Automorphisms of the subspace sum graphs on a vector space

TL;DR

This paper fully characterizes the automorphisms of the subspace sum graph on a finite-dimensional vector space, extending previous structural studies by providing a complete description of its symmetries.

Contribution

It offers the first complete characterization of automorphisms of the subspace sum graph, a problem left open in prior research.

Findings

Automorphisms are fully characterized.

The structure of the automorphism group is described.

Results extend understanding of the graph's symmetries.

Abstract

The subspace sum graph $\mathcal{G}(\mathbb{V})$ on a finite dimensional vector space $\mathbb{V}$ was introduced by Das [Subspace Sum Graph of a Vector Space, arXiv:1702.08245], recently. The vertex set of $\mathcal{G}(\mathbb{V})$ consists of all the nontrivial proper subspaces of $\mathbb{V}$ and two distinct vertices $W_1$ and $W_2$ are adjacent if and only if $W_1+W_2=\mathbb{V}$. In that paper, some structural indices (e.g., diameter, girth, connectivity, domination number, clique number and chromatic number) were studied, but the characterization of automorphisms of $\mathcal{G}(\mathbb{V})$ was left as one of further research topics. Motivated by this, we in this paper characterize the automorphisms of $\mathcal{G}(\mathbb{V})$ completely.

Equations

$${\sigma }_X(\alpha )={\sigma }_X(\sum _{i=1}^n{a}_i{e}_i)=\sum _{i=1}^n(\sum _{j=1}^n{x}_{ij}{a}_j)e_i,\forall \alpha =\sum _{i=1}^n{a}_i{e}_i\in \mathbb{V}.$$

$${\sigma }_f(\beta )={\sigma }_f(\sum _{i=1}^n{b}_i{e}_i)=\sum _{i=1}^n{\sigma }_f(b_i)e_i,\forall \beta =\sum _{i=1}^n{b}_i{e}_i\in \mathbb{V}.$$

$$N_r=\frac{(q^k-1)(q^k-q)\cdots (q^k-q^{r-1})(q^n-q^k)(q^n-q^{k+1})\cdots (q^n-q^{n-1})}{(q^{n-k+r}-1)(q^{n-k+r}-q)\cdots (q^{n-k+r}-q^{n-k+r-1})}.$$

$$W_2+W_3=\mathbb{V}(i.e.,W_2\sim W_3).$$

$$\sigma (W_2)=⟨{\beta }_s⟩\subset ⟨{\beta }_1,\dots ,{\beta }_{s-1}⟩.$$

$$⟨{\beta }_1,\dots ,{\beta }_{s-1}⟩\subset \sigma (W_3).$$

$$\sigma (W_2)\subset \sigma (W_3),$$

$${\sigma }^{-1}(W^{\prime })=⟨\alpha ,{\alpha }_1,\dots ,{\alpha }_k⟩=⟨{\alpha }_1,\dots ,{\alpha }_k⟩=W,$$

$$B_1=P_{1k}(I-\sum _{j=k+1}^n{a}_j{E}_{jk}).$$

$$B_2=P_{2t}(I-b_t^{-1}\sum _{i\ne t}{b}_i{E}_{it}),$$

$$W_k=⟨e_i+a_k{e}_k,e_{i+1},\dots ,e_{k-1},e_{k+1},\dots ,e_n⟩.$$

$${\sigma }_1(W_k)=⟨e_i+f_{ik}(a_k)e_k,e_{i+1},\dots ,e_{k-1},e_{k+1},\dots ,e_n⟩.$$

$$⟨e_i+\sum _{j=i+1}^n{b}_j{e}_j⟩\subset ⟨e_i+f_{ik}(a_k)e_k,e_{i+1},\dots ,e_{k-1},e_{k+1},\dots ,e_n⟩,$$

$$⟨e_1⟩\subset ⟨e_1+f_{1i}(a)e_i+f_{1j}(ab)e_j,e_i+f_{ij}(b)e_j⟩,$$

$$f(a)=\frac{f_{12}(a)}{f_{12}(1)}=\frac{f_{1j}(a)}{f_{1j}(1)}=\frac{f_{ij}(a)}{f_{ij}(1)}.$$

$$f(ab)=\frac{f_{1n}(ab)}{f_{1n}(1)}=\frac{f_{12}(a)f_{2n}(b)}{f_{12}(1)f_{2n}(1)}=f(a)f(b),$$

