Let $v_1, v_2, ..., v_n$ be a spanning set (in particular, a basis) in an inner product space $V$, prove that if $(x,v) = 0$ for all $v \in V$, then $x = 0$.
I feel like I might have messed up a bit at the end of my proof, can you correct me on my answer? Is this the shortest answer?
My proof:
$(x,v) = 0$ implies $$(x, \sum_{k=1}^n\alpha_k v_k) = 0$$ where $\alpha_k$ are scalars.
Then we factor out the scalars by linearity:
$$(x, \alpha_1v_1 + ...+ \alpha_nv_n)$$
$$= \alpha_1(x,v_1) + ... + \alpha_n(x,v_n) = 0$$
Since the system $v_1, ..., v_n$ is linearly independent, $\alpha_k$ will not be all zero for $1\leq k \leq n$.
And since $v_1, ..., v_n$ is spanning, $v_1 \neq ... \neq v_n \neq 0$.
Thus the equality only holds when $x = 0$.
Replies: @Michal: Yes, this is the original question, not asking to prove $(x, v_i)$.