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I have seen in an article that

$ \min_{\mathbf{K}} \hspace{0.2cm} tr[\mathbf{K} \Sigma \mathbf{K}^T]$ s.t. $ \mathbf{KH} = \mathbf{I} $

where $\mathbf{H}$ is of full column rank yields,

$\tilde{\mathbf{K} } = (\mathbf{H}^T\Sigma^{-1}\mathbf{H})^{-1}\mathbf{H}^T\Sigma^{-1}$.

Does anyone aware of some theorem related to this result.

shani
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1 Answers1

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First, let $L=(\sqrt{\Sigma})^{-1}H$ the minimization problem becomes $$ \min_{K:K\sqrt{\Sigma}\cdot L=I}{\rm tr}((K\sqrt{\Sigma})(K\sqrt{\Sigma})^T) =\min_{S:SL=I}\,{\rm tr}(SS^T) =\min_{S:SL=I}\,\Vert S\Vert^2_{HS} $$ Where $\Vert\cdot\Vert_{HS}$ is the Hilbert-Schmidt norm. This minimization problem is well-known, and its solution goes back to Penrose:"On best approximate solutions of linear matrix equations". The minimum is attained at a unique $S$ which is the pseudo-inverse of $L$. In our case : $S^{+}=(L^TL)^{-1}L^T$ because $L$ has full rank. Rolling back we see that the solution of the original minimization problem is given by $\tilde K=(H^T\Sigma^{-1}H)^{-1}H^T\Sigma^{-1}$.

Omran Kouba
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