$$f_{13}(-a)=-f_{12}(-a)f_{23}(-1),$$

$$f(-a)=-f(a),$$

$$⟨e_1+f(a+b)f_{1n}(1)e_n⟩\subset ⟨e_1+f(-a)f_{12}(1)e_2+f(a)f_{1n}(1)e_n,e_2+f(a^{-1}b)f_{2n}(1)e_n⟩,$$

$$\begin{array}[]{rcl}f(a+b)f_{1n}(1)&=&f(a)f_{1n}(1)-f(-a)f_{12}(1)f(a^{-1}b)f_{2n}(1)\\ &=&f(a)f_{1n}(1)+f(a)f_{12}(1)f(a^{-1}b)f_{2n}(1)\\ &=&f(a)f_{1n}(1)+[f(a)f(a^{-1}b)][f_{12}(1)f_{2n}(1)]\\ &=&f(a)f_{1n}(1)+f(b)f_{1n}(1).\end{array}$$

$${\sigma }_1(⟨e_i+\sum _{j=i+1}^n{a}_j{e}_j⟩)=⟨e_i+\sum _{j=i+1}^n{f}_{ij}(a_j)e_j⟩=⟨e_i+\frac{1}{f_{1i}(1)}\sum _{j=i+1}^{n}f(a_j)f_{1j}(1)e_j⟩.$$

$${\sigma }_D\circ {\sigma }_1(⟨e_i+\sum _{j=i+1}^n{a}_j{e}_j⟩)=⟨e_i+\sum _{j=i+1}^{n}f(a_j)e_j⟩,$$

$$\sigma ={\sigma }_{B^{-1}}\circ {\sigma }_{D^{-1}}\circ {\sigma }_f={\sigma }_A\circ {\sigma }_f$$

Full text

Automorphisms of the subspace sum graphs on a vector space

Fenglei Tian,    Dein Wong

School of Mathematics, China University of Mining and Technology, Xuzhou 221116, China. Corresponding author.   E-mail address: wongdein@163.com. Supported by “the Fundamental Research Funds for the Central Universities (No. 2017BSCXB53)”.

Abstract: The subspace sum graph $\mathcal{G}(\mathbb{V})$ on a finite dimensional vector space $\mathbb{V}$ was introduced by Das [Subspace Sum Graph of a Vector Space, arXiv:1702.08245], recently. The vertex set of $\mathcal{G}(\mathbb{V})$ consists of all the nontrivial proper subspaces of $\mathbb{V}$ and two distinct vertices $W_{1}$ and $W_{2}$ are adjacent if and only if $W_{1}+W_{2}=\mathbb{V}$. In that paper, some structural indices (e.g., diameter, girth, connectivity, domination number, clique number and chromatic number) were studied, but the characterization of automorphisms of $\mathcal{G}(\mathbb{V})$ was left as one of further research topics. Motivated by this, we in this paper characterize the automorphisms of $\mathcal{G}(\mathbb{V})$ completely.

Keywords: Automorphisms; Subspace sum graphs; Vector space

AMS classification: 05C25; 05C69; 20H20

1 Introduction

Up to now, there are some abstract graphs defined on certain algebraic objects, which are proposed from the motivation that the properties of those algebraic structures can be revealed by studying the graphs associated with them. For example, the zero-divisor graphs defined on finite rings [4], commutative rings [6] or noncommutative rings [3], regular graphs or total graphs on a commutative ring [2, 5], cozero-divisor graph of a commutative ring [1] and so on. For various graphs defined on finite dimensional vector space, readers can refer to [9, 10, 11, 12] for details.

Let $\mathbb{V}$ be a finite dimensional vector space of dimension greater than one over a field $F$. Recently, the subspace sum graph $\mathcal{G}(\mathbb{V})$ on $\mathbb{V}$ was introduced by Das [8], the vertex set of which is composed of all the nontrivial proper subspaces of $\mathbb{V}$ and two vertices $W_{1},W_{2}$ are adjacent, written as $W_{1}\sim W_{2}$, if and only if $W_{1}+W_{2}=\mathbb{V}$. In [8], the author investigated the diameter, girth, connectivity, domination number, clique number, chromatic number of $\mathcal{G}(\mathbb{V})$ and the properties of $\mathcal{G}(\mathbb{V})$ with the base field $F$ being finite.

