1. Feb 25, 2012

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 [Broken]

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

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,

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

Last edited by a moderator: May 5, 2017
2. Feb 25, 2012

fauboca

Last edited by a moderator: May 5, 2017
3. Feb 25, 2012

Shackleford

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

4. Feb 25, 2012

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

Last edited: Feb 25, 2012
5. Feb 25, 2012

alanlu

What do you mean by T*?

6. Feb 25, 2012

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.

7. Feb 26, 2012

Shackleford

<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.

8. Feb 26, 2012

Shackleford

9. Feb 26, 2012

fauboca

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

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

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

10. Feb 26, 2012

Shackleford

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.

Last edited: Feb 26, 2012
11. Feb 26, 2012

fauboca

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

12. Feb 26, 2012

Shackleford

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.

13. Feb 26, 2012

fauboca

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

14. Feb 26, 2012

Shackleford

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.

15. Feb 26, 2012

fauboca

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>

16. Feb 26, 2012

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?

17. Feb 26, 2012

Shackleford

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

18. Feb 26, 2012

Shackleford

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