Linear Matrix Inequalities in Control

Linear Matrix Inequalities in Control 

Siep Weiland and Carsten Scherer 

Dutch Institute of Systems and Control 

Class 1 

Siep Weiland and Carsten Scherer (DISC) Linear Matrix Inequalities in Control Class 1 1 / 59

Outline of Part I 

1 Course organization 

Material and contact information 

Topics 

Homework and grading 


Outline of Part II 

2 Convex sets and convex functions 

Convex sets 

Convex functions 

3 Why is convexity important? 

Examples 

Ellipsoidal algorithm 

Duality and convex programs 

4 Linear Matrix Inequalities 

Definitions 

LMI’s and convexity 

LMI’s in control 

5 A design example 


Part I 

Course organization 


organization and addresses 

All material and info is posted on website: 

w3.ele.tue.nl/nl/cs/education/courses/DISClmi/ 

Lectures: 

December 15, 2008 

January 5, 2009 



How to contact us 

Siep Weiland 

Department of Electrical Engineering 

Eindhoven University of Technology 

P.O. Box 513; 5600 MB Eindhoven 

Phone: +31.40.2475979; Email: s.weiland@tue.nl 

Carsten Scherer 

Delft Center for Systems and Control 

Delft University of Technology 

Mekelweg 2; 2628 CD Delft 

Phone: +31-15-2785899; Email: c.w.scherer@tudelft.nl 


course topics 

Facts from convex analysis. LMI’s: history, algorithms and software. 

The role of LMI’s in dissipativity, stability and nominal performance. 

Analysis results. 

From LMI analysis to LMI synthesis. State-feedback and 

output-feedback synthesis algorithms 

From nominal to robust stability, robust performance and robust 

synthesis. 

IQC’s and multipliers. Relations to classical tests and to µ-theory. 

Mixed control problems and parametrically-varying systems and 

control design. 


homework and grading 

Exercises 

Grading 

One exercise set is issued for every class. 

All sets have options to choose from 

Please hand in within 2 weeks. 

Your average grade over 4 homework sets. 


Part II 

Class 1 


Merging control and optimization 

In contrast to classical control, H∞ synthesis allows to design optimal 

controllers. However H∞ paradigm is restricted: 

Performance specs in terms of complete closed loop transfer matrix. 

Sometimes only particular channels are relevant. 

One measure of performance only. Often multiple specifications have 

been imposed for controlled system 

No incorporation of structured time-varying/nonlinear uncertainties 

Can only design LTI controllers 

Control vs. optimization 

View controller as decision variable of optimization problem. Desired 

specifications are constraints on controlled closed loop system system. 


Major goals for control and optimization 

Distinguish easy from difficult problems. (Convexity is key!) 

What are consequences of convexity in optimization? 

What is robust optimization? 

How to check robust stability by convex optimization? 

Which performance measures can be dealt with? 

How can controller synthesis be convexified? 

What are limits for the synthesis of robust controllers? 

How can we perform systematic gain scheduling? 


Optimization problems 

Casting optimization problems in mathematics requires 

X : decision set 

S ⊆ X : feasible decisions 

f : S → R: cost function 

f assigns to each decision x ∈ S a cost f(x) ∈ R. 

Wish to select the decision x ∈ S that minimizes the cost f(x) 



1 What is least possible cost? Compute optimal value 

fopt := inf f(x) = inf{f(x) | x ∈ S} ≥ −∞ 

x∈S 

Convention: S = ∅ then fopt = +∞ 

Convention: If fopt = −∞ then problem is said to be unbounded 

2 How to determine almost optimal solutions? For arbitrary ε > 0 find 

xε ∈ S with fopt ≤ f(xε) ≤ fopt + ε. 

