Homework Help: Inner Product Spaces

1. Jun 5, 2010

hitmeoff

1. The problem statement, all variables and given/known data
Let $$\beta$$ be a basis for a finite dimensional inner-product space.

b) Prove that is < x, z > = < y, z> for all z $$\in$$ $$\beta$$, then x = y

2. Relevant equations

3. The attempt at a solution
|< x, z >| $$\leq$$ ||x|| ||z||

then because <x,z> = <y,z>

|< y, z >| $$\leq$$ ||x|| ||z||

so y = x, is this correct?

2. Jun 5, 2010

vela

Staff Emeritus
Nope.

3. Jun 5, 2010

hitmeoff

Any guidance?

4. Jun 5, 2010

Tedjn

What is < x, z> - < y, z>? When is < x, x> = 0?

5. Jun 5, 2010

hitmeoff

<x,z> - <y,z> = 0 and <x,x> = 0 iff x =0

<x,z> + <-y,z> = <x-y, z> = 0

and |<x-y, z>| $$\leq$$ ||x-y|| $$\cdot$$ ||z||

if z $$\neq$$ 0 the ||x-y|| must = 0. and ||x-y|| = 0 iff x-y = 0 so x = y

6. Jun 5, 2010

Tedjn

C.S. gives you an upper bound, via rearrangement, a lower bound for ||x-y|| of 0. That doesn't mean ||x-y|| = 0. Use the fact that <x-y,z> = 0 for every basis vector.

7. Jun 5, 2010

hitmeoff

if z is a basis vector, then x-y is orthogonal to any and every z $$\in$$ $$\beta$$ but im not sure where to go from there.

8. Jun 5, 2010

Tedjn

Hint: x-y is also orthogonal to every linear combination of z's.

9. Jun 6, 2010

hitmeoff

Im completely stuck. I know that x-y $$\in$$ ortho complement of $$\beta$$

But in this section, im not supposed to know that yet, so I dont wanna go down that route.

10. Jun 6, 2010

lanedance

how about assuming you have a suitable orthonormal basis for the space, then consider the product of x and y with each of ths basis vectors...

or if you know <x-y,z> = 0 for all z, how about the case z = x-y?

11. Jun 6, 2010

hitmeoff

in the case z = x-y, then <x-y, x-y> = 0 and thats only if x-y = 0 meaning x = y.

But heres my question, a few posts ago I thought about this, but then I got to thinking why x-y would necessarily be in $$\beta$$. I know x-y must = some linear combination of z's $$\in$$ $$\beta$$ since it is a basis. but why is x-y necessarily in $$\beta$$?

let x' = x-y and x-y = c1z1 +...+ cnzn then

<x', c1z1 +...+ cnzn> = <x', c1z1> +...+ <x', cnzn>

with each term inner product = 0

but c1z1 +...+ cnzn isn't necessarily in $$\beta$$, rather in the span($$\beta$$)

Last edited: Jun 6, 2010
12. Jun 6, 2010

lanedance

Thats a fair point and I think the above is only true if x & y are in B.

As it says innner product space as opposed to a "subspace" its probably fair to assume that x & y are in B. Once x & y are in B, clearly x-y is in B.

As a counter example say B is the subspace of R^3 spanned by [0,0,1]. Then let x = [1,0,1] and y = [0,1,1]

clearly <x,z> = <y,z> = 0 for all z in B, but x != y

note that the projection of x & y onto B will still be equal though...

Last edited: Jun 6, 2010
13. Jun 6, 2010

hitmeoff

Got it... I guess I read the problem wrong. I think you are write. The wya the problem is written, I think its meaning to imply x and y are in the inner product space. Geez, I had thought of this solution long ago, but I never figuered x and y were implied as being in B