Let V be an inner product space. Prove that if v and w are vectors in V such that <v,x> = <w,x> for every x in V, then v = w.
Opalg's solution is obviously nicer, however to complete this one:
Since $\displaystyle <v,x> = <w,x> ~ \forall x \in V$, we get that $\displaystyle <v,v_i> = <w,v_i> ~ \forall 1 \leq i \leq n$.
But, note that:
$\displaystyle w = \sum_{i=1}^n <w,v_i>v_i \Rightarrow w = \sum_{i=1}^n <v,v_i>v_i = v$
But how can you say that <v,v_i> = <w,v_i> for all x in V? If we had
<v,v1>[<v1,x>] - <w,v1>[<v1,x>] +..+ <v,vn>[<vn,x>] - <w,vn>[<vn,x>]=0
then we could factor and get
[<v1,x>][<v,v1>-<w,vi>] +...+ [<vn,x>][<v,vn>-<w,vn>]=0
I get that if <vi,x> were linearly independent, we would have that <w,v_i>=<v,v_i> for all x in V, but how can we show that?
I think you're making this too complicated --
We are given that for every $\displaystyle x \in V$, $\displaystyle <v,x> = <w,x>$.
In particular, if we let $\displaystyle x=v_1$, we get that $\displaystyle <v,v_1> = <w,v_1>$. Now let $\displaystyle x = v_2$, then $\displaystyle <v,v_2> = <w,v_2>$ and so on: for every $\displaystyle 1 \leq i \leq n$, we get that $\displaystyle <v,v_i>=<w,v_i>$.