I have a cost function that is \begin{equation} J=\text{Trace} [\ (\ I-LC)\ KQK^T(\ I-LC)\ ^T ]\ \end{equation}

where $Q$ is a unknown positive definite matrix, $K$ and $C$ are full rank $n\times n$ and $m\times n$ matrices, respectively. Objective is to find $L$ such that minimizes $J$. Because $Q$ is unknown, I can't use it in minimization, so I tried to minimize upper bound of $J$. We can consider two upper bounds \begin{equation} J\leq \lambda_{max}(Q) \text{Trace} [\ (\ I-L_1C)\ KK^T(\ I-L_1C)\ ^T ]\ \end{equation}

\begin{equation} J\leq \lambda_{max}(KQK^T) \text{Trace} [\ (\ I-L_2C)\ (\ I-L_2C)\ ^T ]\ \end{equation} $\lambda_{max}(Q)$ and $\lambda_{max}(KQK^T)$ are unknown, but non-negative, so the only way is to minimizing Trace parts. $L_1$ and $L_2$ that minimize those are \begin{equation} L_1=KK^TC^T(\ CKK^TC^T)\ ^{-1} \end{equation} \begin{equation} L_2=C^T(\ CC^T)\ ^{-1} \end{equation}

Now it makes sense that \begin{equation} \lambda_{max}(Q) \text{Trace} [\ (\ I-L_1C)\ KK^T(\ I-L_1C)\ ^T ]\ \leq \lambda_{max}(KQK^T) \text{Trace} [\ (\ I-L_2C)\ (\ I-L_2C)\ ^T ]\ \end{equation} but I can't prove this.

what do you think?


P.S: A constraint for $Q$ that can be considered is \begin{equation} \begin{split} Q_{ii}&>|Q_{ij}|\qquad&\text{for}\qquad i\not= j\\ Q_{ii}&=Q_{jj}\qquad&\text{for}\qquad i\not= j \end{split} \end{equation} and $n\gg m$, $C=[\ I_m\quad 0]\ $

  • $\begingroup$ What do you mean, $Q$ is unknown? It is another degree of freedom in the optimization? Or you need to minimize something like the supremum of $J$ over all possible $Q$? $\endgroup$ – user7530 Jan 2 '14 at 21:11
  • $\begingroup$ second one, I need to minimize supremum of $J$ over all symmetric positive definite $Q$. $\endgroup$ – user118748 Jan 3 '14 at 4:42
  • 1
    $\begingroup$ Is there no constraint on $Q$ except being SPD? Otherwise, I can't see why such an optimization problem would have a solution except in the very special case that you can zero-out $I-LC$. $\endgroup$ – Algebraic Pavel Jan 3 '14 at 12:21
  • $\begingroup$ Edited question. $\endgroup$ – user118748 Jan 3 '14 at 12:43

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