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.

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- Apr 16th 2010, 05:56 PMmathbugInner Products
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.

- Apr 16th 2010, 06:07 PMDefunkt
- Apr 16th 2010, 06:53 PMmathbug
So then <v,x> = sum <v,vi> [<vi,x>] = sum <w,vj> [<vj,x>] = <w,x> Then

<v,v1>[<v1,x>] - <w,v1>[<v1,x>] +..+ <v,vn>[<vn,x>] - <w,vn>[<vn,x>]=0

So then where am I supposed to go from here? - Apr 16th 2010, 11:50 PMOpalg
- Apr 17th 2010, 02:07 AMDefunkt
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:

$\displaystyle w = \sum_{i=1}^n <w,v_i>v_i \Rightarrow w = \sum_{i=1}^n <v,v_i>v_i = v$ - Apr 17th 2010, 09:44 AMmathbug
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? - Apr 17th 2010, 02:08 PMDefunkt
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 $\displaystyle <v,v_i>=<w,v_i>$. - Apr 17th 2010, 08:34 PMmathbug
Oh! You're right, I do tend to make things too complicated. That answer is very simple! But makes so much sense... I completely understand now. Thank you so much for all your help!