# How to verify this matrix identity?

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.

• The transpose is a companion matrix, whose characteristic and minimal polynomials can be read off from the matrix. I expect you'll find that solves your problem, since $M^{2m}=I_m$ will hold if and only if the minimal/characteristic polynomials divide $x^{2m}-1$. Jun 8 at 20:10
• @ArturoMagidin, many thanks for the hint! I can solve the problem from here. Jun 8 at 20:14

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 $$M^{2m}=I$$.

• Thank you very much! Your answer is crystal clear and correct to me. Jun 8 at 20:41
• @Zuriel You're welcome! I also added an alternative approach that you might find interesting. Jun 8 at 20:47
• @Apass.Jack You could make that into your own separate answer, if you'd like Jun 23 at 14:53