Since automorphisms of graphs can reveal the relationship among their vertices, then they can help analyze the structure of graphs and let us go further to achieve the motivation stated at the beginning. Moreover, automorphisms of graphs are also of importance in algebraic graph theory. Thus, characterizing the automorphisms of graphs also have attracted many attentions. Let $R$ be a ring and $Z(R)\setminus\{0\}$ the nonzero zero divisor set of $R$. The zero-divisor graph $\Gamma(R)$ of $R$ is defined as a graph with vertex set $Z(R)\setminus\{0\}$ and for vertices $x,y$, there is a directed edge from $x$ to $y$ if $xy=0$. Anderson and Livingston [6] proved that the automorphism group of $\Gamma(\mathbb{Z}_{n})$ ($n\geq 4$) is a direct product of some symmetric groups. Wong et al. [15] characterized the automorphisms of the zero-divisor graph, whose vertex set consists of all rank one upper triangular matrices over a finite field. Applying the main results of [15], Wang [13, 14] obtained the automorphisms of the zero-divisor graph on upper triangular matrix ring or full matrix ring with a finite base field, respectively. Inspired by these results, we can ask a natural problem: how about the automorphisms of a subspace sum graph? In addition, this is also an open problem for further research on subspace sum graphs in [8].

Hence, in present paper we focus on this problem and address it completely (see Section 3). Prior to presenting the proof of it, some lemmas are demonstrated in Section 2.

2 Preliminaries and Lemmas

Let $F_{q}$ be a finite field of order $q$ and $q=p^{m}$ with $p$ a prime integer. Throughout, we let $\mathbb{V}$ be a finite dimensional vector space with the base field $F_{q}$. The dimension of a subspace $W$ of $\mathbb{V}$ is denoted by $dim(W)$. Denote by $V$ the vertex set of $\mathcal{G}(\mathbb{V})$. From the definition of graph $\mathcal{G}(\mathbb{V})$, it is clear that if $dim(\mathbb{V})=1$, then the vertex set $V$ is empty. If $dim(\mathbb{V})=2$, one can easily see that $\mathcal{G}(\mathbb{V})$ is a complete graph (also see Theorem 3.1 of [8]). Hence, in the following, we always set $dim(\mathbb{V})=n\geq 3$. Let $e_{i}\in\mathbb{V}$ be a vector with the $i$-th component 1 and the others 0. Then $\{e_{1},e_{2},\ldots,e_{n}\}$ constructs a basis of $\mathbb{V}$ and any vector $\alpha\in\mathbb{V}$ can be uniquely expressed as $\alpha=\sum_{i=1}^{n}a_{i}e_{i}$ with $a_{i}\in F_{q}$.

First, we present two standard automorphisms of $\mathcal{G}(\mathbb{V})$ as follows.

Invertible linear transformation

Let $X=[x_{ij}]$ be an invertible matrix of order $n$ over $F_{q}$. Define a mapping $\sigma_{X}$ from $\mathbb{V}$ to itself as

[TABLE]

Clearly, $\sigma_{X}$ is an invertible linear transformation of $\mathbb{V}$, which naturally generates an automorphism of $\mathcal{G}(\mathbb{V})$, also written as $\sigma_{X}$, such that $\sigma_{X}(W)=\{\sigma_{X}(w),w\in W\}$ for $W\in V$.

Field automorphism

Let $f$ be an automorphism of the base field $F_{q}$ and $\sigma_{f}$ a mapping from $\mathbb{V}$ to itself such that

[TABLE]

Then one can easily check that the mapping on $V$ generated by $\sigma_{f}$, also denoted by $\sigma_{f}$, sending $W\in V$ to $\{\sigma_{f}(w),w\in W\}$ is an automorphism of $\mathcal{G}(\mathbb{V})$.

Next, we give some lemmas which will be used later.

Lemma 2.1**.**

(Theorem 6.1, [8])*  Let $W$ be a $k$-dimensional subspace of $\mathbb{V}$ over the base field $F_{q}$. Then the degree of $W$ in $\mathcal{G}(\mathbb{V})$, denoted by $deg(W)$, is $deg(W)=\sum_{r=0}^{k-1}N_{r}$, where*

[TABLE]

From Lemma 2.1, it is clear that $deg(W)$ only depends on the dimension $k$ of $W$, and if $W_{1},W_{2}\in V$ are of distinct dimensions $k_{1},k_{2}$, then $deg(W_{1})\neq deg(W_{2})$. Thus Lemma 2.2 follows from Lemma 2.1 immediately.

Lemma 2.2**.**

Let $\sigma$ be an automorphism of $\mathcal{G}(\mathbb{V})$. Then $\sigma$ sends each $k$-dimensional subspace of $\mathbb{V}$ to a subspace of equal dimension, where $1\leq k\leq n-1$.

