Normal Operator Proof: Proving ##T \in L(V)## is Normal

[STEMucator]

Normal Operator Proof

Homework Statement

Prove an operator ##T \in L(V)## is normal ##⇔ ||T(v)|| = ||T^*(v)||##.

Homework Equations

(1) ##T \in L(V)## is normal if ##TT^*= T^*T##.

(2) If T is a self-adjoint operator on V such that ##<T(v), v> = 0, \space \forall v \in V##, then ##T=0##.

The Attempt at a Solution

##"\Rightarrow"## Assume ##T## is normal (1) :

##TT^*= T^*T##
##TT^* - T^*T = 0##

Now using (2) we can write :

##<(TT^* - T^*T)(v), v> = 0, \space \forall v \in V##

Using some inner product rules yields :

##<T^*T(v), v> = <TT^*(v), v>, \space \forall v \in V##
##||T(v)||^2 = ||T^*(v)||^2, \space \forall v \in V##
##||T(v)|| = ||T^*(v)||, \space \forall v \in V##


##"\Leftarrow"## : The proof will be exactly as above, except I start at ##||T(v)|| = ||T^*(v)||## and I finish at ##TT^*= T^*T## I believe?

 

[ulyj]

TT ∗ −T ∗ T=0

Now using (2) we can write :

<(TT ∗ −T ∗ T)(v),v>=0, ∀v∈V

You cannot use (2) here since T=0 is not the assumption rather the result.

 

[STEMucator]

You cannot use (2) here since T=0 is not the assumption rather the result.

This confused me a bit, I used (2) because ##TT^* - T^*T## is a self adjoint operator.

Using the fact it is self adjoint, and that ##TT^* - T^*T = 0##, I could choose any vector in ##V## and still get a result of zero.

 

[ulyj]

This confused me a bit, I used (2) because ##TT^* - T^*T## is a self adjoint operator.

Using the fact it is self adjoint, and that ##TT^* - T^*T = 0##, I could choose any vector in ##V## and still get a result of zero.

The result is right, what I'm pointing out is the reason for why the result holds. From what I understand, you didn't use (2) to deduce that ##<(TT^* - T^*T)(v), v> = 0, \space \forall v \in V##