7 Clarified the meaning of the direct sum

Edit: I couldn't resist my predilection for generalizations: Using darij grinberg's simplification, the proof below shows:

Let $k$ be a field, $q \in GL_n(k)$ a matrix of finite exponent $m$ with char$(k) \nmid m$ and $M \subseteq k^n$. Futhermore, let $E$ be the eigenspace of $q$ corresponding to the eigenvalue $1$ and let $U \le k^n$ be the space spanned by the columns of $1-q$. Then the following is true for $A := 1+q+\dots + q^{m-1}$:

• $\lbrace x \in k^n \mid Ax \in M \rbrace = U \oplus + \frac{1}{m}(E \cap M)$
• $U$ and $(1/m)(E \cap M)$ intersect in $0$ iff $0 \in M$, otherwise the intersection is empty
• $A$ is diagonizable with diagonal $(m,...,m,0,...,0)$ where the number of m's equals $\dim E$

(Older formulation)

Let $E \le \mathbb{C}^n$ be the eigenspace of $1$ of the matrix $q$ and let $U \le \mathbb{C}^n$ be the space spanned by the columns of $1-q$.

Set $A := 1+q+\dots + q^{m-1}$ and $X:= \lbrace x \in \mathbb{C}^n \mid A\cdot x \in \mathbb{Z}^n \rbrace$ and $L := E \cap \mathbb{Z}^n$.

Then the following holds:

$X = U \oplus \frac{1}{m}L$.

Proof: Assume $\dim E = d$. Then $\dim U = \text{rank}(1-q) = n-d$.

Since each $x \in E$ satisfies $Ax = mx$, $E$ contains eigenvectors from $A$ of the eigenvalue $m$. From $A \cdot (1-q) = 0$ it follows that $U$ consists of eigenvectors of $A$ of the eigenvalue $0$. Hence $E \cap U = 0$ and for dimensional reasons $$\mathbb{C}^n = U \oplus E.$$ Since $q$ has integral entries, it's possible to chosse a basis of $E$ in $\mathbb{Q}^n$ and by multiplying with a suitable integer it's also possible to choose a basis in $\mathbb{Z}^n$. Therefore $L = E \cap \mathbb{Z}^n$ is a lattice of rank $d$. Let $\lbrace e_1, \dots, e_d \rbrace$ be a basis of $L$. Let $x \in X$ and write $$x = u + \sum_i \alpha_i e_i \text{ with } \alpha_i \in \mathbb{C}.$$ Then $Ax = \sum_i m\alpha_i e_i \in \mathbb{Z}^n$ and $q(Ax) = Ax$. It follows $Ax \in E \cap \mathbb{Z}^n = L = \oplus_i \mathbb{Z}e_i$ and therefore $m\alpha_i \in \mathbb{Z}$. This shows $X \subseteq U \oplus (1/m)L$. The converse inclusion is obvious. qed.

Edit: Also note that the image of $A$ is given by $$Y := \lbrace Ax \mid x \in X \rbrace = L.$$

6 Restriction on m added

Edit: I couldn't resist my predilection for generalizations: Using darij grinberg's simplification, the proof below shows:

Let $k$ be a field, $q \in GL_n(k)$ be a matrix of finite exponent $m$ with char$(k) \nmid m$ and $M \subseteq k^n$. Futhermore, let $E$ be the eigenspace of $q$ corresponding to the eigenvalue $1$ and let $U \le k^n$ be the space spanned by the columns of $1-q$. Then the following is true for $A := 1+q+\dots + q^{m-1}$:

• $\lbrace x \in k^n \mid Ax \in M \rbrace = U \oplus \frac{1}{m}(E \cap M)$
• $A$ is diagonizable with diagonal $(m,...,m,0,...,0)$ where the number of m's equals $\dim E$

(Older formulation)

Let $E \le \mathbb{C}^n$ be the eigenspace of $1$ of the matrix $q$ and let $U \le \mathbb{C}^n$ be the space spanned by the columns of $1-q$.

