I have solved a question but I am not sure the last step of the question. If someone can verify it that would be great.
Let $V$ be a finite dimensional vector space with complex inner product. Let $T: V \rightarrow V$ be a linear transformation. If $\forall v\in V$, $\|Tv\|=\|T^*v\|$ then $T$ is a normal operator.
So what I did is,
Since $\forall v \in V$, $\|Tv\|=\|T^*v\|$, we will have
$\forall v \in V$, $\langle Tv,Tv\rangle =\langle v,T^*Tv\rangle $ on the other hand
$\forall v \in V$, $\langle T^*v,T^*v\rangle =\langle v,TT^*v\rangle $ and therefore
$$\langle v,T^*Tv\rangle =\langle v,TT^*v\rangle \Rightarrow \langle v,(T^*T-TT^*)v\rangle =0\Rightarrow T^*T-TT^*=0 \Rightarrow T^*T=TT^*.$$
And therefore the operator is normal.
But then I started thinking, why can I actually say that the last two implications are correct? Can't be $T^*T-TT^*$ , the operator which sends v to it's perpendicular vector? Why wouldn't that be possible? I tried to answer this question but I couldn't.
I would be happy if someone can help me answer this question. Thanks!
(Reminder: Just in case we have different notations $T^*$ is the adjoint operator)
Best Answer
You have to show that $$ \langle Sv,v\rangle =0 \quad \forall v\in V $$ implies $S=0$.
To prove this one has to use polarization identities. In the complex case this works for a general mapping $S$. In the real case, this holds only for self-adjoint $S$. However, the mapping $T^*T-TT^*$ is self-adjoint.