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When I study the output-based optimal control problem, I meet such a optimization problem as

$\min\limits_{K\in \mathbb{R}^{s\times m}} x^TC^T K^T RKCx+x^TBKCx+x^TC^TK^TB^Tx$

where $x\in \mathbb{R}^n$ can be any vector and $R\in \mathbb{R}^{s\times s}$ is a definite-positive matrix. Moreover, $C\in \mathbb{R}^{m\times n}$ and $B\in \mathbb{R}^{n\times s}$.

In fact, I want to find a matrix $K$ such that $ C^T K^T RKC+BKC+C^TK^TB^T$ is minimized.

peng
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Aug 19 '19 at 08:52
  • What are the dimensions of the matrices? Is the expression even a scalar? And what does "reversible" mean? – mathreadler Aug 19 '19 at 09:00
  • ${\bf K= 0}$ will give $0$ which if we are working with real numbers will be minimum since real squares are always positive. But not if the resulting matrix is not positive definite or semi definite. If resulting matrix is not positive (semi) definite then minimum will be $-\infty$. – mathreadler Aug 19 '19 at 09:04
  • The problem has been updated. Thanks! – peng Aug 19 '19 at 09:14

1 Answers1

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Since $R$ is positive definite, there exists a full rank $L\in \mathbb R^{s\times s}$ such that $L^T R L = I$ (see Cholesky decomposition). Since the linear transformation defined by $L$ is bijective, we can rewrite your minimization problem through the $L$ transformation : \begin{align*} &\min\limits_{K\in \mathbb{R}^{s\times m}} x^TC^T K^T RKCx+x^TBKCx+x^TC^TK^TB^Tx\\ =&\min\limits_{K\in \mathbb{R}^{s\times m}} x^TC^T K^T KCx+x^TBLKCx+x^TC^TK^TL^TB^Tx\\ =&\bigg(\min\limits_{K\in \mathbb{R}^{s\times m}} x^TC^T K^T KCx+x^TBLKCx+x^TC^TK^TL^TB^Tx+x^T BLL^TB^Tx\bigg)-x^T BLL^TB^Tx\\ =&\bigg(\min\limits_{K\in \mathbb{R}^{s\times m}} x^T(KC+L^TB^T)^T(KC+L^TB^T) x \bigg)-x^T BLL^TB^Tx\\ =&\bigg(\min\limits_{K\in \mathbb{R}^{s\times m}} \left\lVert(KC+L^TB^T) x \right\rVert^2 \bigg)-x^T BLL^TB^Tx\\ =&\bigg(\min\limits_{K\in \mathbb{R}^{s\times m}} \left\lVert(KC+L^TB^T) x \right\rVert^2 \bigg)-x^T BR^{-1}B^Tx \end{align*} By observing that $RLL^T=L^{-T}L^T RLL^T = L^{-T} L^T = I$ so that $LL^T=R^{-1}$. To do the last minimization you can observe that if $K=-L^TB^T x(x^TC^TCx)^{-1} x^T C^T$, then \begin{align*} (-v(x^TC^TCx)^{-1} x^T C^TC+L^TB^T) x &= -L^TB^T x(x^TC^TCx)^{-1} x^T C^TC x+L^TB^T x\\ &= 0 \end{align*} which is clearly the minimizer since the quantity is positive, hence \begin{align*} \min\limits_{K\in \mathbb{R}^{s\times m}} x^TC^T K^T RKCx+x^TBKCx+x^TC^TK^TB^Tx = -x^T BR^{-1}B^Tx \end{align*}

The optimal value for $K$ may seem a bit arbitrary but it actually follows from the Moore Pseudo inverse, suppose you are minimizing $\lVert Au+v\rVert$ with respect to $A$, then $u\neq 0$ can be as a full rank $k\times 1$ matrix with pseudo inverse $u^\dagger=(u^T u)^{-1}u^T$ and then it is known that the minimal value of the minimization problem is given by $-v u^\dagger$, which gives $-v u^\dagger u+v = -v(u^T u)^{-1}u^T u + v = 0$

P. Quinton
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  • Thanks a lot for your answer! In the expression of $K$, I do not want use the $x$, because $x$ is time-varying. I want to derive the $K^$ such that $C^TK^{T}RK^C+BK^C+C^TK^{*T}B^T\leq C^TK^{T}RKC+BKC+C^TK^{T}B^T$. – peng Aug 20 '19 at 08:42
  • If so, your solution will not be a minimizer, you may want to reformulate your problem in an other question. – P. Quinton Aug 20 '19 at 08:49