3 Is there an optimal solution (or minimizer)? Does there exist 

xopt ∈ S with fopt = f(xopt) 

4 Can we calculate all optimal solutions? (Non)-uniqueness 

arg min f(x) := {x ∈ S | fopt = f(x)} 





x∈S 











x∈S 




We write: f(xopt) = minx∈S f(x) 








x∈S 








Recap: infimum and minimum of functions 

Infimum of a function 

Any f : S → R has infimum L ∈ R ∪ {−∞} denoted as infx∈S f(x) 

defined by the properties 

L ≤ f(x) for all x ∈ S 

L finite: for all ε > 0 exists x ∈ S with f(x) < L + ε 

L infinite: for all ε > 0 there exist x ∈ S with f(x) < −1/ε 

Minimum of a function 

If exists x0 ∈ S with f(x0) = infx∈S f(x) we say that f attains its 

minimum on S and write L = minx∈S f(x). 

Minimum is uniquely defined by the properties 

L ≤ f(x) for all x ∈ S 

There exists some x0 ∈ S with f(x0) = L 


A classical result 

Theorem (Weierstrass) 

If f : S → R is continuous and S is a compact subset of the normed linear 

space X , then there exists xmin, xmax ∈ S such that for all x ∈ S 

Comments: 

inf 

x∈S f(x) = f(xmin) ≤ f(x) ≤ f(xmax) = sup 

x∈S 

Answers problem 3 for “special” S and f 

No clue on how to find xmin, xmax 

No answer to uniqueness issue 

f(x) 

S compact if for every sequence xn ∈ S a subsequence xnm exists 

which converges to a point x ∈ S 

Continuity and compactness overly restrictive! 


Convex sets 

Definition 

A set S in a linear vector space X is convex if 

x1, x2 ∈ S =⇒ αx1 + (1 − α)x2 ∈ S for all α ∈ (0, 1) 

The point αx1 + (1 − α)x2 with α ∈ (0, 1) is a convex combination of x1 

and x2. 


The point x ∈ X is a convex combination of x1, . . . , xn ∈ X if 

x := 

n 

αixi, αi ≥ 0, 

i=1 

n 

αi = 1 

Note: set of all convex combinations of x1, . . . , xn is convex. 

Siep Weiland and Carsten Scherer (DISC) Linear Matrix Inequalities in Control Class 1 15 / 59 

i=1

Convex sets 


A set S in a linear vector space X is convex if 

x1, x2 ∈ S =⇒ αx1 + (1 − α)x2 ∈ S for all α ∈ (0, 1) 

The point αx1 + (1 − α)x2 with α ∈ (0, 1) is a convex combination of x1 

and x2. 


The point x ∈ X is a convex combination of x1, . . . , xn ∈ X if 

x := 

n 

αixi, αi ≥ 0, 

i=1 

n 

αi = 1 

Note: set of all convex combinations of x1, . . . , xn is convex. 


i=1

Examples of convex sets 

Convex sets Non-convex sets 


Basic properties of convex sets 

Theorem 

Let S and T be convex. Then 

αS := {x | x = αs, s ∈ S} is convex 

S + T := {x | x = s + t, s ∈ S, t ∈ T } is convex 

closure of S and interior of S are convex 

S ∩ T := {x | x ∈ S and x ∈ T } is convex. 

x ∈ S is interior point of S if there exist ε > 0 such that all y with 

x − y ≤ ε belong to S 

x ∈ X is closure point of S ⊆ X if for all ε > 0 there exist y ∈ S with 

x − y ≤ ε 


Examples of convex sets 

With a ∈ R n \{0} and b ∈ R, the hyperplane 

and the half-space 

are convex. 

H = {x ∈ R n | a ⊤ x = b} 

H− = {x ∈ R n | a ⊤ x ≤ b} 

The intersection of finitely many hyperplanes and half-spaces is a 

polyhedron. Any polyhedron is convex and can be described as 

{x ∈ R n | Ax ≤ b, Dx = e} 

for suitable matrices A and D and vectors b, e. 

A compact polyhedron is a polytope. 


The convex hull 


The convex hull of a set S ⊂ X is 

co(S) := ∩{T | T is convex and S ⊆ T } 

co(S) is convex for any set S 

co(S) is set of all convex combinations of points of S 

The convex hull of finitely many points co(x1, . . . , xn) is a polytope. 

Moreover, any polytope can be represented in this way!! 

