# Prove Vector Space Properties Using Vector Space Axioms

## Problem 711

Using the axiom of a vector space, prove the following properties.

Let $V$ be a vector space over $\R$. Let $u, v, w\in V$.

**(a)** If $u+v=u+w$, then $v=w$.

**(b)** If $v+u=w+u$, then $v=w$.

**(c)** The zero vector $\mathbf{0}$ is unique.

**(d)** For each $v\in V$, the additive inverse $-v$ is unique.

**(e)** $0v=\mathbf{0}$ for every $v\in V$, where $0\in\R$ is the zero scalar.

**(f)** $a\mathbf{0}=\mathbf{0}$ for every scalar $a$.

**(g)** If $av=\mathbf{0}$, then $a=0$ or $v=\mathbf{0}$.

**(h)** $(-1)v=-v$.

The first two properties are called the **cancellation law**.

Contents

- Problem 711
- The Axioms of a Vector Space
- Solution.
- (a) If $u+v=u+w$, then $v=w$.
- (b) If $v+u=w+u$, then $v=w$.
- (c) The zero vector $\mathbf{0}$ is unique.
- (d) For each $v\in V$, the additive inverse $-v$ is unique.
- (e) $0v=\mathbf{0}$ for every $v\in V$, where $0\in\R$ is the zero scalar.
- (f) $a\mathbf{0}=\mathbf{0}$ for every scalar $a$.
- (g) If $av=\mathbf{0}$, then $a=0$ or $v=\mathbf{0}$.
- (h) $(-1)v=-v$.

## The Axioms of a Vector Space

Recall the axioms of a vector space:

A set $V$ is said to be a **vector space** over $\R$ if

(1) an addition operation “$+$” is defined between any two elements of $V$, and

(2) a scalar multiplication operation is defined between any element of $K$ and any element in $V$.

Moreover, the following properties must hold for all $u,v,w\in V$ and $a,b\in \R$:

**Closure Properties**

(c1) $u+v\in V$.

(c2) $av\in V$.

**Properties of Addition**

(a1) $u+v=v+u$.

(a2) $u+(v+w)=(u+v)+w$.

(a3) There is an element $\mathbf{0}\in V$ such that $\mathbf{0}+v=v$ for all $v\in V$.

(a4) Given an element $v\in V$, there is an element $-v\in V$ such that $v+(-v)=\mathbf{0}$.

**Properties of Scalar Multiplication**

(m1) $a(bv)=(ab)v$.

(m2) $a(u+v)=au+av$.

(m3) $(a+b)v=av+bv$.

(m4) $1v=v$ for all $v\in V$.

The element $\mathbf{0} \in V$ is called the **zero vector**, and for any $v\in V$, the element $-v\in V$ is called the **additive inverse** of $v$.

## Solution.

### (a) If $u+v=u+w$, then $v=w$.

We know by $(a4)$ that there is an additive inverse $-u\in V$. Then

\begin{align*}

u+v=u+w

&\implies

-u+(u+v)=-u+(u+w)

\\

&\stackrel{(a2)}{\implies}

(-u+u)+v=(-u+u)+w

\\

&\stackrel{(a1)}{\implies}

(u+(-u))+v=(u+(-u))+w

\\

&\stackrel{(a4)}{\implies}

\mathbf{0}+v=\mathbf{0}+w

\\

&\stackrel{(a3)}{\implies}

v=w.

\end{align*}

### (b) If $v+u=w+u$, then $v=w$.

Now suppose that we have $v+u=w+u$. Then by (a1), we see that $u+v=u+w$. Now, it follows from (a) that $v=w$.

(Alternatively, you may prove this just like part (a).)

### (c) The zero vector $\mathbf{0}$ is unique.

Suppose that $\mathbf{0}’$ is another zero vector satisfying axiom (a3). That is, we have $\mathbf{0}’+v=v$ for any $v \in V$. Since $\mathbf{0}$ is also satisfy $\mathbf{0}+v=v$, we have

\[\mathbf{0}’+v=v=\mathbf{0}+v,\]
where $v$ is any fixed vector (for example $v=\mathbf{0}$ is enough).

