Solution vector space property

Solution.

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

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

u+v=u+w⟹−u+(u+v)=−u+(u+w)⟹(a2)(−u+u)+v=(−u+u)+w⟹(a1)(u+(−u))+v=(u+(−u))+w⟹(a4)0+v=0+w⟹(a3)v=w.$$\begin{array}{rl}u+v=u+w& ⟹-u+(u+v)=-u+(u+w)\\ & \stackrel{(a2)}{⟹}(-u+u)+v=(-u+u)+w\\ & \stackrel{(a1)}{⟹}(u+(-u))+v=(u+(-u))+w\\ & \stackrel{(a4)}{⟹}\mathbf{0}+v=\mathbf{0}+w\\ & \stackrel{(a3)}{⟹}v=w.\end{array}$$

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

Now suppose that we have v+u=w+u$v+u=w+u$. Then by (a1), we see that u+v=u+w$u+v=u+w$. Now, it follows from (a) that v=w$v=w$.
(Alternatively, you may prove this just like part (a).)

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

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

0′+v=v=0+v,$${\mathbf{0}}^{\prime }+v=v=0+v,$$

where v$v$ is any fixed vector (for example v=0$v=\mathbf{0}$ is enough).
Now by the cancellation law (see (b)), we obtain 0′=0${\mathbf{0}}^{\prime }=\mathbf{0}$.
Thus, there is only one zero vector 0$\mathbf{0}$.

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

Since −v$-v$ is the additive inverse of v∈V$v\in V$, we have v+(−v)=0$v+(-v)=\mathbf{0}$. (This is just (a4).)
Now, suppose that we have a vector w∈V$w\in V$ satisfying v+w=0$v+w=\mathbf{0}$. So, w$w$ is another element satisfying axiom (a4).
Then we have

v+(−v)=0=v+w.$$v+(-v)=0=v+w.$$

By the cancellation law (see (a)), we have −v=w$-v=w$. Thus, the additive inverse is unique.

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

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

0v=(0+0)v=(m3)0v+0v.$$\begin{array}{rl}0v& =(0+0)v\stackrel{(m3)}{=}0v+0v.\end{array}$$

We also have

0v=(a3)0+0v.$$0v\stackrel{(a3)}{=}0+0v.$$

Hence, combining these, we see that

0v+0v=0+0v,$$0v+0v=0+0v,$$

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

(f) a0=0$a\mathbf{0}=\mathbf{0}$ for every scalar a$a$.

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

a0=a(0+0)=(m2)a0+a0.$$\begin{array}{r}a\mathbf{0}=a(\mathbf{0}+\mathbf{0})\stackrel{(m2)}{=}a\mathbf{0}+a\mathbf{0}.\end{array}$$

We also have

a0=0+a0$$a0=0+a0$$

by (a3). Combining these, we have

a0+a0=0+a0,$$a0+a0=0+a0,$$

and the cancellation law yields a0=0$a\mathbf{0}=\mathbf{0}$.

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

For this problem, we use a little bit logic. Our assumption is av=0$av=\mathbf{0}$. From this assumption, we need to deduce that either a=0$a=0$ or v=0$v=\mathbf{0}$.
Note that if a=0$a=0$, then we are done as this is one of the consequence we want. So, let us assume that a≠0$a\ne 0$. Then we want to prove v=0$v=0$.
Since a$a$ is a nonzero scalar, we have a−1$a^{-1}$. Then we have

a−1(av)=a−10.$$\begin{array}{r}{a}^{-1}(av)=a^{-1}\mathbf{0}.\end{array}$$

The right hand side a−10$a^{-1}\mathbf{0}$ is 0$\mathbf{0}$ by part (f).
On the other hand, the left hand side can be computed as follows:

a−1(av)=(m1)(a−1a)v=1v=(m4)v.$$\begin{array}{r}{a}^{-1}(av)\stackrel{(m1)}{=}(a^{-1}a)v=1v\stackrel{(m4)}{=}v.\end{array}$$

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

(h) (−1)v=−v$(-1)v=-v$.

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

v+(−1)v=(m4)1v+(−1)v=(m3)(1+(−1))v=0v=(e)0.$$\begin{array}{r}v+(-1)v\stackrel{(m4)}{=}1v+(-1)v\stackrel{(m3)}{=}(1+(-1))v=0v\stackrel{(e)}{=}\mathbf{0}.\end{array}$$

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