An interesting way to calculate the powers of a companion matrix

Question

Let $m\geq 3$ be a positive odd number and let $M$ be the $m\times m$ matrix defined by $$M=\begin{bmatrix}0&1&0&0&\cdots&0\\ 0&0&1&0&\cdots &0\\ 0&0&0&1&\cdots &0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&0&\cdots &1\\ 1&-2&2&-2&\cdots&2\end{bmatrix}.$$

It can be proven that $M^{2m}=I_m$, the identity matrix of size $m$. Now I would like to compute the $k$-th power of $M$ for $k=0,1,\ldots,2m-1$. My observation is as follows:

Let $N$ be the $(2m)\times m$ matrix defined by

$$N=\begin{bmatrix} 1&0&0&0&\cdots&0&0\\ 0&1&0&0&\cdots&0&0\\ 0&0&1&0&\cdots &0&0\\ 0&0&0&1&\cdots &0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&0&\cdots &1&0\\ 0&0&0&0&\cdots &0&1\\ 1&-2&2&-2&\cdots&-2&2\\ 2&-3&2&-2&\cdots&-2&2\\ 2&-2&1&-2&\cdots&-2&2\\ 2&-2&2&-3&\cdots&-2&2\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 2&-2&2&-2&\cdots&-3&2\\ 2&-2&2&-2&\cdots&-2&1\\ \end{bmatrix}.$$

We call the top row of $N$ the $0$-th row. Then for $k=0,1,\ldots,2m-1,2m$, $M^k$ is obtained from $N$ by taking the $k\!\pmod{2m}$-th row, $(k+1)\!\pmod{2m}$-th row, etc., up to the $(k+m-1)\!\pmod{2m}$-th row and form a matrix. For example, $$M^2=\begin{bmatrix} 0&0&1&0&\cdots &0&0\\ 0&0&0&1&\cdots &0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&0&\cdots &1&0\\ 0&0&0&0&\cdots &0&1\\ 1&-2&2&-2&\cdots&-2&2\\ 2&-3&2&-2&\cdots&-2&2 \end{bmatrix}.$$

while $$M^{2m-1}=\begin{bmatrix}2&-2&2&-2&\cdots&-2&1\\1&0&0&0&\cdots&0&0\\ 0&1&0&0&\cdots&0&0\\ 0&0&1&0&\cdots &0&0\\ 0&0&0&1&\cdots &0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&0&\cdots &1&0\\\end{bmatrix}$$

This seems to be an interesting way to compute the powers of the matrix $M$, but I am not sure how to verify this method, though some specific examples seem to suggest that this method is valid. Could anyone help with proving this method?

Answer 1

Nice observation.

The key idea is that the matrix product of the $r$-th row of $N$ and $M$ is the $(r+1)$-th row of $N$ (where the $2m$-th row is interpreted as the $0$-th row), which ensures the position of $M^{k+1}$ as a submatrix of $N$ will be one row below the position of $M^k$ as a submatrix of $N$.

The following is a detailed proof.

---

Fix $m$, an odd integer $\ge3$.
Let $P[r]$ denote the $r$-th row of a matrix $P$, $r=0, 1, \cdots$.
Note that $N[0], N[1], \cdots, N[m-1]$ are the standard unit row vectors with $m$ coordinates. Hence for all $m\times m$ matrix $P$ and $0\le r<m$ $$N[r]P=P[r].$$ Lemma: $N[r] M = N[(r+1)\%(2m)]$ for $0\le r<2m$.

Proof: Let $b=(2,-2,2,-2,\cdots,2)$.
$N[r]=b-N[r-m]$ for $m\le r< 2m.$
Verify that $bM=b$. $$N[r] M=\begin{cases} M[r]=N[r+1] &\text{if }0\le r < m -1\\ M[m-1]=N[m] &\text{if }r= m-1\\ (b-N[r-m])M=b-N[r-m+1]=N[r+1] &\text{if } m\le r<2m-1\\ (b-N[m-1])M=b-M[m-1]=N[0] &\text{if } r=2m-1 \end{cases}$$

---

Claim (OP's observation): For all $k\ge0$ and $0\le r\le m-1$, $$M^k[r]=N[(k+r)\% (2m)].$$ Proof: Do induction on $k$.

• It is trivially true when $k=0$.
• Suppose it is true for $k$. $$\begin{aligned} &\quad M^{k+1}[r]\\ \quad&=N[r]M^{k+1}\\ \quad &=(N[r]M^k)M\\ \quad &=(M^k[r])M\\ \text{induction hypothesis}\quad&=N[(k+r)\% (2m)]M\\ \text{the lemma}\quad&=M[(k+r+1)\% (2m)] \end{aligned}$$ So it is true for $k+1$.

Answer 2

$M^T$ is the coefficient matrix for the multiplication with $x$ in the quotient ring defined by $p(x)$ and in the monomial basis.

In other words, each row $i$ of $M$ is the coefficient sequence of $x\cdot x^{i-1}=x^{i}\pmod{p(x)}$, $i=1,2,...,m$. For the lower degrees there is no reduction, and so one just gets the canonical vector that selects the corresponding next-higher power in $[1,x,x^2,\dots,x^{m-1}]^T$. In the last row a simple reduction of $x^m$ is performed, giving the row $[-c_0,-c_1,\dots,-c_{m-1}]$ if $$p(x)=x^m+c_{m-1}x^{m-1}+\dots+c_1x+c_0.$$

The rows of $N$ contain just the reductions of the monomials, the $i$th row of $N$ contains the coefficients of $x^{i-1}\pmod(p(x))$. The power $M^k$ corresponds to the multiplication with $x^k$, expressed in the monomial basis. Thus the row $i$ in $M^k$ corresponds to $x^{k+i-1}\pmod{p(x)}$ which is found in row $(k+i)$ in $N$. Thus $M^k=N_{k+1:k+m}$

The claimed return of $M^{2m}$ to the identity matrix only happens if $p(x)|(x^{2m}-1)$, as was proven in the linked earlier post.

As $p(x)=(x-1)H_m(-x)$ one gets that $$-p(x)p(-x)=(x-1)H(x)(x+1)H(-x)=(x^m-1)(x^m+1)=x^{2m}-1$$ where $H_m(x)=1+x+\dots+x^{m-1}$ is the Haar polynomial.