## Basic Properties of Vector Spaces

**Definition**. Let $\mathbb{F}$ be a field. The collection of vectors, as $n\times 1$ matrices over $\mathbb{F}$, is called a **vector space** over the field of scalars $\mathbb{F}.$ We denote a vector space by $\mathbb{V}$.

Since vectors are just $n\times 1$ matrices, the proof of the next two theorems follow immediately.

**Theorem**. Let $\mathbb{V}$ be a vector space.

(1) For all ${u}, {v}\in V$, ${u}+{v}={v}+{u}.$

(2) For all ${u}, {v}, {w} \in V$, $({u}+{v})+{w}={u}+({v}+{w}).$

(3) For all ${v}\in V$, there exists ${0} \in V$ such that ${v}+{0}={v}.$

(4) For every ${v}\in V$, there exists ${w}\in V$ such that ${v}+{w}={0}.$

**Theorem**. Let $\mathbb{V}$ be a vector space over a field $\mathbb{F}$.

(1) For all ${v}\in V$, $1{v}={v}.$

(2) For all $a, b\in \mathbb{F}$ and ${u}\in V$, $(a b) {v}=a (b {v}).$

(3) For all $a \in \mathbb{F}, u, v\in V$, $a ({u}+{v})=a {u}+ a{v}.$

(4) For all $a, b \in\mathbb{F}$ and ${u}\in V$, $(a+b){u}=a {u}+ b {u}.$

**Theorem**. Let $\mathbb{V}$ be a vector space. There exists a unique additive identity (denoted by ${0}$). Every ${v}\in V$ has a unique additive inverse, (denoted by $-{v}$).

**Proof**. Let ${u}_1$ and ${u}_2$ be additive identities in $\mathbb{V}$, then ${v}+{u}_1={v}$ and ${v}+{u}_2={v}$ for every ${v}\in V.$ Then, ${u}_1={u}_1+{u}_2={u}_2+{u}_1={u}_2$ as desired. Let ${v}_1$ and ${v}_2$ be additive inverses of ${w}$ in $\mathbb{V}$, then ${w}+{v}_1={0}$ and ${w}+{v}_2={0}.$ Then, \begin{align} {v}_1 & ={v}_1+{0} ={v}_1+({w}+{v}_2)\ & =({v}_1+{w})+{v}_2 =({w}+{v}_1)+{v}_2 ={0}+{v}_2={v}_2 \end{align} as desire

**Theorem**. Let $\mathbb{V}$ be a vector space. If ${v}\in V$, then $0\, {v}={0}.$ If $a\in \mathbb{F}$, then $a\, {0}={0}.$

**Proof**. Let ${v}\in V$, then $${v}=1 {v}=(1+0) {v}= 1 {v}+0 {v}= {v}+0{v} $$ which shows that $0 {v}$ is the additive identity of $\mathbb{V}$, namely $0 {v}={0}.$ Let $a\in \mathbb{F}$, then $$ a {0} =a({0}+{0}) =a{0}+a{0} $$ which shows that $a {0}$ is the additive identity of $\mathbb{V}$, namely $a {0}={0}.$

**Theorem**. Let $\mathbb{V}$ be a vector space. If ${v}\in V$, then $-(-{v})={v}.$ If ${v}\in V$, then $(-1)\, {v}=-{v}.$

**Proof**. Let ${v}\in V$, then $$ {v}+(-1){v} =1 {v}+(-1) {v} =(1+(-1)) {v} =0 {v} ={0}$$ which shows that $(-1){v}$ is the unique additive inverse of ${v}$ namely, $$ (-1){v}=-{v}. $$ Since $-{v}$ is the unique additive inverse of ${v}$, ${v}+(-{v})={0}.$ Then $$ (-{v})+{v}={0} $$ shows that ${v}$ is the unique additive inverse of $-{v}$, namely, ${v}=-(-{v})$ as desired.

**Theorem**. Let $\mathbb{V}$ be a vector space with $a\in \mathbb{F}$ and ${v}\in V.$ If $a\,{v}={0}$, then $a=0$ or ${v}={0}.$

**Proof**. Suppose $a\neq 0.$ If $a v ={0}$ then $$ {v}=1 {v} =(a^{-1} a) {v} =a^{-1} (a {v}) =a^{-1} {0} ={0}. $$ Otherwise $a=0$ as desired.