LMIs in Control/Matrix and LMI Properties and Tools/Convexity of LMIs

Definition

A set, ${\mathcal {S}}$, in a real inner product space is convex if for all $x,y\in {\mathcal {S}}$ and $\alpha \in \mathbb {R}$, where $0\leq \alpha \leq 1$, it holds that $\alpha x+(1-\alpha )y\in {\mathcal {S}}$.

Lemma 1.1

The set of solutions to an LMI is convex.

That is, the set ${\mathcal {S}}=\{x\in \mathbb {R} ^{m}\mid F(x)\leq 0\}$ is a convex set, where $F:\mathbb {R} ^{m}\longrightarrow \mathbb {S} ^{n}$ is an LMI.

Lemma 1.2

An LMI, $F:\mathbb {R} ^{m}\longrightarrow \mathbb {S} ^{n}$, in the variable $x\in \mathbb {R} ^{m}$ is an expression of the form

$F(x)=F_{0}+\sum _{i=1}^{m}x_{i}F_{i}\leq 0$

where $x^{T}=[x_{1}\cdots x_{m}]$ and $F_{i}\in \mathbb {S} ^{n}$, $i=0,\ldots ,m$.

Proof

Consider $x,y\in \mathbb {R} ^{m}$ and $\alpha \in [0,1]$, and suppose that $x$ and $y$ satisfy Lemma 1.2.

The LMI $F:\mathbb {R} ^{m}\longrightarrow \mathbb {S} ^{n}$ is convex, since

$F(\alpha x+(1-\alpha )y)=F_{0}+\sum _{i=1}^{m}(\alpha x_{i}+(1-\alpha )y_{i})F_{i}$

${\begin{alignedat}{2}F(\alpha x+(1-\alpha )y)&=F_{0}+\sum _{i=1}^{m}(\alpha x_{i}+(1-\alpha )y_{i})F_{i}\\&=F_{0}-\alpha F_{0}+\alpha F_{0}+\alpha \sum _{i=1}^{m}x_{i}F_{1}+(1-\alpha )\sum _{i=1}^{m}y_{i}F_{i}\\&=\alpha F_{0}+\alpha \sum _{i=1}^{m}x_{i}F_{i}+(1-\alpha )F_{0}+(1-\alpha )\sum _{i=1}^{m}y_{i}F_{i}\\&=\alpha F(x)+(1-\alpha )F(y)\\\end{alignedat}}$

Convexity of LMI

From Lemma 1.1, it is known that an optimization problem with a convex objective function and LMI constraints is convex.

The following is a non-exhaustive list of scalar convex objective functions involving matrix variables that can be minimized in conjunction with LMI constraints to yield a semi-definite programming (SDP) problem.

• ${\mathcal {J}}(x)={\frac {1}{2}}x^{T}PX+q^{T}x+r$, where $x,q\in \mathbb {R} ^{n}$, $P\in \mathbb {S} ^{n}$, $P>0$, and $r\in \mathbb {R}$.

1. Special case when $q=0$ and $r=0:{\mathcal {J}}(x)={\frac {1}{2}}x^{T}Px$, where$x\in \mathbb {R} ^{n}$, $P\in \mathbb {S} ^{n}$, and $P>0$.
2. Special case when $P=$2$\cdot 1$, $q=0$, and $r=0:{\mathcal {J}}(x)=x^{T}x=\|x\|_{2}^{2}$, where $x\in \mathbb {R} ^{n}$.

• ${\mathcal {J}}(X)=tr(X^{T}PX+Q^{T}X+X^{T}R+S)$, where $X$, $Q$, $R\in \mathbb {R} ^{n\times m}$, $P\in \mathbb {S} ^{n}$, $S\in \mathbb {R} ^{n\times n}$, and $P\geq 0$.

1. Special case when $Q=R=0$ and $S=0:{\mathcal {J}}(X)=tr(X^{T}PX)$, where $X\in \mathbb {R} ^{n\times m}$, $P\in \mathbb {S} ^{n}$, and $P\geq 0$.
2. Special case when $P=1$, $Q=R=0$ and $S=0:{\mathcal {J}}(X)=tr(X^{T}X)=\|X\|_{F}^{2}$, where $X\in \mathbb {R} ^{n\times m}$.
3. Special case when $P=0$, $R=0$ and $S=0:{\mathcal {J}}(X)=tr(Q^{T}X)$, where $X$, $Q\in \mathbb {R} ^{n\times m}$.
4. Special case when $P=1$, $Q=R=0$, $S=0$, and $X\in \mathbb {S} ^{n}:{\mathcal {J}}(X)=tr(X^{2})$, where $X\in \mathbb {S} ^{n}$.

• ${\mathcal {J}}(X)=\log(\det(X^{-1}))=\log(\det(X))$, where $X\in \mathbb {S} ^{n}$ and $X>0$.

Relative Definition of a Matrix

The definiteness of a matrix can be found relative to another matrix.

For example,

Consider the matrices $A\in \mathbb {S} ^{n}$ and $B\in \mathbb {S} ^{n}$. The matrix inequality $A<B$ is equivalent to $A-B<0$ or $B-A<0$.

Knowing the relative definiteness of matrices can be useful.

For example,

If in the previous example we have $A<B$ and also know that $A>0$, when we know that $B>0$.

This follows from $0<A<B$.

Strict and Non-strict Matrix Inequalities

A strict matrix inequality can be converted to a non-strict matrix inequality.

For example,

$A>0$ is implied by $A\geq \epsilon 1$, where $\epsilon \in \mathbb {R} _{>0}$. Similarly, $B<0$ is implied by $B\leq -\epsilon 1$, where $\epsilon \in \mathbb {R} _{>0}$

Converting a strict matrix inequality into a non-strict matrix inequality is useful when working with LMI solvers that cannot handle strict constraints.

Concatenation of LMIs

A useful property of LMIs is that multiple LMIs can be concatenated together to form a single LMI.

For example,

satisfying the LMIs $A<0$ and $B<0$ is equivalent to satisfying the concatenated LMI

${\begin{bmatrix}A&0\\0&B\end{bmatrix}}<0$

More generally, satisfying the LMIs $A_{i}<0$, $i=1,\ldots ,n$ is equivalent to satisfying the concatenated LMI $diag\{A_{1},\ldots ,A_{n}\}<0$.

External Links

• LMI Methods in Optimal and Robust Control - A course on LMIs in Control by Matthew Peet.
• LMIs in Systems and Control Theory - A downloadable book on LMIs by Stephen Boyd.
• LMI Properties and Applications in Systems, Stability, and Control Theory - A downloadable book on LMIs by Ryan James Caverly and James Richard Forbes.