# Is the complex symmetric part of a $2\times2$ normal matrix always normal?

I want to prove (or disprove) that, for a normal matrix $A\in\mathbb{C}^{2x2}$ (so that $AA^\ast=A^\ast A$), the sum $A+A^T$ is (or isn't) normal.

I know that for $3\times3$ normal matrices, $A+A^T$ isn't normal, but for the $2\times2$ case, when I try to come with a counterexample I do not manage to make it.

• Is there a difference between $*$ and $T$ here? Are these matrices with complex entries? Commented Dec 1, 2016 at 16:35
• $*$ = transpose and conjugate, $T$ transpose. Yes, they're in $\mathbb{C}^{2x2}$
– plr
Commented Dec 1, 2016 at 16:38

Yes, $A+A^T$ is normal if $A$ is $2\times2$ normal. The reason is that all $2\times2$ complex skew-symmetric matrices commute (because they are scalar multiples of the rotation matrix for angle $\frac\pi2$).
1. Let $A=S+K$, where $S=\frac12(A+A^T)$ is complex symmetric and $K=\frac12(A-A^T)$ is complex skew-symmetric.
2. Using the fact that $K\bar{K}=\bar{K}K$, rearrange the equality $AA^\ast=A^\ast A$ to $[S,\bar{S}] = [S,\bar{K}]+[\bar{S},K]$, where $[X,Y]$ denotes the commutator $XY-YX$.
3. Since $[S,\bar{S}]$ is purely imaginary (because $\overline{[S,\bar{S}]}=[\bar{S},S]=-[S,\bar{S}]$) but $[S,\bar{K}]+[\bar{S},K]$ is real, the rearranged equality in last step implies that $[S,\bar{S}]=0$. As $\bar{S}=S^\ast$, we conclude that $S$ is normal.
• @rdguez In step 2. There was a typo: in the first sentence, $KK^T=K^TK$ should read $K\bar{K}=\bar{K}K$ (now fixed), and the latter is true because $K$ is $2\times2$ complex skew-symmetric. Commented Dec 20, 2016 at 5:29