1
$\begingroup$

Let $A$, $C$, $W$ and $V$ be given (known) matrices with $V$ and $W$ being semidefinite positive. We would like to determine the matrices $X$ and $Y$ by solving the following inequality \begin{equation} \begin{bmatrix}\begin{bmatrix}X & YC\\ 0 & A \end{bmatrix}^{T}\begin{bmatrix}YVY^{T} & 0\\ 0 & W \end{bmatrix}^{-1}\begin{bmatrix}X & YC\\ 0 & A \end{bmatrix} & & \begin{bmatrix}X & YC\\ 0 & A \end{bmatrix}^{T}\begin{bmatrix}YVY^{T} & 0\\ 0 & W \end{bmatrix}^{-1}\\ \\ \begin{bmatrix}YVY^{T} & 0\\ 0 & W \end{bmatrix}^{-1}\begin{bmatrix}X & YC\\ 0 & A \end{bmatrix} & & \begin{bmatrix}YVY^{T} & 0\\ 0 & W \end{bmatrix}^{-1} \end{bmatrix}\succeq0. \end{equation}

Does anyone know how to transform this inequality into linear matrix inequality (LMI) using Schur complements or other methods please? Or does anyone know how to solve this inequality in order to determine the matrices $X$ and $Y$ please? Thanks.

$\endgroup$
  • $\begingroup$ It might be useful to note that we can rewrite the matrix as the product $$ \pmatrix{\pmatrix{X & YC\\0&A}\\ & I}^T \pmatrix{\pmatrix{YVY^T\\ & W}^{-1} \\ & \pmatrix{YVY^T\\ & W}^{-1}} \pmatrix{\pmatrix{X & YC\\0&A}\\ & I} $$ $\endgroup$ – Omnomnomnom Jun 17 '17 at 3:15
1
$\begingroup$

Note that if $P$ is positive semidefinite, then so is $A^TPA$ for any matrix $A$.

By rewriting our matrix as the product $$ \pmatrix{\pmatrix{X & YC\\0&A}\\ & I}^T \pmatrix{\pmatrix{YVY^T\\ & W}^{-1} \\ & \pmatrix{YVY^T\\ & W}^{-1}} \pmatrix{\pmatrix{X & YC\\0&A}\\ & I} $$ We can see that your matrix will be positive semidefinite whenever $V$ and $W$ are positive semidefinite.

$\endgroup$
  • $\begingroup$ @G.Trav edit the question; that's easier than writing a comment $\endgroup$ – Omnomnomnom Jun 17 '17 at 12:58
  • $\begingroup$ Thank you very much for your insightful answer Omnomnomnom. Excuse me really, I was not specific about the question. The real problem is the following: math.stackexchange.com/questions/2326173/… . It would be great to have your opinion on that please. Thanks for all. $\endgroup$ – G. Trav Jun 17 '17 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.