LMIs in Control/Click here to continue/Fundamentals of Matrix and LMIs/LMI for Generalized eigenvalue problem

LMI for Generalized Eigenvalue Problem

Technically, the generalized eigenvalue problem considers two matrices, like $A$ and $B$ , to find the generalized eigenvector, $x$ , and eigenvalues, $\lambda$ , that satisfies $Ax=\lambda Bx$ . If the matrix $B$ is an identity matrix with the proper dimension, the generalized eigenvalue problem is reduced to the eigenvalue problem.

The System

Assume that we have three matrice functions which are functions of variables $x=[x_{1}\quad x_{2}\quad ...\quad x_{n}]^{\text{T}}\in \mathbb {R} ^{n}$ as follows:

$A(x)=A_{0}+A_{1}x_{1}+...+A_{n}x_{n}$

$B(x)=B_{0}+B_{1}x_{1}+...+B_{n}x_{n}$

$C(x)=C_{0}+C_{1}x_{1}+...+C_{n}x_{n}$

where are $A_{i}$ , $B_{i}$ , and $C_{i}$ ( $i=1,2,...,n$ ) are the coefficient matrices.

The Data

The $A(x)$ , $B(x)$ , and $C(x)$ are matrix functions of appropriate dimensions which are all linear in the variable $x$ and $A_{i}$ , $B_{i}$ , $C_{i}$ are given matrix coefficients.

The Optimization Problem

The problem is to find ${\begin{aligned}x=[x_{1}\quad x_{2}...x_{n}]\end{aligned}}$ such that:

$A(x)<\lambda B(x)$ , $B(x)>0$ , and $C(x)<0$ are satisfied and $\lambda$ is a scalar variable.

The LMI: LMI for Schur stabilization

A mathematical description of the LMI formulation for the generalized eigenvalue problem can be written as follows:

${\begin{aligned}&{\text{min}}\quad \lambda \\&{\text{s.t.}}\quad A(x)<\lambda B(x)\\&\quad \quad B(x)>0\\&\quad \quad C(x)<0\end{aligned}}$

Conclusion:

The solution for this LMI problem is the values of variables $x$ such that the scalar parameter, $\lambda$ , is minimized. In practical applications, many problems involving LMIs can be expressed in the aforementioned form. In those cases, the objective is to minimize a scalar parameter that is involved in the constraints of the problem.