Proving T is Normal from \left\|T^{*}v\right\|\leq\left\|Tv\right\|

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In summary, the conversation discusses a proof that if the norm of T* is less than or equal to the norm of T, then T is normal in an inner product space. Two attempts at a solution are presented, with one being questionable due to the definition of an inner product on L(V) and the justification for a certain step. The other attempt shows that TT*-T*T is a positive, self-adjoint operator with all eigenvalues equal to 0, leading to the conclusion that T is the zero map.
  • #1
Quantumpencil
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Homework Statement


Suppose V is an inner product space and [tex]T\in L(V)[/tex]. Prove that if [tex]\left\|T^{*}v\right\|[/tex][tex]\leq[/tex][tex]\left\|Tv\right\|[/tex] then T is Normal.

Sorry for being bothersome; this is the first year I've ever written proofs so I'm a bit sketchy about them.

Homework Equations


The Attempt at a Solution

Define an Inner Product on V [tex]\left\langle Sv, Tv\right\rangle[/tex] = [tex]\ Trace(ST^{*})[/tex].
Then [tex]\left\|T^{*}v\right\|[/tex][tex]\leq[/tex][tex]\left\|Tv\right\| = Trace (TT^{*}) \leq Trace( T^{*}T)[/tex].
However, [tex]\ Trace (TT^{*}) = Trace (T^{*}T)[/tex]. Which implies [tex]\left\|T^{*}v\right\|[/tex] = [tex]\left\|Tv\right\|[/tex], and therefore T is normal

I worked up another solution, which I'm almost sure is correct, but if this one does indeed check out I'd prefer to use it ;p. It just seems wrong to me for some reason; because it implies that basically all linear maps are normal; but I can't place what I did incorrectly.
 
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  • #2


Quantumpencil said:

Homework Statement


Suppose V is an inner product space and [tex]T\in L(V)[/tex]. Prove that if [tex]\left\|T^{*}v\right\|[/tex][tex]\leq[/tex][tex]\left\|Tv\right\|[/tex] then T is Normal.

Sorry for being bothersome; this is the first year I've ever written proofs so I'm a bit sketchy about them.


Homework Equations





The Attempt at a Solution

Define an Inner Product on V [tex]\left\langle Sv, Tv\right\rangle[/tex] = [tex]\ Trace(ST^{*})[/tex].
That is not an inner product on V, it is an inner product on L(V).

Then [tex]\left\|T^{*}v\right\|[/tex][tex]\leq[/tex][tex]\left\|Tv\right\| = Trace (TT^{*}) \leq Trace( T^{*}T)[/tex].
What justification do you have for [tex]\left\|T^{*}v\right\|\leq\left\|Tv\right\|[/tex]?

However, [tex]\ Trace (TT^{*}) = Trace (T^{*}T)[/tex]. Which implies [tex]\left\|T^{*}v\right\|[/tex] = [tex]\left\|Tv\right\|[/tex], and therefore T is normal

I worked up another solution, which I'm almost sure is correct, but if this one does indeed check out I'd prefer to use it ;p
 
  • #3


We assumed that T*v was less than or equal to Tv; in order to show that T had to be normal.

An inner product on L(V), right. I think that destroys my "proof." Since it doesn't make sense to take the "Trace" of a vector.

I have another brief question; If all the eigenvalues of a particular operator are equal to 0, does that imply the operator is equal to 0? or just Nilpotent? The other proof I worked on shows that TT*-T*T is positive, and since Trace of that operator is 0, It must have all zeros along the diagonal. Then I sort of hand-wavily assumed I could say TT*-T*T = 0, and therefore TT* = T*T and T is a normal operator. Is there any solid justification for that?

EDIT: Nevermind I think I figured it out. T is Positive, so it's also self-adjoint; which means V has a basis of eigenvectors, but since all the eigenvalues are 0, TT*-T*T must be the zero map.

Hurray!
 
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1. What does it mean for a transformation to be "normal"?

In mathematics, a transformation is considered normal if it commutes with its adjoint. This means that the transformation and its adjoint have the same eigenvectors.

2. How can I prove that a transformation is normal?

One way to prove that a transformation is normal is by showing that it commutes with its adjoint. This can be done by applying the definition of the adjoint and the transformation to a general vector and using properties of inner products.

3. What is the significance of the inequality \left\|T^{*}v\right\|\leq\left\|Tv\right\| in proving normality?

This inequality, also known as the Cauchy-Schwarz inequality, is a necessary condition for a transformation to be normal. It ensures that the transformation and its adjoint have the same eigenvectors, which is a key characteristic of normal transformations.

4. Can a transformation be normal if it does not satisfy the inequality \left\|T^{*}v\right\|\leq\left\|Tv\right\|?

No, a transformation cannot be normal if it does not satisfy this inequality. The Cauchy-Schwarz inequality is a necessary condition for normality, and if it is not satisfied, the transformation and its adjoint will not have the same eigenvectors.

5. Are there any other methods for proving normality besides using the inequality \left\|T^{*}v\right\|\leq\left\|Tv\right\|?

Yes, there are other methods for proving normality, such as using the spectral theorem or showing that the transformation has a diagonalizable matrix representation. However, the inequality \left\|T^{*}v\right\|\leq\left\|Tv\right\| is a common and relatively simple method for proving normality.

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