# Normal proof

1. Jul 20, 2013

### Zondrina

Normal Operator Proof

1. The problem statement, all variables and given/known data

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

2. Relevant 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 $T=0$.

3. 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?

Last edited: Jul 20, 2013
2. Jul 22, 2013

### ulyj

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

3. Jul 22, 2013

### Zondrina

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.

4. Jul 22, 2013

### ulyj

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