Prove using axioms for a vector space

Question

Please help. I have no idea how to go about answering this question.

Question: Prove the following using only the axioms for a vector space and its associated field, explaining which axioms are used at each step of your proof. $(\forall x \in V)(\forall \lambda \in F)(-\lambda)x=-(\lambda x)=\lambda (-x)$

Any help will be greatly appreciated. Thank you

Answer 1

To prove that $\;(-\lambda )x=-(\lambda x)$, you simply have to prove that $$(-\lambda )x+(\lambda x)=0,$$ and similarly, for the other equality, $$\lambda(-x )+(\lambda x)=0,$$ using the relevant axioms.

Answer 2

Lemma: in any group the inverse is unique; so this also applies to $(V,+,-,0)$.

SO to see $-(\lambda x)$ (the inverse of $\lambda x$) equals $(-\lambda)x$ the latter must also satisfy the inverse property:

$$(-\lambda)x + (\lambda x) = (-\lambda + \lambda)x \text{ (distributivity) } = 0x \text { (in the scalar field) } = 0$$

where the last is usually not an axiom but proven from the the fact that

$$\forall z: (z+z=z) \to z=0$$ in the addition group (by adding $-z$ to both sides and associativity)

$$0x=(0+0)x=0x + 0x \text{ (distributivity) } \text{ so } 0x=0$$.

So it's a mini-series of lemmata leading to it, the main ingredient being the distributivity axiom

$$\forall \lambda,\mu, \forall x: (\lambda+\mu)x=\lambda x + \mu x$$

plus standard facts on the addition Abelian group.

$\lambda(-x)$ also satisfies the property of being inverse to $\lambda x$:

$$\lambda(-x) + \lambda x = \lambda(x + (-x))=\lambda 0 = 0$$

using another distributivity fact and the fact (separately to be shown ) that $$\forall \lambda: \lambda 0=0$$