Proof of normal operator and self-adjoint operator 1) Let $T∈L(V,V)$ be a normal operator. Prove that $||T(v)||=||T^*(v)||$ for every $v∈V$.
($T^*$ is  the adjoint of $T$)
2) Let $T$ be an operator on the finite dimensional inner product space $(V,<,>)$ and assume that $TT^*=T^2$. Prove that T is self-adjoint. (Can I simple get $T=T^*$ from $TT^*=T^2$? So there is nothing to prove)
Thank you for this two questions.
 A: 1>
$||T(v)||^2 = <Tv,Tv> = <v,T^*Tv>=<v,TT^*v>=<T^*v,T^*v> =<T^*v,T^*v>=||T^*v||^2$
A: For 1:
$$
\|Tv\|^2=\langle Tv,Tv\rangle =\langle T^*Tv,v\rangle=\langle TT^*v,v\rangle=\|T^*v\|^2.
$$
For 2, what you say would work if $T$ is invertible, but no one is saying it is. And you wouldn't be using the finite-dimension hypothesis. 
If you look at the Schur decomposition of $T$, you have $T=VXV^*$, with $V$ a unitary and $X$ upper triangular. The equality $TT^*=T^2$ implies $XX^*=X^2$. 
The diagonal entries of $XX^*$ are non-negative, and they agree with the diagonal entries of $X^2$, which are $X_{kk}^2$ (since $X$ is triangular). So the numbers $X_{kk}^2$ are non-negative, which implies that $X_{kk}$ is real for all $k$. The diagonal entries of $X^2$ are
$$
X_{11}^2,X_{22}^2,\ldots,X_{nn}^2;
$$
and the diagonal entries of $XX^*$ are 
$$
X_{11}^2,|X_{12}|^2+X_{22}^2, |X_{13}|^2+|X_{23}|^2+X_{33}^2,\ldots,|X_{11}|^2+\cdots+|X_{1,n-1}|^2+X_{nn}^2.
$$
So the equality $XX^*=X^2$ implies that $X_{kj}=0$ if $j>k$. That is, $X$ is diagonal with real diagonal, so it is selfadjoint. Then $T$ is selfadjoint. 
