Primitive roots of unity occuring as eigenvalues of a product

Question

I am currently trying to understand the proof of Benson's Lemma (1.9.1) in Generalized Quadrangles by Payne and Thas.

Background

We have two $k × k$ matrices $Q$ and $M$. We want to determine some formula for $\operatorname{tr}(QM)$. The following information has been proven thus far:

The matrix $Q$ is a permutation matrix with order $n$. Hence, the eigenvalues $\xi$ of $Q$ are all $n$-th roots of unity.

and

The matrix $M$ is a real symmetric matrix. It has eigenvalues $\lambda_0, \lambda_1,$ and $\lambda_2$ with multiplicity $m_0, m_1,$ and $m_2$ respectively. We know that each $\lambda_j$ is an integer, and furthermore that $\lambda_0 = 0$ and $m_1 = 1$.

We also know that $QM = MQ$, and that $\operatorname{tr}(QM)$ is an integer. Lastly, we know that there is some eigenvalue $\theta_1$ of $QM$ such that $\theta_1 = \lambda_1$.

Deductions

Since both $Q$ and $M$ are normal matrices, they are both diagonalizable. Moreover, since they commute, they are simultaneously diagonalizable (by $S$, say). Then

$$ S(QM)S^{-1} = (SQS^{-1})(SMS^{-1}),$$

and so the eigenvalues $\theta$ of $QM$ have the form $\theta = \xi\lambda$ where $\xi$ is an eigenvalue of $Q$ and $\lambda$ is an eigenvalue of $M$. Since the eigenvalue $\theta_1$ agrees with $\lambda_1$ (which has a multiplicity of $1$ as an eigenvalue of $M$), it follows that $\theta_1$ has a multiplicity of $1$ as an eigenvalue of $QM$. Moreover, since $\lambda_0 = 0$ has a multiplicity of $m_0$ in $M$, it follows that $0$ is an eigenvalue of $QM$ also with multiplicity of $m_0$.

Therefore, all other eigenvalues of $QM$ have the form $\xi\lambda_2$. We know that there are exactly $m_2$ such eigenvalues.

The Problem

Payne and Thas claim the following fact:

For each divisor $d$ of $n$, let $U_d$ denote the sum of all primitive $d$-th roots of unity. Then $U_d$ is an integer. For each divisor $d$ of $n$, the primitive $d$-th roots of unity all contribute the same number of times to eigenvalues $\theta$ of $QM$ with $|\theta| = \lambda_2$. Let $a_d$ denote the multiplicity of $\xi_d\lambda_2$ as an eigenvalue of $QM$, with $d \mid n$ and $\xi_d$ a primitive $d$-th root of unity, then we have

$$\operatorname{tr}(QM) = \lambda_1 + \sum_{d \mid n}(a_dU_d)\lambda_2.$$

This bolded statement is what I am having trouble understanding.

• If $\xi_1$ and $\xi_2$ are two different primitive $d$-th roots of unity, why should the multiplicity of $\xi_1\lambda_2$ and $\xi_2\lambda_2$ be equal?

• If $\xi$ is a $d$-th root of unity, and there exists an eigenvalue $\xi\lambda_2$ of $QM$, why do all other $d$-th roots of unity appear as well?

Answer 1

$\def\i{\mathrm{i}}$First, by definition, for any $d \in \mathbb{N}_+$, the $d$-th cyclotomic polynomial is$$ {\mit Φ}_d(x) = \prod_{\substack{0 \leqslant l \leqslant d - 1\\(l, d) = 1}} ( x - ω_d^l), $$ where $ω_d = \exp\left( \dfrac{2π\i}{d} \right)$. Note that ${\mit Φ}_d(x) \in \mathbb{Z}[x] \subseteq \mathbb{Q}[x]$ and ${\mit Φ}_d(x)$ is irreducible in $\mathbb{Q}[x]$, thus for any $0 \leqslant l \leqslant d - 1$ such that $(l, d) = 1$, the minimal polynomial of $ω_d^l$ over $\mathbb{Q}$ is ${\mit Φ}_d(x)$.