Latter property allows explicit representation of polytopes. For example 

{x ∈ R n | a ≤ x ≤ b} consists of 2n inequalities and requires 2 n 

generators for its representation as convex hull! 


The convex hull 


The convex hull of a set S ⊂ X is 

co(S) := ∩{T | T is convex and S ⊆ T } 

co(S) is convex for any set S 

co(S) is set of all convex combinations of points of S 

The convex hull of finitely many points co(x1, . . . , xn) is a polytope. 

Moreover, any polytope can be represented in this way!! 

Latter property allows explicit representation of polytopes. For example 

{x ∈ R n | a ≤ x ≤ b} consists of 2n inequalities and requires 2 n 

generators for its representation as convex hull! 




A function f : S → R is convex if 

S is convex and 

for all x1, x2 ∈ S, α ∈ (0, 1) there holds 

We have 

f(αx1 + (1 − α)x2) ≤ αf(x1) + (1 − α)f(x2) 

f : S → R convex =⇒ {x ∈ S | f(x) ≤ γ} 

 

Sublevel sets 

Derives convex sets from convex functions 

Converse ⇐= is not true! 

f is strictly convex if < instead of ≤ 

convex for all γ ∈ R 


Examples of convex functions 


f(x) = ax 2 + bx + c convex if 

a > 0 

f(x) = |x| 

f(x) = x 

f(x) = sin x on [π, 2π] 

Non-convex functions 

f(x) = x 3 on R 

f(x) = −|x| 

f(x) = √ x on R+ 

f(x) = sin x on [0, π] 


Affine sets 


A subset S of a linear vector space is affine if x = αx1 + (1 − α)x2 

belongs to S for every x1, x2 ∈ S and α ∈ R 

Geometric idea: line through any two points belongs to set 

Every affine set is convex 

S affine iff S = {x | x = x0 + m, m ∈ M} with M a linear subspace 


Affine functions 


A function f : S → T is affine if 

f(αx1 + (1 − α)x2) = αf(x1) + (1 − α)f(x2) 

for all x1, x2 ∈ S and for all α ∈ R 

Theorem 

If S and T are finite dimensional, then f : S → T is affine if and only if 

f(x) = f0 + T (x) 

where f0 ∈ T and T : T → T a linear map (a matrix). 


Affine functions 


A function f : S → T is affine if 

f(αx1 + (1 − α)x2) = αf(x1) + (1 − α)f(x2) 

for all x1, x2 ∈ S and for all α ∈ R 

Theorem 

If S and T are finite dimensional, then f : S → T is affine if and only if 

f(x) = f0 + T (x) 

where f0 ∈ T and T : T → T a linear map (a matrix). 


Cones and convexity 


A cone is a set K ⊂ R n with the property that 

x1, x2 ∈ K =⇒ α1x1 + α2x2 ∈ K for all α1, α2 ≥ 0. 

Since x1, x2 ∈ K implies αx1 + (1 − α)x2 ∈ K for all α ∈ (0, 1) every 

cone is convex. 

If S ⊂ X is an arbitrary set, then 

K := {y ∈ R n | 〈x, y〉 ≥ 0 for all x ∈ S} 

is a cone. Also denoted K = S ∗ and called dual cone of S. 


Why is convexity interesting ??? 

Reason 1: absence of local minima 


f : S → R. Then x0 ∈ S is a 

local optimum if ∃ε > 0 such that 

f(x0) ≤ f(x) for all x ∈ S with x − x0 ≤ ε 

global optimum if f(x0) ≤ f(x) for all x ∈ S 

Theorem 

If f : S → R is convex then every local optimum x0 is a global optimum 

of f. If f is strictly convex, then the global optimum x0 is unique. 



Reason 1: absence of local minima 


f : S → R. Then x0 ∈ S is a 

local optimum if ∃ε > 0 such that 

f(x0) ≤ f(x) for all x ∈ S with x − x0 ≤ ε 

global optimum if f(x0) ≤ f(x) for all x ∈ S 

Theorem 

If f : S → R is convex then every local optimum x0 is a global optimum 

of f. If f is strictly convex, then the global optimum x0 is unique. 



