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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.

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

1 Answer 1

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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.
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  • $\begingroup$ I'm interested whether in 3x3 it can be proved that skew-symmetric don't commute? $\endgroup$
    – Widawensen
    Commented Dec 2, 2016 at 9:07
  • $\begingroup$ @Widawensen 3x3 real skew-symmetric matrices are cross product matrices, which do not commute in general. $\endgroup$
    – user1551
    Commented Dec 2, 2016 at 19:26
  • $\begingroup$ I was thinking so that they must be somehow linked with cross product.. $\endgroup$
    – Widawensen
    Commented Dec 2, 2016 at 20:02
  • $\begingroup$ Where do you use that the matrix is 2x2? $\endgroup$
    – plr
    Commented Dec 19, 2016 at 21:50
  • $\begingroup$ @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. $\endgroup$
    – user1551
    Commented Dec 20, 2016 at 5:29

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