Proving the Bi-Implication of Inner Product and Norm in Linear Operators

[Shackleford]

I don't know how to start this problem. Since it's a bi-implication, I need to show each statement implies the other. I started playing around with the definitions of inner product and norm directly, but it's not going anywhere.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20120225_142910.jpg

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20120225_142929.jpg

If I expand the inner product on the right,

<T(x),T(y)> = (2/4)<T(x),T(y)> + (2/4)<T(y),T(x)> = <x,y>.

Of course,

$\|\vec{T(x)}\| = \sqrt{<T(x),T(x)>} = \sqrt{<x,x>}$.

 

[fauboca]

I don't know how to start this problem. Since it's a bi-implication, I need to show each statement implies the other. I started playing around with the definitions of inner product and norm directly, but it's not going anywhere.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20120225_142910.jpg

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20120225_142929.jpg

If I expand the inner product on the right,

<T(x),T(y)> = (2/4)<T(x),T(y)> + (2/4)<T(y),T(x)> = <x,y>.

Look up isometries it should help you with this problem.

 

[Shackleford]

Look up isometries it should help you with this problem.

Two sections ahead in the book they talk about isometry and list the two implications in a theorem.

 

[Shackleford]

<x,y> = <T*T(x),y> = <T(x),T**y> = <T(x),T(y)>

<x,x> = <T*T(x),x> = <T(x),T**x> = <T(x),T(x)> = $\|\vec{T(x)}\|^{2}$.

 

[alanlu]

What do you mean by T*?

 

[fauboca]

Here is a hint:

$\langle T(x),T(y)\rangle = \frac{1}{4}\left(||T(x)+T(y)||^2-||T(x)-T(y)||^2\right)=\cdots$

That is by the polarization identity.

 

[Shackleford]

Here is a hint:

$\langle T(x),T(y)\rangle = \frac{1}{4}\left(||T(x)+T(y)||^2-||T(x)-T(y)||^2\right)=\cdots$

That is by the polarization identity.

I already wrote that out.

<T(x),T(y)> = (2/4)<T(x),T(y)> + (2/4)<T(y),T(x)> = <x,y>.

The <T(x),T(x)> and <T(y),T(y)> terms cancel.

 

[Shackleford]

What do you mean by T*?

T adjoint.

 

[fauboca]

I already wrote that out.

<T(x),T(y)> = (2/4)<T(x),T(y)> + (2/4)<T(y),T(x)> = <x,y>.

The <T(x),T(x)> and <T(y),T(y)> terms cancel.

If you use the polar identity, your next step should be

$$\frac{1}{4}\left(||T(x+y)||^2-||T(x-y)||^2\right)$$

$$=\frac{1}{4}\left(||x+y||^2-||x-y||^2\right)=\langle x,y\rangle$$

 

[Shackleford]

If you use the polar identity, your next step should be

$$\frac{1}{4}\left(||T(x+y)||^2-||T(x-y)||^2\right)$$

$$=\frac{1}{4}\left(||x+y||^2-||x-y||^2\right)=\langle x,y\rangle$$

How did you get to the second line from the first one, specifically T(x+y) = (x+y)? It looks like you used the left statement in the problem to do that.

In that last line, if you set x = y, then it looks like you get the relation you want.

 

[fauboca]

How did you get to the second line from the first one, specifically T(x+y) = (x+y)? It looks like you used the left statement in the problem to do that.

In that last line, if you set x = y, then it looks like you get the relation you want.

Your question says, "Let T be a linear operator..." What does that mean?

 

[Shackleford]

Your question says, "Let T be a linear operator..." What does that mean?

T(cx + y) = cT(x) + T(y).

The transformation disappeared. I'm wondering where it went there. You went from the norm squared of T(x+y) to the norm squared of just (x+y). If you use the first property listed, then it makes sense to me.

 

[fauboca]

T(cx + y) = cT(x) + T(y).

The transformation disappeared. I'm wondering where it went there. You went from the norm squared of T(x+y) to the norm squared of just (x+y).

Because once you prove ||T(x)|| = ||x||, what does ||T(x+y)|| = ??

 

[Shackleford]

Because once you prove ||T(x)|| = ||x||, what does ||T(x+y)|| = ??

That's what I just said twice. In order to make the step you did, you have to use that property. I suppose that satisfies the iff in the proposition.

 

[fauboca]

That's what I just said twice. In order to make the step you did, you have to use that property. I suppose that satisfies the iff in the proposition.

Then why did you ask if you knew that part?

What you have as a proof doesn't look like you used the polar identity. I am not really sure what you have done to claim <T(x),T(y)>=<x,y>

 

[Deveno]

suppose <T(x),T(y)> = <x,y> for all x and y. what happens when x = y?

on the other hand, suppose ||T(x)|| = ||x|| for all x.

what happens when you use x+y and x-y instead of x?

 

[Shackleford]

Then why did you ask if you knew that part?

What you have as a proof doesn't look like you used the polar identity. I am not really sure what you have done to claim <T(x),T(y)>=<x,y>

I just wanted to be sure. I wasn't getting anywhere with my work.

 

[Shackleford]

suppose <T(x),T(y)> = <x,y> for all x and y. what happens when x = y?

on the other hand, suppose ||T(x)|| = ||x|| for all x.

what happens when you use x+y and x-y instead of x?

For <T(x),T(y)> = <x,y>, if x = y, you can get the norm squared of x.

I thought we already already plugged in x+y and x-y.

<T(x+y),T(x+y)> = <T(x),T(x)> + <T(y),T(y)> + <T(x),T(y)> + <T(y),T(x)>

<T(x-y),T(x-y)> = <T(x),T(x)> + <T(y),T(y)> - <T(x),T(y)> - <T(y),T(x)>

 

[Shackleford]

Okay, guys, I think I have figured it out. I'll finish it up on my paper and be finished with this assignment. Thanks for the help.