Determine which of the axioms holds for $\lambda v$ = $-v$

Question

Let V be a vector space over field F such that V contains two distinct vectors (so V $\neq \{0\}$). Define a new structure on V by keeping the same addition, but now define a new 'scalar multiplication' on V by $\lambda v$ = $-v$, for all $v \in $V and all $\lambda \in$ F (where $-v$ denotes the negative of $v$). Determine which of the eight vector space axioms holds for this new structure and which don't. For those that do not, explain why not.

ok this is the problem i'm working with and i'm struggling with understanding vector spaces and proofs. theres a question thats similar but with $\lambda v$ = $v$, but the anwser doesn't really explain rather just points out one of the axiom(distributivity) does not hold. Any hint or advice how I could approach this or break it down.

i know that the 6 it needs to be closed under these conditions

1. v+w=w+v
2. v+(u+w)=(v+w)+u
3. v+0=0+v=v
4. v+(−v)=0
5. 1*v=v
6. $\lambda$(v+w)=$\lambda$v+$\lambda$w

Answer 1

A good strategy would be too look at the entire list of axioms. As far as I know, there are only $6$ axioms and not $8$. Being closed under addition and scalar multiplication are not axioms. Any vector space is already closed under addition and scalar multiplication by definition of the functions of addition and scalar multiplication. The following are the axioms.

$1.$ For all $u ,v\in V$, we have $u+v=v+u$

$2.$ For all $u, v, w\in V$, we have $(u+w)+v=u+(v+w)$. For all $a, b\in F$, we have $(ab)v=a(bv)$.

$3.$ There exists an element $0\in V$ such that for all $v\in V$, we have $0+v=v$

$4.$ For every $v\in V$, there exists a $w\in V$ such that $v+w=0$.

$5.$ Let $1$ be the multiplicative identity of $F$. Then we have $1v=v$

$6 $ For all $a, b\in F$ and $u, v\in V$, we have $a(u+v)=au+av$ and $(a+b)v=av+bv$

Now you are playing around with the definition of scalar multiplication. So you should be looking at those axioms which use scalar multiplication. In the list above, those axioms are number $5$ and $6$.

You have to show that your new definition of scalar multiplication doesn't satisfy one or both of the axioms above. Try doing that.