Inner product property proof

Question

Let $V$ be a vector space. Prove that if $\forall\beta \in V,\langle\alpha,\beta\rangle=0$ then $\alpha=0$. I couldn't think of any way to prove this directly, so I tried the contrapositive. Which would transform it to: Prove that if $\alpha \neq 0$ then $\exists\beta\in V,\langle\alpha,\beta\rangle\neq0$.

Proof: Since $\alpha$ is non-zero and is in $V$ then we have that $\langle\alpha,\alpha\rangle>0$.

Is this proof valid? Also is there a way to prove this directly?

Answer 1

A vector space per se has no inner product. If you are working on a vector space equipped with an inner product, then it depends on how the properties of the inner product are stated: if you say that $\langle\alpha,\alpha\rangle>0$ for every nonzero $\alpha$, then the natural way is as you did. If, on the other hand (and maybe more commonly) the property is stated that $\langle\alpha,\alpha\rangle=0$ implies $\alpha=0$, then you can just take $\beta=\alpha$.

Answer 2

This is the easiest proof. Let $\beta = \alpha$. Then $\langle \alpha,\alpha \rangle = 0$, so by the axioms of inner product, $a = 0$. You don't need to use the contrapositive.

Answer 3

Your proof is fine, but here's a direct proof (which is essentially the same). Suppose $\langle \alpha,\beta\rangle=0$ for all $\alpha$, then $\langle\alpha,\alpha\rangle=0$. One of the axioms of an inner product is $\langle x,x\rangle=0$ iff $x=0$, so w conclude $\alpha=0$.