When I'm trying to solve a matrix inequality set of the following:

\begin{equation} \begin{array}{l} A-BC^{-1}B^{T}>0\\ C>0 \end{array} \end{equation}

Where $A$ is a given $p\times p$ positive-definite matrix, $B$ is an unknown $p\times q$ matrix and $C$ is an unknown $q\times q$ matrix.

At first look, I cannot say that this problem can be represented in a linear form (in the matrix variables $B,C$) and that the convex combination of a given two solutions for the inequalities is also a solution. But Schur complement provides a simple tool to change the inequalities into an equivalent form which is linear in the variables and thus also convex:

\begin{equation} \begin{bmatrix} A & B\\ B^{T} & C \end{bmatrix}>0 \end{equation}

My question is: If there some analog transformation that produce a linear (in the variables) form of the following:

\begin{equation} \begin{array}{l} A-BC^{-1}DC^{-1}B^{T}>0\\ C>0 \end{array} \end{equation}

Where $A$ is a given $p\times p$ positive-definite matrix, $B$ is an unknown $p\times q$ matrix, $C$ is an unknown $q\times q$ matrix and $D$ is a given $q\times q$ positive-definite matrix.

In some sense, if Schur complement help with 2'nd order multiplication, what can help with the 4'th order?

Thanks, Y.

  • $\begingroup$ Wouldn't $B=0$ always be a solution? $\endgroup$ Jul 24, 2018 at 6:50

1 Answer 1


I will assume that $B$ is given, otherwise $B = 0$ would be a trivial solution in which case $C$ can be any positive definite matrix.

Since $C$ has to be positive definite means that it and its inverse is full rank and symmetric, therefore $C^{-1}D\,C^{-1}$ has to be positive definite as well if $D$ is positive definite. Therefore you could use the substitution $P^{-1} = C^{-1}D\,C^{-1}$, so

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

and solve this LMI for $P$. A corresponding positive definite solution for $C$ can then be found from $P$ by using the continuous time algebraic Riccati equation (CARE). Namely the relation between $P$ and $C$ can also be written as $P = C\,D^{-1}C$, which is a very simplified version of a CARE, which is normally defined as

$$ A^\top X + X\,A - X\,B\,R^{-1}B^\top X + Q = 0, $$

but when using $A=0$, $X = C$, $B = I$, $R = D$ and $Q = P$ we again get $P = C\,D^{-1}C$. This should have a solution, since a CARE requires that $R$ is positive definite and $Q$ is semi-positive definite, which is the case since $P$ and $D$ are positive definite.

  • $\begingroup$ Thanks for the Answer Kwin! But actually, the LMI I presented is only part of a larger set of LMI's so choosing $B=0$ isn't possible. I'm trying to find some transformation that keeps all degrees of freedom. I'm getting the feeling that there is no transformation but I'll keep looking. Thanks anyway, Y. $\endgroup$
    – Yoav
    Jul 30, 2018 at 12:35
  • $\begingroup$ @Yoav Are the other LMI's only in $B$ or also in $C$? $\endgroup$ Jul 30, 2018 at 18:25
  • $\begingroup$ Yes, $B$ and $C$ are the only variables here, $A,D$ are given.\\ The other LMI is for the state-feedback control problem if you are familiar with control theory. $\endgroup$
    – Yoav
    Jul 31, 2018 at 7:08
  • $\begingroup$ @Yoav Can you maybe do some clever substitution for $B\,C^{-1}$ or do $B$ and $C^{-1}$ appear separate from each other in the other LMI's? If they do then an approach that works might depend on your other LMI's. $\endgroup$ Jul 31, 2018 at 7:49

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