# LA - Trace - Is this an elegant solution or an incorrect one? (Breif glance needed)

## Homework Statement

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

## The Attempt at a Solution

Define an Inner Product on V $$\left\langle Sv, Tv\right\rangle$$ = $$\ Trace(ST^{*})$$.
Then $$\left\|T^{*}v\right\|$$$$\leq$$$$\left\|Tv\right\| = Trace (TT^{*}) \leq Trace( T^{*}T)$$.
However, $$\ Trace (TT^{*}) = Trace (T^{*}T)$$. Which implies $$\left\|T^{*}v\right\|$$ = $$\left\|Tv\right\|$$, 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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HallsofIvy
Homework Helper

## Homework Statement

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

## The Attempt at a Solution

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

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

However, $$\ Trace (TT^{*}) = Trace (T^{*}T)$$. Which implies $$\left\|T^{*}v\right\|$$ = $$\left\|Tv\right\|$$, 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

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 breif 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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