# Inner Products

• Apr 16th 2010, 05:56 PM
mathbug
Inner 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 PM
Defunkt
Quote:

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:

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

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

now you finish.
• Apr 16th 2010, 06:53 PM
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?
• Apr 16th 2010, 11:50 PM
Opalg
Quote:

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 $x = v-w$ ... .
• Apr 17th 2010, 02:07 AM
Defunkt
Quote:

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 $ = ~ \forall x \in V$, we get that $ = ~ \forall 1 \leq i \leq n$.

But, note that:

$w = \sum_{i=1}^n v_i \Rightarrow w = \sum_{i=1}^n v_i = v$
• Apr 17th 2010, 09:44 AM
mathbug
Quote:

Originally Posted by Defunkt
Opalg's solution is obviously nicer, however to complete this one:

Since $ = ~ \forall x \in V$, we get that $ = ~ \forall 1 \leq i \leq n$.

But, note that:

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

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?
• Apr 17th 2010, 02:08 PM
Defunkt
Quote:

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 $x \in V$, $ = $.
In particular, if we let $x=v_1$, we get that $ = $. Now let $x = v_2$, then $ = $ and so on: for every $1 \leq i \leq n$, we get that $=$.
• Apr 17th 2010, 08:34 PM
mathbug
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!