How might I prove the following simple facts: $$ 0v= \vec 0\\ (−1)v=−v\\ −(v+w)=(−v)+(−w)\\ a\vec 0=\vec0 $$ where $v$ and $w$ are vectors in vector space $V$ and $a$ is a scalar.
Taken from Week 1's notes here.
I'm having difficulty with deriving these from the 8 axioms:
- $v+w=w+v$
- $u+(v+w)=(u+v)+w$
- $\vec0+v=v$
- $-v+v=\vec 0$
- $1v=v$
- $a(bv)=(ab)v$
- $a(v+w)=av+aw$
- $(a+b)v=av+bv$
I guess my problem is, because the proofs can at times seem trivial (as with the proof of the vector cancellation law given in the notes, simply by adding the additive inverse of $u$ to both sides of $u+v=u+w$) I'm not sure whether I've done them correctly or not.
So for the first of these facts, $0v=\vec 0$, I rewrite $\vec 0$ as $(-v + v)$ using axiom III, open the brackets and end up with $v^2=v^2$. However, I'm not sure whether I've proved the algebaic fact, whether there's more work to be done, or whether my approach is completely wrong.
Note, the final fact is discussed here.
Any hints would be appreciated!
Note that it is a general property of $0$ (and $\vec 0$, see axiom 3) that $0+0 = 0$. Applying that to $0v$, we get $$\begin{align} 0v = (0+0)v &= 0v+0v &\text{(axiom 8)}\\ 0v+ (-0v) &= 0v+0v+ (-0v) & \text{(adding $-(0v)$)}\\ \vec 0 &= 0v+\vec 0&\text{(4)}\\ \vec0 &= 0v&\text{(3)} \end{align}$$ For the second property, note that $1 + (-1) = 0$. Thus $$ \begin{align} \vec 0 = 0v = (1+(-1))v &= 1v + (-1)v = v + (-1)v&\text{(above, 8, 5)}\\ \vec 0 + (-v) &= v + (-1)v + (-v)&\text{(adding $-v)$}\\ -v &= \vec 0 + (-1)v&\text{(3, 4)}\\ -v &= (-1)v&\text{(3)} \end{align} $$ Third property uses the second: $$\begin{align} -(u+v) &= (-1)(u+v) &\text{(above)}\\ &= (-1)u+(-1)v&\text{(7)}\\ &= -u-v&\text{(above)} \end{align} $$ And finally, the fourth property, we have $$\begin{align} a\vec 0 = a(\vec 0 + \vec 0) &= a\vec 0 + a \vec 0&\text{(3, 7)}\\ a\vec 0 + (-a\vec 0) &= a\vec 0 + a\vec0 + (-a\vec0)&\text{(adding $-(a\vec0)$)}\\ \vec 0 &= \vec 0 + a\vec 0&\text{(4)}\\ \vec 0 &= a\vec 0&\text{(3)} \end{align}$$