Let $S$ be a subset of $\mathbb{V}$ and the subspace spanned by $S$ be $\langle S\rangle$. The $1$-dimensional subspace spanned by a nonzero vector $\alpha$ is denoted by $\langle\alpha\rangle$.

Lemma 2.3**.**

Let $\sigma$ be an automorphism of $\mathcal{G}(\mathbb{V})$ and $W$ a $k$-dimensional subspace ($2\leq k\leq n-1$). Suppose $\{\alpha_{1},\alpha_{2},\ldots,\alpha_{k}\}$ is a basis of $W$ and $\sigma(\langle\alpha_{i}\rangle)=\langle\beta_{i}\rangle$ for $1\leq i\leq k$. Then we have the following conclusions.

1. (i)

$\langle\beta_{i}\rangle\subset\sigma(W)$* for $1\leq i\leq k$*; 2. (ii)

$\beta_{1},\beta_{2},\ldots,\beta_{k}$* are linearly independent*; 3. (iii)

$\sigma(W)=\langle\beta_{1},\ldots,\beta_{k}\rangle$.

Proof.  First, we show the proof of (i). Assume on the contrary that there exists $\langle\beta_{i}\rangle$ which is not a subspace of $\sigma(W)$. From Lemma 2.2, we see $dim(\sigma(W))=k$, then let $\{\gamma_{1},\gamma_{2},\ldots,\gamma_{k}\}$ be a basis of $\sigma(W)$. From the assumption, we obtain that $\beta_{i},\gamma_{1},\gamma_{2},\ldots,\gamma_{k}$ are linearly independent. Extend $\{\beta_{i},\gamma_{1},\gamma_{2},\ldots,\gamma_{k}\}$ to a basis $\{\beta_{i},\gamma_{1},\gamma_{2},\ldots,\gamma_{k},\ldots,\gamma_{n-1}\}$ of $\mathbb{V}$. Let $W_{1}=\langle\gamma_{1},\gamma_{2},\ldots,\gamma_{k},\ldots,\gamma_{n-1}\rangle$ be a $(k-1)$-dimensional subspace, then $\sigma(W)\subset W_{1}$ and $\sigma(W)+W_{1}=W_{1}\neq\mathbb{V}$. Thus $\sigma(W)$ is not adjacent to $W_{1}$, i.e., $\sigma(W)\nsim W_{1}$. Furthermore, $\langle\beta_{i}\rangle+W_{1}=\mathbb{V}$, and thus $\langle\beta_{i}\rangle\sim W_{1}$. Since $\sigma$ is an automorphism, so is $\sigma^{-1}$. Then applying $\sigma^{-1}$ to $\sigma(W)\nsim W_{1}$ and $\langle\beta_{i}\rangle\sim W_{1}$, we get $W\nsim\sigma^{-1}(W_{1})$ and $\langle\alpha_{i}\rangle\sim\sigma^{-1}(W_{1})$, which contradicts with $\langle\alpha_{i}\rangle\subset W$. As a result, the assumption does not hold, and the conclusion follows.

Second, We will apply induction on $k$ to complete the proof of (ii). First, if $k=2$ and $\beta_{1},\beta_{2}$ are not linearly independent, then $\beta_{1}=a\beta_{2}$ with $0\neq a\in F_{q}$. Thus $\langle\beta_{1}\rangle=\langle\beta_{2}\rangle$, that is, $\sigma(\langle\alpha_{1}\rangle)=\sigma(\langle\alpha_{2}\rangle)$, a contradiction.

Now suppose the conclusion holds for any $k=s-1\geq 2$ and we will show the case when $k=s$. Let $\alpha_{1},\alpha_{2},\ldots,\alpha_{s}$ ($s>2$) are linearly independent vectors and $\sigma(\langle\alpha_{i}\rangle)=\langle\beta_{i}\rangle$ for $1\leq i\leq s$. Extend $\{\alpha_{1},\alpha_{2},\ldots,\alpha_{s}\}$ to a basis $\{\alpha_{1},\alpha_{2},\ldots,\alpha_{s},\alpha_{s+1},\ldots,\alpha_{n}\}$ of $\mathbb{V}$. Let $W_{2}=\langle\alpha_{s}\rangle$ and $W_{3}=\langle\alpha_{1},\ldots,\alpha_{s-1},\alpha_{s+1},\ldots,\alpha_{n}\rangle$, then

[TABLE]

