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Let $P$ be an unknown complex $m \times n$ tall matrix ($m\geq n$, full column-rank), and $D$ a known real $m \times m$ diagonal matrix with positive entries. Is there a solution for $P$ from the following equation:

$P^+ = tr\left[\left(P^HP\right)^{-1}\right] P^H D P P^H$

where $P^+ = \left(P^HP\right)^{-1} P^H$ is the pseudoinverse of $P$, $(\cdot)^H$ denotes complex conjugate transpose, and $tr[\cdot]$ is the matrix trace operator?

If there is not a closed-form solution, any ideas on how can I solve for $P$ numerically?

Thank you.

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Since $P^*$ is onto, the equation reduces to $(P^*P)^{-1}=tr((P^*P)^{-1})P^*DP$.

We seek a particular solution in the form $P=\begin{pmatrix}\Delta_{n,n}\\0_{m-n,n}\end{pmatrix}$ where $\Delta$ is real diagonal. Let $D=diag(E_n,F_{m-n})$. Then $\Delta^{-2}=tr(\Delta^{-2})\Delta^2E$, that is $I=tr(\Delta^{-2})\Delta^4E$, an equation in $M_n$. Put $\Delta=diag(x_i),E=diag(e_i)$; then, for every $i$, $1=sx_i^4e_i$, where $s=\sum_j\dfrac{1}{x_j^2}$. It remains to calculate $s$ as a function of the $(e_i)$; a positive solution satisfies $x_i^2=\dfrac{1}{\sqrt{se_i}}$, implies that $s=\sqrt{s}\sum_i\sqrt{e_i}$ and $s=(\sum_i\sqrt{e_i})^2$.

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  • $\begingroup$ Great, thank you. You assume from the beginning that $P$ is a real rectangular diagonal matrix. Any solution to the equation should have this form? Why? Thanks... $\endgroup$ – CViteri Mar 19 '16 at 1:31
  • $\begingroup$ @ CViteri , we give an explicit solution. Unfortunately, it may exist solutions that are not real rectangular diagonal matrices. $\endgroup$ – user91684 Mar 19 '16 at 11:01

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