The Adjoint of a Linear Operator: When is ||T(x)|| equal to ||x||?

[Wildcat]

Homework Statement

Let T be a linear operator on an inner product space V. Prove that
||T(x)|| = ||x|| for all xεV iff <T(x),T(y)> = <x,y> for all x,yεV

Homework Equations

The Attempt at a Solution

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

This seems too simple. What else do I need to show?

 

[Dick]

Homework Statement

Let T be a linear operator on an inner product space V. Prove that
||T(x)|| = ||x|| for all xεV iff <T(x),T(y)> = <x,y> for all x,yεV

Homework Equations

The Attempt at a Solution

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

This seems too simple. What else do I need to show?

How does that show anything?

 

[Wildcat]

Thats what I was afraid of. The book does not cover this very well so I was trying to go by an example my professor did. I will have to research in another book I have.

 

[Dick]

Thats what I was afraid of. The book does not cover this very well so I was trying to go by an example my professor did. I will have to research in another book I have.

Do you know that ||x|| is defined by ||x||^2=<x,x>? One direction of your 'iff' is really easy. Do that one. The other direction uses a trick called the 'polarization identity'.

 

[Wildcat]

Yes, I do know that about the norm and I saw the polarization but wasn't sure how to use it. Thank you that will give me a starting point.

 

[Wildcat]

The trick on the polar identities do I use the def of the polar identity of <x,y> and show it is equal to the polar identity of <T(x), T(y)>??

 

[Dick]

The trick on the polar identities do I use the def of the polar identity of <x,y> and show it is equal to the polar identity of <T(x), T(y)>??

Sure. That would show <x,y>=<T(x),T(y)>, right?

 

[Wildcat]

Yes and also show ||T(x)||^2 = ||x||^2??

 

[Dick]

Yes and also show ||T(x)||^2 = ||x||^2??

I'm not following your reasoning from these little out-of-context snippets of your proof. You'll have to state the whole proof before I can make a meaningful comment.

 

[Wildcat]

Sure. That would show <x,y>=<T(x),T(y)>, right?

i think/know I'm missing the trick with the polar identities

I start out with <x,y> = ¼||x+y||² - ¼||x-y||²=
¼[<x+y,x+y> - <x-y,x-y>] =
¼[x² + 2xy +y² - x² +2xy -y²]=
¼[4xy]=
xy = <x,y>
I missed the trick and if I start out with <T(x),T(y)> I miss it that way too.
Can you give me a hint?

 

[Office_Shredder]

If you know <T(x),T(y)>=<x,y> for all x and y, it should be really simple to prove that ||x||=||T(x)|| for all x. Why don't you try that direction first. There aren't any tricks involved

 

[Dick]

i think/know I'm missing the trick with the polar identities

I start out with <x,y> = ¼||x+y||² - ¼||x-y||²=
¼[<x+y,x+y> - <x-y,x-y>] =
¼[x² + 2xy +y² - x² +2xy -y²]=
¼[4xy]=
xy = <x,y>
I missed the trick and if I start out with <T(x),T(y)> I miss it that way too.
Can you give me a hint?

You don't multiply anything out. Are you assuming ||T(x)||=||x|| and trying to show <Tx,Ty>=<x,y>, I hope? Express the inner product <Tx,Ty> in terms of the polarization identity and then use that T doesn't change norms. State your assumption and try to go step by step, please?

 

[Wildcat]

Assume ||T(x)|| = ||x||
<T(x), T(y)> = ¼||T(x) + T(y)||² - ¼||T(x) - T(y)||²=
(then since T doesn't change norms) ¼||x+y||² - ¼||x-y||² = <x,y>

Assume <T(x),T(y)>=<x,y>
||x||² = <x,x> = <T(x),T(x)> = ||T(x)||²
hence ||x|| = ||T(x)||

 

[Dick]

Assume ||T(x)|| = ||x||
<T(x), T(y)> = ¼||T(x) + T(y)||² - ¼||T(x) - T(y)||²=
(then since T doesn't change norms) ¼||x+y||² - ¼||x-y||² = <x,y> ??

You skipped a step. T(x)+T(y)=T(x+y). Now you can say ||T(x)+T(y)||=||T(x+y)||=||x+y||. The other direction is the REALLY EASY one.

 

[Wildcat]

You skipped a step. T(x)+T(y)=T(x+y). Now you can say ||T(x)+T(y)||=||T(x+y)||=||x+y||. The other direction is the REALLY EASY one.

'



Yes, that makes sense, I was wondering about that.

This is probably a stupid question, but is the reason the polar identity is used is because it uses norms? I'm a novice at proofs and am trying to figure out where the ideas come from.

 

[Dick]

'



Yes, that makes sense, I was wondering about that.

This is probably a stupid question, but is the reason the polar identity is used is because it uses norms? I'm a novice at proofs and am trying to figure out where the ideas come from.

Well yes, <Tx,Ty>=<x,y> expresses the invariance of inner products under T. ||T(x)||=||x|| expresses the invariance of norms under T. If you can express an inner product in terms of norms, then they just be the same, right? Now that's got to make sense.