LMIs in Control/Click here to continue/Fundamentals of Matrix and LMIs/Dualization Lemma

Dualization Lemma
Consider $$P_{i}\in \text{S}^{n} $$ and the subspaces $$U, V$$, where $$P$$ is invertible and $$U +V = \text{R}^{n} $$. The following are equivalent.

$$ X^{T}PX<0$$ for all $$X\in \text{U}$$ \ $$\left\{0\right\}$$ and  $$ X^{T}PX \ge 0$$ for all $$X\in \text{V} $$.

$$ X^{T}P^{-1}X>0$$ for all $$X\in \text{U}^{\bot}$$ \ $$\left\{0\right\}$$ and  $$ X^{T}P^{-1}X \le 0$$ for all $$X\in \text{V}^{\bot} $$.

Example
Consider the matrices $$Q\in \text{S}^{n},S\in \text{R}^{n \times m},R\in \text{S}^{m},M\in \text{R}^{m \times n} $$ where $$ R\ge 0 ,$$ which define the quadratic matrix inequality

$$ \begin{align} \qquad \begin{bmatrix} 1 & M\\ \end{bmatrix} \begin{bmatrix} Q & S\\ S^{T} & R\\ \end{bmatrix} \begin{bmatrix} 1\\ M\\ \end{bmatrix} <0. \qquad (1) \end{align} $$

Define $$ \begin{align} P= \begin{bmatrix} Q & S\\ S^{T} & R\\ \end{bmatrix} ,U=R( \begin{bmatrix} 0\\ 1\\ \end{bmatrix} ) \end{align} $$ where $$U+V=R^{n+m}$$. Notice that $$(1)$$ is equivalent to $$X^{T}PX<0$$ for all $$X\in \text{U}$$ \ $$\left\{0\right\}$$.Additionally, $$X^{T}PX<0 \ge$$ for all $$X\in \text{V} $$ is euaivalent to

$$ \begin{align} \qquad \begin{bmatrix} 0 & 1\\ \end{bmatrix} \begin{bmatrix} Q & S\\ S^{T} & R\\ \end{bmatrix} \begin{bmatrix} 0\\ 1\\ \end{bmatrix} =R \ge 0, \end{align} $$

which is satisfied based on the definition of $$R$$. By the dualization lemma, $$(1)$$ is satisfied with $$R \ge 0$$ if and only if

$$ \begin{align} \qquad \begin{bmatrix} -M^{T} & 1\\ \end{bmatrix} \begin{bmatrix} \tilde{Q} & \tilde{S}\\ \tilde{S}^{T} & \tilde{R}\\ \end{bmatrix} \begin{bmatrix} -M^{T}\\ 1\\ \end{bmatrix} >0,\qquad \tilde{Q}\le 0, \end{align} $$

where $$ \begin{align} \qquad \begin{bmatrix} \tilde{Q} & \tilde{S}\\ \tilde{S}^{T} & \tilde{R}\\ \end{bmatrix} = \begin{bmatrix} Q & S\\ S^{T} & R\\ \end{bmatrix}^{-1}, U^{\bot}=N([1\quad M^{T}])=R( \begin{bmatrix} -M^{T}\\ 1\\ \end{bmatrix} )

\end{align} $$, and $$ \begin{align}V^{\bot}=N([0\quad 1])=R( \begin{bmatrix} 1\\ 0\\ \end{bmatrix} )

\end{align} $$.