Vector Spaces and Basic Theorems

Do you know what a field is? In this article, I go over some basic properties of a vector space over a field. These theorems are good exercises for someone, just learning how to write proofs.

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.

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.