Reason 2: uniform bounds 

Theorem 

Suppose S = co(S0) and f : S → R is convex. Then equivalent are: 

f(x) ≤ γ for all x ∈ S 

f(x) ≤ γ for all x ∈ S0 

Very interesting if S0 consists of finite number of points, i.e, 

S0 = {x1, . . . , xn}. A finite test!! 



Reason 3: subgradients 


A vector g = g(x0) ∈ R n is a subgradient of f at x0 if 

for all x ∈ S 

f(x) ≥ f(x0) + 〈g, x − x0〉 

Geometric idea: graph of affine function x ↦→ f(x0) + 〈g, x − x0〉 tangent 

to graph of f at (x0, f(x0)). 

Theorem 

A convex function f : S → R has a subgradient at every interior point x0 

of S. 



Reason 3: subgradients 


A vector g = g(x0) ∈ R n is a subgradient of f at x0 if 

for all x ∈ S 

f(x) ≥ f(x0) + 〈g, x − x0〉 

Geometric idea: graph of affine function x ↦→ f(x0) + 〈g, x − x0〉 tangent 

to graph of f at (x0, f(x0)). 

Theorem 

A convex function f : S → R has a subgradient at every interior point x0 

of S. 


Examples and properties of subgradients 

f differentiable, then g = g(x0) = ∇f(x0) is subgradient 

So, for differentiable functions every gradient is a subgradient 

the non-differentiable function f(x) = |x| has any real number 

g ∈ [−1, 1] as its subgradient at x0 = 0. 

f(x0) is global minimum of f if and only if 0 is subgradient of f at x0. 

Since 

〈g, x − x0〉 > 0 =⇒ f(x) > f(x0), 

all points in half space H := {x | 〈g, x − x0〉 > 0} can be discarded in 

searching for minimum of f. 

Used explicitly in ellipsoidal algorithm 



Aim: Minimize convex function f : R n → R 

Step 0 Let x0 ∈ R n and P0 > 0 such that all minimizers of f are 

located in the ellipsoid 

E0 := {x ∈ R n | (x − x0) ⊤ P −1 

0 (x − x0) ≤ 1}. 

Set k = 0. 

Step 1 Compute a subgradient gk of f at xk. If gk = 0 then stop, 

otherwise proceed to Step 2. 

Step 2 All minimizers are contained in 

Hk := Ek ∩ {x | 〈gk, x − xk〉 ≤ 0}. 

Step 3 Compute xk+1 ∈ R n and Pk+1 > 0 with minimal 

determinant det Pk+1 such that 

Ek+1 := {x ∈ R n | (x − xk+1) ⊤ P −1 

k+1 (x − xk+1) ≤ 1} 

contains Hk. 

Step 4 Set k to k + 1 and return to Step 1. 



Remarks ellipsoidal algorithm: 

Convergence f(xk) → infx f(x). 

Exist explicit equations for xk, Pk, Ek such that volume of Ek 

decreases with e −1/2n . 

Simple, robust, easy to implement, but slow convergence. 



Reason 4: Duality and convex programs 

Set of feasible decisions often described by equality and inequality 

constraints: 

S = {x ∈ X | gk(x) ≤ 0, k = 1, . . . , K, hℓ(x) = 0, ℓ = 1, . . . , L} 

Primal optimization: 

Popt = inf 

x∈S f(x) 

One of index sets K or L infinite: semi-infinite optimization 

Both index sets K and L finite: nonlinear program 



Examples: saturation constraints, safety margins, physically 

meaningful variables, constitutive and balance equations all assume 

the form S. 

linear program: 

f(x) = c ⊤ x, g(x) = g0 + Gx, h(x) = h0 + Hx 

quadratic program: 

f(x) = x ⊤ Qx, g(x) = g0 + Gx, h(x) = h0 + Hx 

quadratically constraint quadratic program: 

f(x) = x ⊤ Qx+2s ⊤ x+r, gj(x) = x ⊤ Qjx+2s ⊤ j x+rj, h(x) = h0+Hx 





the form S. 











the form S. 











the form S. 








