Proving Inner Product Spaces: Proving x=y if <x,z> = <y,z>

[hitmeoff]

Homework Statement

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

Homework Equations

The Attempt at a Solution

start with the Cauchy-Schwarz:
|< x, z >| \leq ||x|| ||z||

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

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

so y = x, is this correct?

 

[vela]

Nope.

 

[hitmeoff]

Nope.

Any guidance?

 

[Tedjn]

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

 

[hitmeoff]

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

<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

 

[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.

 

[hitmeoff]

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.

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

 

[Tedjn]

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

 

[hitmeoff]

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

But in this section, I am not supposed to know that yet, so I don't want to go down that route.

 

[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?

 

[hitmeoff]

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?

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

But here's 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 +...+ c_nz_n then

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

with each term inner product = 0

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

 

[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...

 

[hitmeoff]

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

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