Next, for any $f(x) \in \mathbb{Q}[x]$, if $\dfrac{f(x)}{x - ω_d^l} \in \mathbb{C}[x]$ for some $0 \leqslant l \leqslant d - 1 $ and $(l, d) = 1$, i.e. $f(ω_d^l) = 0$, because the minimal polynomial of $ω_d^l$ over $\mathbb{Q}$ is ${\mit Φ}_d(x)$, then $\dfrac{f(x)}{{\mit Φ}_d(x)} \in \mathbb{Q}[x]$. This further implies that if $\dfrac{f(x)}{(x - ω_d^l)^m} \in \mathbb{C}[x]$ for some $m \in \mathbb{N}_+$, then $\dfrac{f(x)}{({\mit Φ}_d(x))^m} \in \mathbb{Q}[x]$.

Now, because all eigenvalues of $QM$ are of the form $ξλ$, where $ξ$ and $λ$ are eigenvalues of $Q$ and $M$, respectively, and the multiplicity of $λ_0 = 0$ as an eigenvalue of $M$ is $m_0$, and that of $λ_1$ as an eigenvalue of $M$ is $1$, and $θ_1 = λ_1 \in \mathbb{Z}$, then$$ |xI - QM| = x^{m_0} (x - λ_1) f_0(x), $$ where$$ f_0(x) = \prod_{j = 1}^{k - m_0 - 1} (x - λ_2 ξ_j) \in \mathbb{C}[x], \quad ξ_j^n = 1. $$ Note that $QM \in \mathrm{M}_{k × k}(\mathbb{Z})$ implies $|xI - QM| \in \mathbb{Z}[x]$, then $f_0(x) \in \mathbb{Z}[x]$. Define$$ f(x) = \frac{1}{λ_2^{k - m_0 - 1}} f_0(λ_2 x) = \prod_{j = 1}^{k - m_0 - 1} (x - ξ_j) \in \mathbb{Q}[x]. $$ For any $d \mid n$, suppose the multiplicity of $ω_d^l$ ($0 \leqslant l \leqslant d - 1$ and $(l, d) = 1$) in $f(x)$ is $a_{d, l}$ and $a_d = \max\limits_{\substack{0 \leqslant l \leqslant d - 1\\(l, d) = 1}} a_{d, l}$. Because $\dfrac{f(x)}{(x - ω_d^l)^{a_d}} \in \mathbb{C}[x]$ for some $l$, then $\dfrac{f(x)}{({\mit Φ}_d(x))^{a_d}} \in \mathbb{Q}[x]$, which implies $a_{d, l} = a_d$ for all $0 \leqslant l \leqslant d - 1$ and $(l, d) = 1$.

Answer 2

This is a geometric fact—the roots of unity must all appear equally often because otherwise the sum of all the eigenvalues of $QM$ (the trace of $QM$) won't be an integer—it'll be a complex number.

Note that the eigenvalues of $M$ are integers, and the eigenvalues of $Q$ are roots of unity, and the eigenvalues of $QM$ are products of an eigenvalue of $M$ and an eigenvalue of $Q$.

From a purely geometric perspective, if $\lambda$ is an integer and $u_1, \ldots, u_d$ are the $d$th roots of unity, then $\lambda u_1, \ldots, \lambda u_d$ are equally-spaced complex numbers in the plane with equal magnitude. Their sum is an integer $(\lambda)$ but if we take any unevenly-weighted linear combination of them, the result is necessarily a complex number.

Because the trace of a matrix is equal to the sum of its eigenvalues (counting multiplicities repeatedly), and because the trace of $QM$ is known to be an integer, there cannot be any uneven weights. It must be that for each integer eigenvalue $\lambda$ of $M$, the values $\lambda u_1, \ldots \lambda u_d$ occur equally often as eigenvalues of $QM$ (i.e. have the same multiplicity.)