Let $m\geq 3$ be a positive odd number and let $M$ be the $m\times m$ matrix defined by
$$M=\begin{bmatrix}0&1&0&0&\cdots&0\\
0&0&1&0&\cdots &0\\
0&0&0&1&\cdots &0\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&0&\cdots &1\\
1&-2&2&-2&\cdots&2\end{bmatrix}.$$
Is it true that $M^{2m}=I_m$, the identity matrix of size $m$?
I verified the case $m=3,5,7,9$ but am not sure how to prove this in general. Any help/hint will be appreciated.
Best Answer
As Arturo's comment notes, this matrix is the transpose of the companion matrix associated with a certain polynomial. In particular, we have $M = C(p)^\top$ with $$ p(x) = x^m - 2x^{m-1} + 2x^{m-2} - \cdots -2x^2 + 2x - 1. $$ With that, we find that $M^{2m} = I$ will hold if and only if $p(x)$ divides $x^{2m} - 1$. We find that this is indeed the case. In particular, we find that $x^{2m - 1} = p(x)q(x)$, where $$ q(x) = -p(-x) = x^{m} + 2x^{m-1} + \cdots + 2x + 1. $$ We can see this as follows. \begin{align} q(x)p(x) &= (x^{m} + 2x^{m-1} + \cdots + 2x + 1) (x^m - 2x^{m-1} + 2x^{m-2} - \cdots -2x^2 + 2x - 1) \\ & = [(x^m + x^{m-1}) + (x^{m-1}+x^{m-2}) + \cdots + (x^1 + x^0)] \\ &\qquad \cdot [(x^m - x^{m-1}) - (x^{m-1} - x^{m-2}) + \cdots +(x^1 - x^0)] \\ & = (x+1)[x^{m-1} + \cdots + x + 1] \ \cdot\ (x-1)[x^{m-1} - x^{m} + \cdots - x + 1] \\ & = (x-1)[x^{m-1} + \cdots + x + 1] \ \cdot\ (x+1)[x^{m-1} - x^{m} + \cdots - x + 1] \\ & = (x^m - 1)(x^m + 1) = x^{2m} - 1. \end{align}
Alternatively, we can use complex numbers. Let $\omega = e^{\pi i/m}$, i.e. an elementary $2m$-th root of unity. We find that $$ q(x) = (x + 1) \prod_{k=1}^{m-1}(x - \omega^{2m}), $$ so that its roots are $-1$ along with $\omega^{2k}$ for $k = 1,\dots,m-1$. On the other hand, the roots of $p(x) = -q(-x)$ are $-1$ together with $-\omega^{2k}$ for $k = 1,\dots,m-1$. Because $-\omega^{2k} = \omega^{2k + m} = \omega^{2k - m}$, we can conclude that the roots of $p(x)q(x)$ are equal to all of the $2m$-th roots of unity, each with multiplicity $1$.
Or, we could use these facts to see that the roots of $p$ are indeed roots of $x^{2m} - 1$, which implies that $p$ divides $x^{2m} - 1$.
Another proof can be given by identifying the eigenspaces of $M^m$, where $M$ is considered acting on the column vectors with $m$ coordinates by multiplying from the left.
Let $v_1, v_2, \cdots, v_m$ be the column $1,2,\cdots, m$ of the following matrix respectively, where hidden entries are zeros. $$\pmatrix{ & & & & &1&1\\ & & & &1&1& \\ & & &\vdots&1& & \\ & &1&\vdots& & & \\ &1&1& & & & \\ 1&1& & & & & \\ 1& & & & & &-1\\ }$$
Verify that $Mv_1=v_2$, $Mv_2=v_3$, $\cdots$, $Mv_{m-2}=v_{m-1}$ as well as $Mv_{m-1}=v_m$ and $Mv_m=-v_1$.
So $M^mv_1=-v_1$.
Hence $M^mv_i=-v_i$ for all $i$.
Hence $M^{2m}v_i=v_i$ for all $i$.
Let $u=\pmatrix{1\\1\\\vdots\\1}$. Verify that $Mu=u$.
Hence $M^{2m}u=u$.
Since $\pmatrix{v_1, v_2,, \cdots, v_{m-1}, u-v_1-v_3-\cdots-v_{m-2}}$ is the following triangular matrix, which is nonsingular, $$\pmatrix{ & & & & &1&1\\ & & & &1&1& \\ & & &\vdots&1& & \\ & &1&\vdots& & & \\ &1&1& & & & \\ 1&1& & & & & \\ 1& & & & & &\\ }$$ $v_1, v_2, \cdots, v_{m-1}, u$ are linearly independent vectors. Since there are $m$ of them, $M^{2m}$ fixes the entire space of column vectors with $m$ coordinates. Hence $M^{2m}=I$.