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The classical Orthogonal Procrustes Problem is

$$\begin{array}{ll} \text{minimize} & \|A\Omega-B\|_{F}\\ \text{subject to} & \Omega'\Omega=I\end{array}$$

where $A$ and $B$ are known matrices.

Suppose $A$ is the identity matrix. I would like to solve the less strict problem

$$\begin{array}{ll} \text{minimize} & \|\Omega-B\|_{F}\\ \text{subject to} & \Omega'\Omega = \mbox{diag}(d_1,d_2,d_3,\dots)\end{array}$$

i.e., I want to relax the orthonormality constraint to mere orthogonality. For my purposes it is only important that $\Omega' \Omega$ be diagonal. With a less strict constraint than orthonormality, my hope is that $\Omega$ can get closer to $B$. Any ideas on how to do this?

Rodrigo de Azevedo
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Mael
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    the wikipedia page lists several variants, I suspect the one about orthogonal but not orthonormal is yours... http://en.wikipedia.org/wiki/Orthogonal_Procrustes_problem#Generalized.2Fconstrained_Procrustes_problems – Will Jagy Nov 30 '13 at 22:28
  • yes, orthogonal would be $\Omega'\Omega = D$ diagonal and orthonormal require all $d_k = 1$ – mathreadler Sep 07 '16 at 22:15
  • Is $\Omega$ square? It would be nice to have the matrices' dimensions. – Rodrigo de Azevedo Mar 28 '18 at 09:50
  • Isn't the classical Orthogonal Procrustes Problem about minimizing the Frobenius norm of $\Omega A - B$ rather than the Frobenius norm of $A \Omega - B$? – Rodrigo de Azevedo Mar 28 '18 at 10:40

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