Let $A, B \in \mathbb{C}^{n\times n}$ be selfadjoint, such that $[A,B] := AB − BA = 0$ . Show that $C := A + iB$ is normal matrix.
Could someone give me a hint on this problem ? I think that as selfadjoint matrices are also normal, I could first prove that $A$ and $B$ are normal, so therefore $A$ + $iB$ gives us a normal matrix. However I am not sure if that is the right way to prove that.
Your approach won't work, because the sum of two normal matrices is, in general, not normal.
Instead, simply check the definition of normality:
\begin{align*}[C,C^*] &= CC^*-C^*C \\&= (A+iB)(A+iB)^*-(A+iB)^*(A+iB) \\&=(A+iB)(A^*+(iB)^*)-(A^*+(iB)^*)(A+iB) \\&= AA^*+iBA^*+A(iB)^* + (iB)(iB)^* - A^*A-(iB)^*A-A^*iB-(iB)^*(iB) \\&= (AA^*-A^*A) + (A^*iB-iBA^*)+(A\bar{i}B^*-\bar{i}B^*A)+(iB\bar{i}B^*-\bar{i}B^*iB) \\&= [A,A^*] + i[A^*,B] + \bar{i}[A,B^*] + i\bar{i}[B,B^*] \\&= [A,A] + i[A,B] + \bar{i}[A,B] + [B,B]\mbox{ since $A$ and $B$ are self-adjoint} \\&= 0\mbox{ since [A,B] = 0, and $[X,X] = 0$ for all $X$}\end{align*}