Upper and lower bounds for convex programs 

Primal optimization problem 


x∈X 

f(x) 

subject to g(x) ≤ 0, h(x) = 0 

Can we obtain bounds on optimal value Popt ? 

Upper bound on optimal value 

For any x0 ∈ S we have 

Popt ≤ f(x0) 

which defines an upper bound on Popt. 



Primal optimization problem 


x∈X 

f(x) 

subject to g(x) ≤ 0, h(x) = 0 

Can we obtain bounds on optimal value Popt ? 

Upper bound on optimal value 

For any x0 ∈ S we have 

Popt ≤ f(x0) 

which defines an upper bound on Popt. 



Lower bound on optimal value 

Let x ∈ S. Then for arbitrary y ≥ 0 and z we have 

L(x, y, z) := f(x) + 〈y, g(x)〉 + 〈z, h(x)〉 ≤ f(x) 

and, in particular, 

so that 

ℓ(y, z) := inf L(x, y, z) ≤ inf L(x, y, z) ≤ inf f(x) = Popt. 

x∈X x∈S x∈S 

Dopt := sup ℓ(y, z) = sup 

y≥0, z 

y≥0, z 

defines a lower bound for Popt. 


x∈X 

L(x, y, z) ≤ Popt 



Lower bound on optimal value 

Let x ∈ S. Then for arbitrary y ≥ 0 and z we have 

L(x, y, z) := f(x) + 〈y, g(x)〉 + 〈z, h(x)〉 ≤ f(x) 

and, in particular, 

so that 

ℓ(y, z) := inf L(x, y, z) ≤ inf L(x, y, z) ≤ inf f(x) = Popt. 

x∈X x∈S x∈S 

Dopt := sup ℓ(y, z) = sup 

y≥0, z 

y≥0, z 

defines a lower bound for Popt. 


x∈X 

L(x, y, z) ≤ Popt 


Duality and convex programs 

Some terminology: 

Remarks: 

Lagrange function: L(x, y, z) 

Lagrange dual cost: ℓ(y, z) 

Lagrange dual optimization problem: 

Dopt := sup ℓ(y, z) 

y≥0, z 

ℓ(y, z) computed by solving an unconstrained optimization problem. 

Is concave function. 

Dual problem is concave maximization problem. Constraints are 

simpler than in primal problem 

Main question: when is Dopt = Popt? 


Example of duality 

Primal Linear Program 

Lagrange dual cost 

Dual Linear Program 


x 

c ⊤ x 

subject to x ≥ 0, b − Ax = 0 

ℓ(y, z) = inf c 

x ⊤ x − y ⊤ x + z ⊤ (b − Ax) 

 

b 

= 

⊤z if c − A⊤z − y = 0 

−∞ otherwise 

Dopt = sup 

z 

b ⊤ z 

subject to y = c − A ⊤ z ≥ 0 


Example of duality 

Primal Linear Program 

Lagrange dual cost 

Dual Linear Program 


x 

c ⊤ x 

subject to x ≥ 0, b − Ax = 0 

ℓ(y, z) = inf c 

x ⊤ x − y ⊤ x + z ⊤ (b − Ax) 

 

b 

= 

⊤z if c − A⊤z − y = 0 

−∞ otherwise 

Dopt = sup 

z 

b ⊤ z 

subject to y = c − A ⊤ z ≥ 0 


Karush-Kuhn-Tucker and duality 

We need the following property: 


Suppose f, g convex and h affine. 

(g, h) satisfy the constraint qualification if ∃x0 in the interior of X with 

g(x0) ≤ 0, h(x0) = 0 such that gj(x0) < 0 for all component functions gj 

that are not affine. 

Example: (g, h) satisfies constraint qualification if g and h are affine. 



Theorem (Karush-Kuhn-Tucker) 

If (g, h) satisfies the constraint qualification, then we have strong duality: 

Dopt = Popt. 