Set $A := 1+q+\dots + q^{m-1}$ and $X:= \lbrace x \in \mathbb{C}^n \mid A\cdot x \in \mathbb{Z}^n \rbrace$ and $L := E \cap \mathbb{Z}^n$.

Then the following holds:

$X = U \oplus \frac{1}{m}L$.

Proof: Assume $\dim E = d$. Then $\dim U = \text{rank}(1-q) = n-d$.

Since each $x \in E$ satisfies $Ax = mx$, $E$ contains eigenvectors from $A$ of the eigenvalue $m$. From $A \cdot (1-q) = 0$ it follows that $U$ consists of eigenvectors of $A$ of the eigenvalue $0$. Hence $E \cap U = 0$ and for dimensional reasons $$\mathbb{C}^n = U \oplus E.$$ Since $q$ has integral entries, it's possible to chosse a basis of $E$ in $\mathbb{Q}^n$ and by multiplying with a suitable integer it's also possible to choose a basis in $\mathbb{Z}^n$. Therefore $L = E \cap \mathbb{Z}^n$ is a lattice of rank $d$. Let $\lbrace e_1, \dots, e_d \rbrace$ be a basis of $L$. Let $x \in X$ and write $$x = u + \sum_i \alpha_i e_i \text{ with } \alpha_i \in \mathbb{C}.$$ Then $Ax = \sum_i m\alpha_i e_i \in \mathbb{Z}^n$ and $q(Ax) = Ax$. It follows $Ax \in E \cap \mathbb{Z}^n = L = \oplus_i \mathbb{Z}e_i$ and therefore $m\alpha_i \in \mathbb{Z}$. This shows $X \subseteq U \oplus (1/m)L$. The converse inclusion is obvious. qed.

Edit: Also note that the image of $A$ is given by $$Y := \lbrace Ax \mid x \in X \rbrace = L.$$

5 added 290 characters in body

Edit: I couldn't resist my predilection for generalizations: Using darij grinberg's simplification, the proof below shows:

Let $k$ be a field, $q \in GL_n(k)$ be a matrix of finite exponent $m$ and $M \subseteq k^n$. Futhermore, let $E$ be the eigenspace of $q$ corresponding to the eigenvalue $1$ and let $U \le k^n$ be the space spanned by the columns of $1-q$. Then the following is true for $A := 1+q+\dots + q^{m-1}$:

• $\lbrace x \in k^n \mid Ax \in M \rbrace = U \oplus \frac{1}{m}(E \cap M)$
• $A$ is diagonizable with diagonal $(m,...,m,0,...,0)$ where the number of m's equals $\dim E$
• (Older formulation)

Let $E \le \mathbb{C}^n$ be the eigenspace of $1$ of the matrix $q$ and let $U \le \mathbb{C}^n$ be the space spanned by the columns of $1-q$.

Then $Ax = \sum_i m\alpha_i e_i \in \mathbb{Z}^n$ and $q(Ax) = Ax$. It follows $Ax \in E \cap \mathbb{Z}^n = L = \oplus_i \mathbb{Z}e_i$ and therefore $m\alpha_i \in \mathbb{Z}$. This shows $X \subseteq U \oplus (1/m)L$. The converse inclusion is obvious. qed.

Remark: The requirement $q \in GL(n,\mathbb{Z})$ can be replaced by $q \in GL(n,\mathbb{Q})$ and the result still holds (same proof).

Edit: The proof also shows that for every matrix $q \in GL_n(k)$ of finite exponent $m$ ($k$ any field), the matrix $1+q+\dots + q^{m-1}$ is diagonizable with diagonal $(m,\dots,m,0,\dots,0)$ where the number of m's equals the dimension of the eigenspace corresponding to the eigenvalue $1$ of $q$.

4 Result also holds for matrices with rational entries
3 added 303 characters in body
2 Added the image of the matrix $A$
1