Inner Products

**mathbug** · April 16th 2010, 05:56 PM

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.

**Defunkt** · April 16th 2010, 06:07 PM

Originally Posted by mathbug

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.

Let $\{v_1,...,v_n\}$ be an orthonormal basis of V, then:

$v = \sum_{i=1}^n <v,v_i>v_i$

$w = \sum_{j=1}^n <w,v_j>v_j$

now you finish.

**mathbug** · April 16th 2010, 06:53 PM

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?

**Opalg** · April 16th 2010, 11:50 PM

Originally Posted by mathbug

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.

So you know that $\langle v-w,x\rangle = 0$ for every x. In particular, you could put $x = v-w$ ... .

**Defunkt** · April 17th 2010, 02:07 AM

Originally Posted by mathbug

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?

Opalg's solution is obviously nicer, however to complete this one:

Since $<v,x> = <w,x> ~ \forall x \in V$ , we get that $<v,v_i> = <w,v_i> ~ \forall 1 \leq i \leq n$ .

But, note that:

$w = \sum_{i=1}^n <w,v_i>v_i \Rightarrow w = \sum_{i=1}^n <v,v_i>v_i = v$

**mathbug** · April 17th 2010, 09:44 AM

Originally Posted by Defunkt

Opalg's solution is obviously nicer, however to complete this one:

Since $<v,x> = <w,x> ~ \forall x \in V$ , we get that $<v,v_i> = <w,v_i> ~ \forall 1 \leq i \leq n$ .

But, note that:

$w = \sum_{i=1}^n <w,v_i>v_i \Rightarrow w = \sum_{i=1}^n <v,v_i>v_i = v$

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?

**Defunkt** · April 17th 2010, 02:08 PM

Originally Posted by mathbug

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 $x \in V$ , $<v,x> = <w,x>$ .
In particular, if we let $x=v_1$ , we get that $<v,v_1> = <w,v_1>$ . Now let $x = v_2$ , then $<v,v_2> = <w,v_2>$ and so on: for every $1 \leq i \leq n$ , we get that $<v,v_i>=<w,v_i>$ .

**mathbug** · April 17th 2010, 08:34 PM

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!

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