From the induction hypothesis, we know that $\beta_{1},\beta_{2},\ldots,\beta_{s-1}$ are linearly independent. If $\beta_{1},\ldots,\beta_{s-1},\beta_{s}$ are not linearly independent, then $\beta_{s}$ can be linearly expressed by $\beta_{1},\ldots,\beta_{s-1}$, that is, $\beta_{s}=\sum_{i=1}^{s-1}a_{i}\beta_{i}$ with $a_{i}\in F_{q}$ and $\sum_{i}|a_{i}|\neq 0$. Then

[TABLE]

In addition, by the result of (i), $\langle\beta_{i}\rangle\subset\sigma(W_{3})$ for $1\leq i\leq s-1$. Hence

[TABLE]

Consequently, we derive that

[TABLE]

which implies that $W_{2}\nsim W_{3}$, a contradiction. So, we say $\beta_{1},\ldots,\beta_{s-1},\beta_{s}$ are linearly independent.

At last, we prove the result of (iii). Combining (i) and (ii), we see that $\langle\beta_{1},\ldots,\beta_{k}\rangle\subset\sigma(W)$ and $dim(\langle\beta_{1},\ldots,\beta_{k}\rangle)=k=dim(W)$, which complete the proof. $\square$

The following lemma is a more general result of Lemma 2.3 (i).

Lemma 2.4**.**

Let $\sigma$ be an automorphism of $\mathcal{G}(\mathbb{V})$ and $W=\langle\alpha_{1},\alpha_{2},\ldots,\alpha_{k}\rangle$ ($2\leq k\leq n-1$) a $k$-dimensional subspace. Suppose $\alpha$ is a nonzero vector, we obtain

1. (i)

if $\langle\alpha\rangle\subset W$, then $\sigma(\langle\alpha\rangle)\subset\sigma(W)$; 2. (ii)

if $\langle\alpha\rangle$ is not a subspace of $W$, then $\sigma(\langle\alpha\rangle)$ is not a subspace of $\sigma(W)$.

Proof.  Here we mainly give the proof of (i) and the other can be proved similarly. Since $\langle\alpha\rangle\subset W$, then $\alpha$ can be linearly expressed by $\alpha_{1},\alpha_{2},\ldots,\alpha_{k}$. Let $\sigma(\langle\alpha\rangle)=\langle\beta\rangle$ and $\sigma(\langle\alpha_{i}\rangle)=\langle\beta_{i}\rangle$ for $1\leq i\leq k$. From Lemma 2.3 (iii), $\sigma(W)=\langle\beta_{1},\ldots,\beta_{k}\rangle$. If $\sigma(\langle\alpha\rangle)$ is not a subspace of $\sigma(W)$, i.e., $\langle\beta\rangle$ is not a subspace of $\langle\beta_{1},\ldots,\beta_{k}\rangle$, then $\beta,\beta_{1},\ldots,\beta_{k}$ are linearly independent. Let $W^{\prime}=\langle\beta,\beta_{1},\ldots,\beta_{k}\rangle$ be a $(k+1)$-dimensional subspace. As $\sigma^{-1}$ is also an automorphism of $\mathcal{G}(\mathbb{V})$, applying $\sigma^{-1}$ to $W^{\prime}$ together with Lemma 2.3, we obtain that

[TABLE]

which contradicts with Lemma 2.2. $\square$

By $I$ and $E_{ij}$, we respectively denote the identity matrix of order $n$ and the matrix unit of order $n$ with the $(i,j)$-entry 1 and others 0. We use $P_{kt}$ to denote the permutation matrix obtained from $I$ by permuting the $k$-th row and the $t$-th row (particularly, $P_{kk}=I$). For an 1-dimensional subspace $\langle\alpha\rangle$, since $\langle\alpha\rangle=\langle a\alpha\rangle$ with $0\neq a\in F_{q}$, the expression can be not unique, but if we let the first nonzero component of $\alpha$ be 1, then the expression become unique. For convenience, in the following, we always suppose that each 1-dimensional subspace is of this form.

Lemma 2.5**.**

Let $\sigma$ be an automorphism of $\mathcal{G}(\mathbb{V})$. Then there exists an invertible matrix $B$ such that the automorphism $\sigma_{B}\circ\sigma$ fixes each $\langle e_{i}\rangle$ ($1\leq i\leq n$).