There exist yopt ≥ 0 and zopt, such that Dopt = ℓ(yopt, zopt). 

Moreover, xopt is an optimal solution of the primal optimization problem 

and (yopt, zopt) is an optimal solution of the dual optimization problem, if 

and only if 

1 g(xopt) ≤ 0, h(xopt) = 0, 

2 yopt ≥ 0 and xopt minimizes L(x, yopt, zopt) over all x ∈ X and 

3 〈yopt, g(xopt)〉 = 0. 





Dopt = Popt. 




and only if 

1 g(xopt) ≤ 0, h(xopt) = 0, 


3 〈yopt, g(xopt)〉 = 0. 





Dopt = Popt. 




and only if 

1 g(xopt) ≤ 0, h(xopt) = 0, 


3 〈yopt, g(xopt)〉 = 0. 



Remarks: 

Very general result, strong tool in convex optimization 

Dual problem simpler to solve, (yopt, zopt) called Kuhn Tucker point. 

The triple (xopt, yopt, zopt) exist if and only if it defines a saddle point 

of the Lagrangian L in that 

L(xopt, y, z) ≤ L(xopt, yopt, zopt) 

 

for all x, y ≥ 0 and z. 

=Popt=Dopt 

≤ L(x, yopt, zopt) 


Linear Matrix Inequalities 


A linear matrix inequality (LMI) is an expression 

where 

F (x) = F0 + x1F1 + . . . + xnFn ≺ 0 

x = col(x1, . . . , xn) is a vector of real decision variables, 

Fi = F ⊤ 

i 

are real symmetric matrices and 

≺ 0 means negative definite, i.e., 

F (x) ≺ 0 ⇔ u ⊤ F (x)u < 0 for all u = 0 

⇔ all eigenvalues of F (x) are negative 

⇔ λmax (F (x)) < 0 

F is affine function of decision variables 


Recap: Hermitian and symmetric matrices 


For a real or complex matrix A the inequality A ≺ 0 means that A is 

Hermitian and negative definite. 

A is Hermitian is A = A ∗ = Ā⊤ . If A is real this amounts to A = A ⊤ 

and we call A symmetric. 

Set of n × n Hermitian or symmetric matrices: H n and S n . 

All eigenvalues of Hermitian matrices are real. 

By definition a Hermitian matrix A is negative definite if 

u ∗ Au < 0 for all complex vectors u = 0 

A is negative definite if and only if all its eigenvalues are negative. 

A B, A ≻ B and A B defined and characterized analogously. 


Simple examples of LMI’s 

1 + x < 0 

1 + x1 + 2x2 < 0 

 

1 

0 

 

0 2 

+ x1 

1 −1 

 

−1 1 

+ x2 

2 0 

 

0 

≺ 0. 

0 

All the same with ≻ 0, 0 and 0. 

Only very simple cases can be treated analytically. 

Need to resort to numerical techniques! 


Main LMI problems 

1 LMI feasibility problem: Test whether there exists x1, . . . , xn such 

that F (x) ≺ 0. 

2 LMI optimization problem: Minimize f(x) over all x for which the 

LMI F (x) ≺ 0 is satisfied. 

How is this solved? 

F (x) ≺ 0 is feasible iff minx λmax(F (x)) < 0 and therefore involves 

minimizing the function 

x ↦→ λmax (F (x)) 

Possible because this function is convex! 

There exist efficient algorithms (Interior point, ellipsoid). 


Main LMI problems 

1 LMI feasibility problem: Test whether there exists x1, . . . , xn such 

that F (x) ≺ 0. 

2 LMI optimization problem: Minimize f(x) over all x for which the 

LMI F (x) ≺ 0 is satisfied. 

How is this solved? 

F (x) ≺ 0 is feasible iff minx λmax(F (x)) < 0 and therefore involves 

minimizing the function 

x ↦→ λmax (F (x)) 

Possible because this function is convex! 

There exist efficient algorithms (Interior point, ellipsoid). 


Linear Matrix Inequalities 


(More general:) A linear matrix inequality is an inequality 

F (X) ≺ 0 

where F is an affine function mapping a finite dimensional vector space 

X to the set H of Hermitian matrices. 