Now by the cancellation law (see (b)), we obtain $\mathbf{0}’=\mathbf{0}$.

Thus, there is only one zero vector $\mathbf{0}$.

### (d) For each $v\in V$, the additive inverse $-v$ is unique.

Since $-v$ is the additive inverse of $v\in V$, we have $v+(-v)=\mathbf{0}$. (This is just (a4).)

Now, suppose that we have a vector $w \in V$ satisfying $v+w=\mathbf{0}$. So, $w$ is another element satisfying axiom (a4).

Then we have

\[v+(-v)=0=v+w.\]
By the cancellation law (see (a)), we have $-v=w$. Thus, the additive inverse is unique.

### (e) $0v=\mathbf{0}$ for every $v\in V$, where $0\in\R$ is the zero scalar.

Note that $0$ is a real number and $\mathbf{0}$ is the zero vector in $V$. For $v\in V$, we have

\begin{align*}

0v&=(0+0)v\stackrel{(m3)}{=} 0v+0v.

\end{align*}

We also have

\[0v\stackrel{(a3)}{=} \mathbf{0}+0v.\]
Hence, combining these, we see that

\[0v+0v=\mathbf{0}+0v,\]
and by the cancellation law, we obtain $0v=\mathbf{0}$.

### (f) $a\mathbf{0}=\mathbf{0}$ for every scalar $a$.

Note that we have $\mathbf{0}+\mathbf{0}=\mathbf{0}$ by (a3).

Thus, we have

\begin{align*}

a\mathbf{0}=a(\mathbf{0}+\mathbf{0})\stackrel{(m2)}{=}a\mathbf{0}+a\mathbf{0}.

\end{align*}

We also have

\[a\mathbf{0}=\mathbf{0}+a\mathbf{0}\]
by (a3). Combining these, we have

\[a\mathbf{0}+a\mathbf{0}=\mathbf{0}+a\mathbf{0},\]
and the cancellation law yields $a\mathbf{0}=\mathbf{0}$.

### (g) If $av=\mathbf{0}$, then $a=0$ or $v=\mathbf{0}$.

For this problem, we use a little bit logic. Our assumption is $av=\mathbf{0}$. From this assumption, we need to deduce that either $a=0$ or $v=\mathbf{0}$.

Note that if $a=0$, then we are done as this is one of the consequence we want. So, let us assume that $a\neq 0$. Then we want to prove $v=0$.

Since $a$ is a nonzero scalar, we have $a^{-1}$. Then we have

\begin{align*}

a^{-1}(av)=a^{-1}\mathbf{0}.

\end{align*}

The right hand side $a^{-1}\mathbf{0}$ is $\mathbf{0}$ by part (f).

On the other hand, the left hand side can be computed as follows:

\begin{align*}

a^{-1}(av) \stackrel{(m1)}{=}(a^{-1}a)v =1v \stackrel{(m4)}{=} v.

\end{align*}

Therefore, we have $v=0$.

Thus, we conclude that if $av=\mathbf{0}$, then either $a=0$ or $v=\mathbf{0}$.

### (h) $(-1)v=-v$.

Note that $(-1)v$ is the scalar product of $-1$ and $v$. On the other hand, $-v$ is the additive inverse of $v$, which is guaranteed to exist by (a4).

We show that $(-1)v$ is also the additive inverse of $v$:

\begin{align*}

v+(-1)v\stackrel{(m4)}{=}1v+(-1)v \stackrel{(m3)}{=}(1+(-1))v=0v \stackrel{(e)}{=} \mathbf{0}.

\end{align*}

So $(-1)v$ is the additive inverse of $v$. Since by part (d), we know that the additive inverse is unique, it follows that $(-1)v=-v$.

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