I have a complex matrix [A], where

$$\left[A\right] = \left[D\right] \overline{\left[A\right]}$$

Where [D] is an arbitrary diagonal matrix with complex entries, and the over-bar indicates element-wise complex conjugation.

What can be said about the matrix [A]? Does it even have any determinable properties?


Notice that $$ A=D\bar A = D\overline{D\bar A} = \underbrace{D\bar D}_{=:S} A$$ where $S$ is real, diagonal, positive semidefinite. Let $d_1, \dotsc, d_n$ be the diagonal elements of $D$. Then, for the $k$-th row of $A$ we have $$ a_k = |d_k|^2 a_k. $$

  1. If $|d_k| \ne 1$, then it follows $a_k = 0$.
  2. If $|d_k| = 1$, then the argument (the angle in polar coordinate) of non-zero $a_{k,l}$ is half the argument of $d_k$ as $$ \arg( a_{k,l} ) = \arg( d_k \bar a_{k,l} ) = \arg(d_k) + \arg(\bar a_{k,l}) = \arg(d_k) - \arg(a_{k,l}). $$ In particular:
    • If $d_k = 1$, then $a_k = \bar a_k$ is real.
    • If $d_k = -1$, then $a_k = -\bar a_k$ is imaginary.
    • If $d_k = \pm i$ (the imaginary unit), the the real part of $a_k$ equals the $\pm$ imaginary part of $a_k$.
  • $\begingroup$ I totally missed that, thanks! If we're looking at row $k$, then $a_k$ would be a vector -- are you saying that all of the entries would either be 0, real, or imaginary, simultaneously? $\endgroup$ – Stuart Barth Nov 24 '16 at 1:53
  • $\begingroup$ @StuartBarth I added all the cases. $\endgroup$ – user251257 Nov 24 '16 at 4:03

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