_i=\begin{pmatrix} a_{(m-1)k+i} & 0 & \ldots & 0 \\ a_{(m-2)k+i} & 0 & \ldots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ a_i & 0 & \ldots & 0\end{pmatrix}$$ for $i=1,\ldots,k-1$ and $$L_0=\begin{pmatrix}a_{(m-1)k} & -1 & 0 & \ldots & 0\\ a_{(m-2)k} & 0 & -1 &\ldots & 0\\ \vdots & \vdots & \ddots & \ddots & \vdots\\ a_{k} & 0 & \ldots & 0 & -1 \\ a_0 & 0 & \ldots & \ldots & 0\end{pmatrix}$$ That is, $L_0$ follows the same rule as the rest of the $L_i$ but has additionally $-1$'s on the diagonal above the main diagonal. For example, if $n=12,k=3,m=4$, we get $$s^3I_4+\sum_{i=0}^{k-1}s^iL_i=\begin{pmatrix} s^3+a_{11}s^2+a_{10}s+a_9&-1&0&0\\ a_8s^2+a_7s+a_6&s^3&-1&0\\a_5s^2+a_4s+a_3&0&s^3&-1\\a_2s^2+a_1s+a_0&0&0&s^3 \end{pmatrix}$$ from which one maybe sees the pattern more easily than from the definition. The point of choosing matrices with that many zeroes is that the determinant contains very few summands, and is thus very easily controllable.
To prove that these matrices will always work, we fix $k\in\mathbb N$ and proceed by induction on $m$. For $m=1$, we have $L_i=(a_i)$ and the claim follows trivially. Now for the induction step $m-1\to m$, let $n=mk$, pick $A\in\mathbb R^{n\times n}$ and choose the $L_i$ as above. We calculate $$ \det\left(s^kI_m+\sum_{i=0}^{k-1}s^iL_i\right)=\det\begin{pmatrix} s^k+a_{n-1}s^{k-1}+\ldots+a_{n-k}&-1&0&\ldots&0\\ a_{n-k-1}a^{k-1}+\ldots+a_{n-2k}&s^k&-1&\ddots&0\\ \vdots&\vdots&\ddots&\ddots&\vdots\\ a_{2k-1}s^{k-1}+\ldots+a_k&0&0&s^k&-1\\ a_{k-1}s^{k-1}+\ldots+a_0&0&\ldots&0&s^k \end{pmatrix} $$ and Laplace expansion by the last row gives \begin{align}&(-1)^{m-1}(a_{k-1}s^{k-1}+\ldots+a_0)\det\begin{pmatrix}-1&0&\ldots&0\\s^k&-1&\ddots&0\\\vdots&\ddots&\ddots&\vdots\\0&\ldots&s^k&-1\end{pmatrix}\\&+(-1)^{2m-2}s^k\det\begin{pmatrix} s^k+a_{n-1}s^{k-1}+\ldots+a_{n-k}&-1&0&\ldots\\ a_{n-k-1}a^{k-1}+\ldots+a_{n-2k}&s^k&-1&\ddots\\ \vdots&\vdots&\ddots&\ddots\\ a_{2k-1}s^{k-1}+\ldots+a_k&0&0&s^k \end{pmatrix}\end{align} where the first determinant is $(-1)^{m-1}$ and the second, by the induction hypothesis, simplifies to $\sum_{i=0}^{(m-1)k}a_{i+k}s^i$. Plugging that in results exactly in $p_A(s)$, finishing the proof. $\square$
Now, obviously calculating these $L_i$ involves calculating $p_A$ first, so this does not really give a new way of calculating $p_A$, which would have been nice. In fact, this approach doesn't use the matrix $A$ at all, it really starts with an arbitrary polynomial.
On the other hand, there should usually be many possible choices for the $L_i$, since you can fill $k$ matrices of type $m\times m$, i.e. you have $km^2=mn$ free variables to choose, but the characteristic polynomial is determined by its $n$ coefficients. Maybe someone can find different choices for the $L_i$ that depend more directly on the matrix $A$.