Proof.  Let $\sigma(\langle e_{1}\rangle)=\langle\alpha\rangle$ and $\alpha=\sum_{i=k}^{n}a_{i}e_{i}$ with $a_{k}=1$ and $1\leq k\leq n$. Then we take

[TABLE]

It is obvious that $\sigma_{B_{1}}\circ\sigma(\langle e_{1}\rangle)=\langle e_{1}\rangle$. Set $\sigma_{1}=\sigma_{B_{1}}\circ\sigma$. Further, suppose $\sigma_{1}(\langle e_{2}\rangle)=\langle\beta\rangle$. Since $e_{1}$ and $e_{2}$ are linearly independent, so are $e_{1}$ and $\beta$ from Lemma 2.3 (ii). Then we can write $\beta=\sum_{i=1}^{n}b_{i}e_{i}$ and $\sum_{i=2}^{n}|b_{i}|\neq 0$. If $b_{t}\neq 0$ ($t\geq 2$), then we take

[TABLE]

and one can easily check that $\sigma_{B_{2}}\circ\sigma_{1}$ fixes $\langle e_{2}\rangle$. Set $\sigma_{2}=\sigma_{B_{2}}\circ\sigma_{1}$. Note that $\sigma_{2}$ also fixes $\langle e_{1}\rangle$.

Let $W=\langle e_{1},e_{2}\rangle$ and $\sigma_{2}(\langle e_{3}\rangle)=\langle\gamma\rangle$. It follows from Lemma 2.3 that $\sigma_{2}(W)=W$. Since $e_{3}\notin W$, by Lemma 2.4 (ii), we have $\langle\gamma\rangle$ is not a subspace of $W$ (i.e., $\gamma\notin W$). Thus we can write $\gamma=\sum_{i=1}^{n}c_{i}e_{i}$ and $\sum_{i=3}^{n}|c_{i}|\neq 0$. Similarly, we can take an invertible matrix $B_{3}$ such that $\sigma_{B_{3}}\circ\sigma_{2}$ fixes $\langle e_{3}\rangle$ and $\langle e_{1}\rangle$, $\langle e_{2}\rangle$. Proceeding in this method, we can find a matrix sequence $B_{1},B_{2},\ldots,B_{n}$ such that $\sigma_{B_{n}}\circ\sigma_{B_{n-1}}\circ\cdots\circ\sigma_{B_{1}}\circ\sigma$ fixes every $\langle e_{i}\rangle$ ($1\leq i\leq n$). Let $B=B_{n}B_{n-1}\cdots B_{1}$, then $\sigma_{B}\circ\sigma$ is what we want. $\square$

3 Automorphisms of the subspace sum graph $\mathcal{G}(\mathbb{V})$

In this section, we characterize the automorphisms of the subspace sum graph $\mathcal{G}(\mathbb{V})$. Let $\mathbb{V}$ be a finite dimensional vector space over a finite base field $F_{q}$ with $dim(\mathbb{V})\geq 3$.

Theorem 3.1**.**

A mapping $\sigma$ on the vertex set $V$ of $\mathcal{G}(\mathbb{V})$ is an automorphism of $\mathcal{G}(\mathbb{V})$ if and only if $\sigma$ can be uniquely decomposed as $\sigma=\sigma_{A}\circ\sigma_{f}$, where $\sigma_{A}$ and $\sigma_{f}$ are two standard automorphisms defined as above.

Proof.  The proof for the sufficiency part is clear. In what follows, we present the proof for the necessity part. From Lemma 2.5, we can find an invertible matrix $B$ such that $\sigma_{B}\circ\sigma$ fixes every $\langle e_{i}\rangle$ ($1\leq i\leq n$). Let $\sigma_{1}=\sigma_{B}\circ\sigma$. The remaining proof is divided into the following claims.

Claim 1.  For a nonzero vector $\alpha=\sum_{i=1}^{n}a_{i}e_{i}$ with $a_{i}\in F_{q}$, assume $\sigma_{1}(\langle\alpha\rangle)=\langle\beta\rangle$, where $\beta=\sum_{i=1}^{n}b_{i}e_{i}$ with $b_{i}\in F_{q}$. Then $a_{i}=0$ if and only if $b_{i}=0$ ($1\leq i\leq n$).

