Normal and self adjoint operator on vector space. Let $V$ a $\mathbb F-$vector space of finite dimension with a inner product $<,>$ and let $T\in L(V)$.
1) If $\mathbb F=\mathbb C$ and $T$ normal, prove that $T$ is self adjoint $\iff$ all eigenvalues are real.
2) Show that if $T^*T=-T$ (and $T^*$ the adjoint of $T$), then $T$ is self adjoint and $Spec(T)\subset \{-1,0\}$.

Attempts
1) 


*

*For $\implies $, let $T(v)=\lambda v$. $$\lambda \left<v,v\right> =\left<\lambda v,v\right>=\left<Tv,v\right>=\left<v,T^*v\right> =\lambda \left<v,T v\right>=\left<v,\lambda v\right>=\bar\lambda \left<v,v\right>.$$
Therefore $\lambda =\bar \lambda $ and thus $\lambda \in\mathbb R$.

*For the converse I don't really know. Suppose $T$ has real eigenvalue. I know that $$\forall v\in V,\left<Tv,v\right>=0\implies T=0$$ and thus I tried to prove that $\left<(T-T^*)v,v\right>=0$ for all $v\in V$. Let $v\in V$. Since $T$ normal,
$$\left<(TT^*-T^*T)v,v\right>=0\implies \left<TT^*v,v\right>=\left<T^*Tv,v\right>\iff \|T(v)\|=\|T^*(v)\|,$$
but I can't conclude. Any idea ?
2) $$\left<Tv,v\right>=-\left<T^*Tv,v\right>=\left<Tv,Tv\right>=-\|Tv\|^2.$$
But not very conclusive. For eigenvalue, I really don't know. Any idea ?
 A: Let $T$ be normal with all real eigenvalues. We have
$$Tv = \lambda v \iff \|(T-\lambda I)v\| = 0 \iff \|(T^* - \lambda I)v\|= \|(T-\lambda I)^*v\| = 0 \iff T^*v = \lambda v$$
Let $\{e_1, \ldots, e_n\}$ be an orthonormal basis of eigenvectors for $T$, with $Te_i = \lambda_i e_i$. We have
$$Tv = \sum_{i=1}^n \langle Tv, e_i\rangle e_i = \sum_{i=1}^n \langle v, T^*e_i\rangle e_i = \sum_{i=1}^n \langle v, \lambda_ie_i\rangle e_i = \sum_{i=1}^n \langle v, Te_i\rangle e_i = \sum_{i=1}^n \langle T^*v, e_i\rangle e_i = T^*v$$
so $T^* = T$.

For $(2)$, as Babelfish suggests, we have
$$-T = T^*T = (T^*T)^* = -T^*$$
so $T^* = T$. Furthermore, $-T = T^*T = T^2$ so the polynomial $x^2+x = x(x+1)$ annihilates $T$.
Hence the eigenvalues of $T$ are contained in $\{0,-1\}$.
A: Part 1
$T$ is normal iff $\|Tx\|=\|T^*x\|$ holds for all $x\in X$. This has interesting consequences, such as $T^2x=0$ iff $Tx=0$, which follows from
\begin{align}
   T^2x=0 &\implies 0=\|T^2x\|=\|T^*Tx\| \\
     &\implies 0=\langle T^*Tx,x\rangle=\|Tx\|^2 \\
     &\implies Tx=0.
\end{align}
Therefore, the minimal polynomial for a normal $T$ has no repeated factors, which is why a normal $T$ is always diagonalizable. The eigenvalues of $T^*$ are the conjugates of the eigenvalues of $T$ because $T-\lambda I$ is normal with $(T-\lambda I)^*=T^*-\overline{\lambda}I$, which gives
$$
     (T-\lambda I)x=0 \iff (T^*-\overline{\lambda}I)x=0.
$$
Therefore, if $T$ has only real eigenvalues, then $T^*$ has the same eigenvalues and corresponding eigenvectors, and the action of $T$ on this basis is the same as the action of $T^*$ on this basis, leading to the conclusion that $T=T^*$.
Part 2
$T^*T=-T$ iff $(T^*+I)T=0$ iff $(T+I)T=0$ because $T$ is normal. So the eigenvalues of $T$ are a subset of $\{0,1\}$. By Part 1, $T=T^*$.
