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Let $\odot$ represent Hadamard or pointwise multiplication. If $\pmb{Y}$ is a $\underline{given}$ positive definite square matrix, can i estimate a positive semi-definite $\pmb{X}$ matrix such that \begin{equation} \pmb{Y} = \pmb{X} \odot \pmb{z}\pmb{z}^H \end{equation} where $\pmb{z}$ is a complex vector with phases only (i.e. each element of $\pmb{z}$ is of magnitude $1$). Also $(.)^H$ stands for conjuate-transposition, or Hermitian (Thanks @igael). I really don't know where to start. Maybe, i have formulated a problem that does not make sense.

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  • $\begingroup$ what is $\pmb{z}^H$ in your notation ? Hadamard product didn't need homogeneous elements ? $\endgroup$
    – user354674
    Commented Sep 11, 2016 at 17:42
  • $\begingroup$ Hi @igael $\pmb{z}^H$ is conjugate-transposition. $\endgroup$ Commented Sep 11, 2016 at 17:43
  • $\begingroup$ I do not know what you mean by homogenous elements. $\endgroup$ Commented Sep 11, 2016 at 17:44
  • $\begingroup$ Let $m=(z_i^{-1})_{i=1}^n$. Notice that $Y \odot mm^H$ is a positive semi-definite Hermitian matrix by Schur theorem and $Y= (Y \odot mm^H)\odot zz^H$. Let $X=Y \odot mm^H$. $\endgroup$
    – Daniel
    Commented Sep 14, 2016 at 18:21

1 Answer 1

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$\mathbf Y = diag(\mathbf z)\mathbf X diag(\mathbf z^H)$. Since $diag(\mathbf z)diag(\mathbf z^H)=\mathbf I$, $\mathbf X$ can be estimated as $\mathbf X = diag(\mathbf z^H)\mathbf Y diag(\mathbf z)$, where $\mathbf z$ is any vector with your constraint.

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  • $\begingroup$ Could you kindly explain how $diag(\pmb{z})diag(\pmb{z}^H) = \pmb{I}$. $\endgroup$ Commented Sep 11, 2016 at 19:36
  • $\begingroup$ each element of $\pmb{z}$ contains entries of magnitude $1$, but they are complex. $\endgroup$ Commented Sep 11, 2016 at 19:36
  • $\begingroup$ $\mathbf z=(exp(j\phi_1),\dots,exp(j\phi_n))$, isn't it? $\endgroup$ Commented Sep 11, 2016 at 19:39
  • $\begingroup$ Oh oh you are right, my bad. I thought you wrote $diag(\pmb{z}\pmb{z}^H)$. Again, my bad. Apoogies. $\endgroup$ Commented Sep 11, 2016 at 19:44
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    $\begingroup$ Or just, take any $\mathbf z$ with your constraint and set $\mathbf X=$ as in my answer, i.e., there exist infinitely many solutions. $\endgroup$ Commented Sep 11, 2016 at 19:52

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