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
  • $\begingroup$ Do you mean "and all $\lambda \in F$"? Beyond that, what have you tried? A good start would be to go through the axioms one by one and check them. $\endgroup$
    – lulu
    Nov 14, 2022 at 21:01

1 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.

  • $\begingroup$ yes sorry i meant 6, confused myself there, fixed it $\endgroup$
    – beepboop
    Nov 14, 2022 at 21:16
  • $\begingroup$ @beepboop There is no need to apologies. Some people put two axioms together in one sentence and some list them differently. I thought you were including closure under addition and scalar multiplication as axioms and just wanted to correct that. $\endgroup$
    – Seeker
    Nov 14, 2022 at 21:19

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