Let $U_{j}=\langle e_{1},\ldots,e_{j-1},e_{j+1},\ldots,e_{n}\rangle$ for $1\leq j\leq n$. Since $\sigma_{1}$ fixes every $\langle e_{i}\rangle$ for $1\leq i\leq n$, then from Lemma 2.3 (iii), $\sigma_{1}(U_{j})=U_{j}$ for $1\leq j\leq n$. If $a_{j}=0$, then $\langle\alpha\rangle\subset U_{j}$. Further, from Lemma 2.4 (i), $\langle\beta\rangle\subset U_{j}$, which implies that $b_{j}=0$. By applying $\sigma_{1}^{-1}$ to the above, we obtain $a_{j}=0$ whenever $b_{j}=0$.

It follows from Claim 1 that $\sigma_{1}(\langle e_{i}+ae_{j}\rangle)=\langle e_{i}+be_{j}\rangle$ with $a,b\in F_{q}$ and $1\leq i<j\leq n$. Thus we define a function $f_{ij}$ over $F_{q}$ such that $\sigma_{1}(\langle e_{i}+ae_{j}\rangle)=\langle e_{i}+f_{ij}(a)e_{j}\rangle$ and $f_{ij}(x)=0$ if and only if $x=0$. Now we investigate the properties of $f_{ij}$.

Claim 2.  $\sigma_{1}(\langle e_{i}+\sum_{j=i+1}^{n}a_{j}e_{j}\rangle)=\langle e_{i}+\sum_{j=i+1}^{n}f_{ij}(a_{j})e_{j}\rangle$* for $1\leq i<j\leq n$.*

From Claim 1, suppose $\sigma_{1}(\langle e_{i}+\sum_{j=i+1}^{n}a_{j}e_{j}\rangle)=\langle e_{i}+\sum_{j=i+1}^{n}b_{j}e_{j}\rangle$. For $i<k\leq n$, let

[TABLE]

Then from Lemmas 2.3 and 2.5,

[TABLE]

Since $\langle e_{i}+\sum_{j=i+1}^{n}a_{j}e_{j}\rangle\subset W_{k}$, then after applying $\sigma_{1}$, we derive from Lemma 2.4(i) that

[TABLE]

which indicates that $b_{k}=f_{ik}(a_{k})$. Notice that $i<k\leq n$, then the conclusion follows.

Claim 3.  Suppose $2\leq i<j\leq n$ and $\forall a,b\in F_{q}$, then

1. (i)

$f_{1j}(ab)=f_{1i}(a)f_{ij}(b)$; 2. (ii)

$f_{1j}(a)=f_{1i}(a)f_{ij}(1)=f_{1i}(1)f_{ij}(a)$; 3. (iii)

$\frac{f_{1j}(a)}{f_{1j}(1)}=\frac{f_{1i}(a)}{f_{1i}(1)}=\frac{f_{ij}(a)}{f_{ij}(1)}$.

Applying $\sigma_{1}$ on $\langle e_{1}\rangle\subset\langle e_{1}+ae_{i}+abe_{j},e_{i}+be_{j}\rangle$, together with Lemmas 2.4 and 2.5 and Claim 2, we obtain

[TABLE]

which points out that $f_{1j}(ab)=f_{1i}(a)f_{ij}(b)$. Taking respectively $a=1$, $b=1$ and $a=b=1$ in (i), one can easily deduce the results of (ii) and (iii).

Claim 4.  Let $f=\frac{f_{12}}{f_{12}(1)}$, then for $\forall a,b\in F_{q}$ and $0\neq c\in F_{q}$,

1. (i)

$f(ab)=f(a)f(b)$; 2. (ii)

$f(1)=1$* and $f(c^{-1})=f(c)^{-1}$*; 3. (iii)

$f(-1)=-1$* and $f(-a)=-f(a)$*; 4. (iv)

$f(a+b)=f(a)+f(b)$; 5. (v)

$f$* is an automorphism of $F_{q}$*.

From Claim 3 (iii), we see for $2\leq i<j\leq n$ that

[TABLE]

Then

[TABLE]

which proves (i).

It follows from the definition of $f$ that $f(1)=1$. Then $1=f(1)=f(cc^{-1})=f(c)f(c^{-1})$, and thus $f(c^{-1})=f(c)^{-1}$, which establishes (ii).

Since $\langle e_{1}-ae_{3}\rangle\subset\langle e_{1}-ae_{2},e_{2}-e_{3}\rangle$, after applying $\sigma_{1}$ together with Lemma 2.4 and Claim 2, it follows that $\langle e_{1}+f_{13}(-a)e_{3}\rangle\subset\langle e_{1}+f_{12}(-a)e_{2},e_{2}+f_{23}(-1)e_{3}\rangle$. Thus we obtain

[TABLE]

which, together with Claim 3 (i), imply that $f_{13}(-a)=-f_{13}(a)$. Further, we obtain

[TABLE]

and $f(-1)=-f(1)=-1$ if $a=1$. The proof of (iii) is finished.

