I have the following matrix inequality that I need to express as an LMI

$ A^T Q^{-1} A - Q + \sum_i A_i^T Q^{-1} A_i < 0$

($Q > 0$, $A = XQ+YB$, $A_i = X_i Q+Y_i B$)

which I rewrote to

$ A^T Q^{-1} A - Q < - \sum_i A_i^T Q^{-1} A_i$

Now I want to use Schur's complement and express this inequality as

$ \begin{bmatrix} -Q & A^T \\ A & -Q \end{bmatrix} < \sum_i \begin{bmatrix} 0 & A_i^T \\ A_i & Q \end{bmatrix} $

and therefore

$ \begin{bmatrix} -Q & A^T \\ A & -Q \end{bmatrix} - \sum_i \begin{bmatrix} 0 & A_i^T \\ A_i & Q \end{bmatrix} < 0 $

which is a form my solver, solving for Q and B, should be able to handle.

Can I use the Schur complement in this way and are these two inequalities equivalent?

  • $\begingroup$ Which variables are known and which variables do you want to solve for? $\endgroup$ – Kwin van der Veen Dec 1 '18 at 23:48
  • $\begingroup$ @KwinvanderVeen I have edited my question. X and Y are known, Q and B are unknown $\endgroup$ – Alex bGoode Dec 2 '18 at 1:51
  • $\begingroup$ The matrices $A_i$ are also known as well? $\endgroup$ – Kwin van der Veen Dec 2 '18 at 2:55
  • $\begingroup$ Yes, they have the same structure, $X_i$ and $Y_i$ are known $\endgroup$ – Alex bGoode Dec 2 '18 at 12:38

You can't use Schur's complement in that way. Namely when you have the matrix inequalities

$$ A - B^\top C^{-1} B \succ 0, \quad C \succ 0, $$

then you can combine this into the following one matrix inequality

$$ M = \begin{bmatrix} A & B^\top \\ B & C \end{bmatrix} \succ 0. $$

This is because $M$ can be decomposed as follows

$$ M = \underbrace{ \begin{bmatrix} I & B^\top C^{-1} \\ 0 & I \end{bmatrix} }_{T^\top} \begin{bmatrix} A - B^\top C^{-1} B & 0 \\ 0 & C \end{bmatrix} \underbrace{ \begin{bmatrix} I & 0 \\ C^{-1} B & I \end{bmatrix} }_{T}. $$

Namely $M \succ 0$ means that $x^\top M\,x > 0\ \forall\, x\neq0$. So when using $x=T\,y$ one gets $x^\top M\,x = y^\top T^\top M\,T\,y$. It holds that $x\neq0$ iff $y\neq0$, since $T$ is full rank (lower triangular with ones on the diagonal), thus it also has to hold that $T^\top M\,T \succ 0$.

But remember that this equivalent matrix inequality formulation only works because the matrix $T$ transforms $M$ into a block diagonal matrix with the blocks the original matrix inequalities. So your proposition only works if there exists a transformation $T$ such that

$$ T^\top \left( \begin{bmatrix} Q & -A^\top \\ -A & Q \end{bmatrix} + \sum_i \begin{bmatrix} 0 & A_i^\top \\ A_i & Q \end{bmatrix} \right) T $$

gives back the original matrix inequalities as blocks on the diagonal.

  • $\begingroup$ Thank you, I will need to take some time and understand your answer. Do you have any pointers on how I can transform my original problem into an LMI or should I post another question for this? $\endgroup$ – Alex bGoode Dec 2 '18 at 12:40
  • $\begingroup$ @AlexbGoode what are the sizes of $Y$ and $B$? If $Y$ has a rank equal to the size of $Q$ then you could define $A$ as a variable and later use that to solve for $B$. $\endgroup$ – Kwin van der Veen Dec 2 '18 at 12:50
  • $\begingroup$ Nevermind, you only have one $B$ and multiple $A$/$A_i$. $\endgroup$ – Kwin van der Veen Dec 2 '18 at 13:14

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