I'm stuck in this problem, and is the last one!.
I have a hermitian operator A with its eigen-everything and I have to prove that it can be writen as $UDU^+$ where U is a unitary transformation, $U^+$ is its adjoint and D is a diagonal matrix.
I tried proving that $UDU^+$ could have the same eigen vectors but I couldn't, I really have no idea of how to tackle this problem.
There is a decomposition known as the Schur Decomposition or Schur Triangulation.
if $ A \in \mathbb{C}^{n \times n}$ then
$$ A = QTQ^{*} \tag{1} $$
you can transform this via hessenberg reduction. The proof works like this
$$ U^{*}AU = \begin{bmatrix} \lambda & B \\ 0 & C\end{bmatrix} \tag{2} $$
by induction there is a Schur factorization $VTV^{*} $ of $C$
$$ Q = U\begin{bmatrix} 1 & 0 \\ 0 & V\end{bmatrix} \tag{3} $$
$$ Q^{*}AQ = \begin{bmatrix} \lambda & BV \\ 0 & T\end{bmatrix} \tag{4}$$