Given T is normal trasformation, and $T^2 = \frac12 (T+T^*)$, prove that $T^2=T$. if T is normal, than there exists a unitary matrix Q such that:
$Q^*TQ=diag(\lambda_1, \lambda_2, ... , \lambda_k)$
where $\lambda_1, \lambda_2, ... , \lambda_k$ are the Eigenvalues of T.
$Q^*TQ Q^*TQ= Q^*T^2Q = diag(\lambda_1, \lambda_2, ... , \lambda_k)diag(\lambda_1, \lambda_2, ... , \lambda_k) = diag(\lambda_1^2, \lambda_2^2, ... , \lambda_k^2)$.
but im stuck trying to show that $\lambda_i = \lambda_i^2$ for $i=1...k$.
 A: From the condition 
$$T^2 = \frac{1}{2}(T+T^*) \qquad (\ast)$$ it follows that for each $i$ you have $\lambda_i^2 = \frac{1}{2} (\lambda_i + \bar\lambda_i)$ . Because the expression $\lambda_i + \bar\lambda_i$ is definitely real, so is $\lambda_i^2$ and hence either $\lambda_i$ is real or $\lambda_i$ is (purely) imaginary. If $\lambda_i \in \mathbb{R}$, then the equation tells you that $\lambda_i^2 = \lambda_i$, which is precisely what you need. Else, if $\lambda_i \in i \mathbb{R}$ then $\lambda_i ^2 = 0$ so $\lambda_i = 0$, and it is also true that $\lambda_i^2 = \lambda_i$.

Or, if you want to avoid diagonalisation, note first that the condition $(\ast)$  implies $$T^{*2} = \frac{1}{2} (T^*+T) = T^2.$$
Then, if you multiply $(\ast)$ by $T$ you get:
$$T^3 = \frac{T^2 + TT^*}{2}$$
Similarly, if you square $(\ast)$ you get:
$$T^4 = \frac{T^2 + 2 TT^* + T^{*2}}{4} = \frac{T^2 + TT^*}{2} = T^3.$$
Thus, $T^3(T-I) = 0$. Because of normality, this implies $T(T-I) = 0$, or $T^2 = T$. This might be preferable if you're not working in finite dimension.
