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.

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    $\begingroup$ Hint: Just multiply $CC^*$ and $C^*C$ out, and compare. (No, the normality of two matrices does not automatically imply the normality of their sum.) $\endgroup$ – darij grinberg Jan 11 at 16:01
  • $\begingroup$ (A+iB)(A* + i B*) = (A*+iB*)(A+i B) But does that prove that $C$ is normal? $\endgroup$ – Kai Jan 11 at 16:07
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    $\begingroup$ You are incorrectly computing $C^\ast$; keep in mind that $i$ becomes $\overline{i} = -i$. $\endgroup$ – darij grinberg Jan 11 at 16:09
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    $\begingroup$ Also, recall that $C$ is normal if and only if $CC^\ast = C^\ast C$. This is the very definition of "normal". $\endgroup$ – darij grinberg Jan 11 at 16:09

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*}


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