# LMIs in Control/Matrix and LMI Properties and Tools/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 $X^{T}PX\geq 0$ for all $X\in {\text{V}}$ .

$X^{T}P^{-1}X>0$ for all $X\in {\text{U}}^{\bot }$ \$\left\{0\right\}$ and $X^{T}P^{-1}X\leq 0$ 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 $R\geq 0,$ which define the quadratic matrix inequality

{\begin{aligned}\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{aligned}} Define {\begin{aligned}P={\begin{bmatrix}Q&S\\S^{T}&R\\\end{bmatrix}},U=R({\begin{bmatrix}0\\1\\\end{bmatrix}})\end{aligned}} 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, $X^{T}PX<0\geq$ for all $X\in {\text{V}}$ is euaivalent to

{\begin{aligned}\qquad {\begin{bmatrix}0&1\\\end{bmatrix}}{\begin{bmatrix}Q&S\\S^{T}&R\\\end{bmatrix}}{\begin{bmatrix}0\\1\\\end{bmatrix}}=R\geq 0,\end{aligned}} which is satisfied based on the definition of $R$ . By the dualization lemma, $(1)$ is satisfied with $R\geq 0$ if and only if

{\begin{aligned}\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}}\leq 0,\end{aligned}} where {\begin{aligned}\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{aligned}} , and {\begin{aligned}V^{\bot }=N([0\quad 1])=R({\begin{bmatrix}1\\0\\\end{bmatrix}})\end{aligned}} .