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.

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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