How to prove: If $A$ and $B$ are normal matrices and AB = BA, $A+B$ and $AB$ are also normal, and $A$ and $B$ are simultaneously diagonalizable. If $A$ and $B$ are normal matrices and they commute $(AB=BA)$, then:

*

*$A+B $  is normal

*$AB $  is normal

*$A$ and $B$ are simultaneously diagonalizable: there is a unitary matrix $U$ such that both $U^*AU$ and $U^*BU$ are diagonal.

How can I prove the above statements?
I found the proposition above on Wikipedia, https://en.wikipedia.org/wiki/Normal_matrix#Consequences. I could have used it to prove another theorem, but I could not prove the proposition itself.
 A: Part 3 follows from the following theorem (proven at https://math.stackexchange.com/a/56491/434895):
Theorem. Two diagonalizable matrices $A$ and $B$ commute ($AB=BA$) if and only if they are simultaneously diagonalizable, that is, there exists an invertible matrix $P$ such that both $P^{-1}AP$ and $P^{-1}BP$ are diagonal matrices.
This theorem can be specialized to our case because, being normal, $A$ and $B$ are diagonalizable with $P=U$, where $U$ is unitary, whence $P^{-1}=U^*$:
Theorem. Two normal matrices $A$ and $B$ commute ($AB=BA$) if and only if they are simultaneously diagonalizable, that is, there exists a unitary matrix $U$ such that both $U^*AP$ and $U^*BP$ are diagonal matrices.
In what follows, we use the previous theorem to prove, first, part 2 and, second, part 1.
Let us define two diagonal matrices $D = U^* A U$ and $E = U^* B U$, whence $A = U D U^*$ and $B = U E U^*$.
$ AB = U D U^* U E U^* $
$ AB = U D E U^* $
$ (AB)^* = U D^* E^* U^* $, where $D^*$ and $E^*$ are the conjugate of $D$ and $E$
$ AB (AB)^* = U D E U^* U D^* E^* U^* = U D D^* E E^* U^* $
$ (AB)^* AB = U D^* E^* U^* U D E U^* = U D D^* E E^* U^* $
Therefore, $AB$ is normal, which proves Part 2.
$ A+B = U D U^* + U E U^* $
$ A+B = U (D U^* + E U^*) $
$ A+B = U (D+E) U^* $
$ (A+B)^* = U (D+E)^* U^* $
$ (A+B) (A+B)^* = U (D+E) U^* U (D+E)^* U^* = U (D+E)(D+E)^* U^* $
$ (A+B)^* (A+B) = U (D+E)^* U^* U (D+E) U^* = U (D+E)(D+E)^* U^* $
Therefore, $A+B$ is normal, which proves Part 1.
