# About Schur complement in a non-linear matrix inequality

1. Mar 29, 2013

### Ronaystein

I have the following matrix inequality which is nonlinear due to M^TM. In order to transform into an LMI, I apply the Schur complement, however I am not sure about the result. Can you tell me if my deduction is correct?. It is possible to linearize the inequality by some heuristic methods?
Before hand thanks

Given the Following matrix inequality

\begin{align}
\begin{bmatrix}C^{T}M^{T}MC-\Phi_{11} & -\Phi_{12} & C^{T}M^{T}MD_{f}-\Phi_{13}\\
\star & -\Phi_{22} & -\mathscr{P}_{2}B_{fi}\\
\star & \star & D_{f}^{T}M^{T}MD_{f}-\beta^{2}I
\end{bmatrix} & \ge0
\end{align}

Equivalently

\begin{align*}
\begin{bmatrix}-\Phi_{11} & -\Phi_{12} & -\Phi_{13}\\
\star & -\Phi_{22} & -\mathscr{P}_{2}B_{fi}\\
\star & \star & -\beta^{2}I
\end{bmatrix}+\begin{bmatrix}(MC)^{T}MC & 0 & (MC)^{T}MD_{f}\\
\star & 0 & 0\\
\star & \star & (MD_{f})^{T}MD_{f}
\end{bmatrix} & \ge0
\end{align*}

Since

\begin{align*}
\begin{bmatrix}(MC)^{T}MC & 0 & (MC)^{T}MD_{f}\\
\star & 0 & 0\\
\star & \star & (MD_{f})^{T}MD_{f}
\end{bmatrix} & =\begin{bmatrix}(MC)^{T}\\
0\\
(MD_{f})^{T}
\end{bmatrix}\begin{bmatrix}MC & 0 & MD_{f}\end{bmatrix}\geq0
\end{align*}
Applying Schur complement

\begin{align*}
\begin{bmatrix}0 & 0 & 0 & (MC)^{T}\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & (MD_{f})^{T}\\
\star & \star & \star & -I
\end{bmatrix} & \geq0
\end{align*}

The LMI becomes

\begin{align}
\begin{bmatrix}-\Phi_{11} & -\Phi_{12} & -\Phi_{13} & (MC)^{T}\\
\star & -\Phi_{22} & -P_{2}B_{fi} & 0\\
\star & \star & -\beta^{2}I & (MD_{f})^{T}\\
\star & \star & \star & -I
\end{bmatrix} & \geq0,\,\,\forall i,j\in\{1,2,...,h\}\label{eq:TheoH_}
\end{align}

Last edited: Mar 29, 2013
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