Note that $f(0)=0$, then if $a=0$ in (iv), the result is clear. Now suppose $a\neq 0$. Applying $\sigma_{1}$ on $\langle e_{1}+(a+b)e_{n}\rangle\subset\langle e_{1}-ae_{2}+ae_{n},e_{2}+a^{-1}be_{n}\rangle$, together with Lemma 2.4 and Claim 2, we deduce

[TABLE]

which indicates that

[TABLE]

As a consequence, $f(a+b)=f(a)+f(b)$ holds.

Combining (i)-(iv), (v) follows immediately.

Claim 5. There exists a diagonal matrix $D$ such that $\sigma_{f}^{-1}\circ\sigma_{D}\circ\sigma_{1}$ fixes each 1-dimensional subspace.

From Claims 3 and 4, $f_{ij}(a_{j})=f(a_{j})f_{ij}(1)=f(a_{j})\frac{f_{1j}(1)}{f_{1i}(1)}$. Then the result of Claim 2 can be rewritten as

[TABLE]

Let $f_{11}(1)=1$ and $D=diag(1,f_{12}^{-1}(1),\ldots,f_{1n}^{-1}(1))$, then

[TABLE]

and further it is clear that $\sigma_{f}^{-1}\circ\sigma_{D}\circ\sigma_{1}$ fixes each 1-dimensional subspace. Set $\sigma_{2}=\sigma_{f}^{-1}\circ\sigma_{D}\circ\sigma_{1}$.

Now we are in a position to complete the proof.

Let $W\in V$ be any $k$-dimensional nontrivial proper subspace of $\mathbb{V}$ and $\{w_{1},w_{2},\ldots,w_{k}\}$ a basis of $W$ with $k\geq 2$. From Claim 5, $\sigma_{2}$ fixes every 1-dimensional subspace, then $\sigma_{2}$ fixes $\langle w_{i}\rangle$ ($1\leq i\leq k$). Using Lemma 2.3, we obtain $\sigma_{2}$ fixes $W$, i.e., $\sigma_{2}$ fixes every vertex of $\mathcal{G}(\mathbb{V})$. As a result, $\sigma_{2}=\sigma_{f}^{-1}\circ\sigma_{D}\circ\sigma_{B}\circ\sigma$ is the identity mapping. Thus

[TABLE]

with $A=B^{-1}D^{-1}$. Next, we show the decomposition of $\sigma$ is unique. Suppose there are two distinct decompositions, $\sigma_{A_{1}}\circ\sigma_{f_{1}}=\sigma_{A_{2}}\circ\sigma_{f_{2}}$, which implies that $\sigma_{A_{2}^{-1}A_{1}}=\sigma_{f_{2}f_{1}^{-1}}$. Since $\sigma_{f_{2}f_{1}^{-1}}$ fixes each $\langle e_{i}\rangle$, so does $\sigma_{A_{2}^{-1}A_{1}}$. Then we see $A_{2}^{-1}A_{1}$ is a diagonal matrix. Similarly, $\sigma_{A_{2}^{-1}A_{1}}$ also fixes every $\langle e_{i}+e_{j}\rangle$ for $1\leq i<j\leq n$, from which it follows that $A_{2}^{-1}A_{1}$ is a scalar matrix. So, $A_{2}$ is a nonzero scalar multiple of $A_{1}$, and thus $\sigma_{A_{1}}=\sigma_{A_{2}}$. As a consequence, $\sigma_{f_{1}}=\sigma_{f_{2}}$.

The proof of Theorem 3.1 is completed. $\square$

Note that the automorphism group of $F_{q}$ with $q=p^{m}$ is a cyclic group of order $m$. Then the following result follows from Theorem 3.1 immediately.

Corollary 3.2**.**

The automorphism group of $\mathcal{G}(\mathbb{V})$ is isomorphic to $PGL_{n}(F_{q})\times\mathbb{Z}_{m}$, where $PGL_{n}(F_{q})$ is the quotient group of all the $n\times n$ invertible matrices over $F_{q}$ to the normal subgroup of all the nonzero scalar matrices over $F_{q}$.

Acknowledgement

The authors thank the anonymous referees for a careful and thorough reading of the paper and for valuable comments, and also thank the support of the Fundamental Research Funds for the Central Universities (No. 2017BSCXB53).