Allows defining matrix valued LMI’s. 

F affine means F (X) = F0 + T (X) with T a linear map (a matrix). 

With Xj basis of X , any X ∈ X can be expanded as 

X = n 

j=1 xjXj so that 

F (X) = F0 + T (X) = F0 + 

which is the standard form. 

n 

j=1 

xjFj 

with Fj = T (Xj) 


Why are LMI’s interesting? 

Reason 1: LMI’s define convex constraints on x, i.e., 

S := {x | F (x) ≺ 0} is convex. 

Indeed, F (αx1 + (1 − α)x2) = αF (x1) + (1 − α)F (x2) ≺ 0. 

Reason 2: Solution set of multiple LMI’s 

F1(x) ≺ 0, . . . , Fk(x) ≺ 0 

is convex and representable as one single LMI 

⎛ 

F1(x) 0 . . . 0 

⎜ 

F (x) = ⎝ 

. 

0 .. 0 

⎞ 

⎟ 

⎠ ≺ 0 

0 . . . 0 Fk(x) 

Reason 3: Incorporate affine constraints such as 

F (x) ≺ 0 and Ax = b 

F (x) ≺ 0 and x = Ay + b for some y 

F (x) ≺ 0 and x ∈ S with S an affine set. 




S := {x | F (x) ≺ 0} is convex. 


F1(x) ≺ 0, . . . , Fk(x) ≺ 0 


⎛ 

F1(x) 0 . . . 0 

⎜ 

F (x) = ⎝ 

. 

0 .. 0 

⎞ 

⎟ 

⎠ ≺ 0 

0 . . . 0 Fk(x) 

Allows to combine LMI’s! 


F (x) ≺ 0 and Ax = b 






S := {x | F (x) ≺ 0} is convex. 


F1(x) ≺ 0, . . . , Fk(x) ≺ 0 


⎛ 

F1(x) 0 . . . 0 

⎜ 

F (x) = ⎝ 

. 

0 .. 0 

⎞ 

⎟ 

⎠ ≺ 0 

0 . . . 0 Fk(x) 


F (x) ≺ 0 and Ax = b 




Affinely constrained LMI is LMI 

Combine LMI constraint F (x) ≺ 0 and x ∈ S with S affine: 

Write S = x0 + M with M a linear subspace of dimension k 

Let {ej} k j=1 

be a basis for M 

Write F (x) = F0 + T (x) with T linear 

Then for any x ∈ S: 

⎛ 

⎞ 

k 

k 

F (x) = F0 + T ⎝x0 + xjej ⎠ = F0 + T (x0) + xjT (ej) 

 

j=1 

constant 

j=1 

 

linear 

= G0 + x1G1 + . . . + xkGk = G(x) 

Result: x unconstrained and x has lower dimension than x !! 



Reason 4: conversion nonlinear constraints to linear ones 

Theorem (Schur complement) 

Let F be an affine function with 

 

F11(x) 

F (x) = 

F21(x) 

 

F12(x) 

, 

F22(x) 

F11(x) is square. 

Then 

F (x) ≺ 0 ⇐⇒ 

⇐⇒ 

 

F11(x) ≺ 0 

F22(x) − F21(x) [F11(x)] −1 F12(x) ≺ 0. 

 

F22(x) ≺ 0 

F11(x) − F12(x) [F22(x)] −1 F21(x) ≺ 0 


.

First examples in control 

Example 1: Stability 

Verify stability through feasibility 

˙x = Ax asymptotically stable ⇐⇒ 

 

−X 0 

0 A⊤ 

≺ 0 feasible 

X + XA 

Here X = X ⊤ defines a Lyapuov function V (x) := x ⊤ Xx for the flow 

˙x = Ax 



Example 2: Joint stabilization 

Given (A1, B1), . . . , (Ak, Bk), find F such that 

(A1 + B1F ), . . . , (Ak + BkF ) asymptotically stable. 

Equivalent to finding F , X1, . . . , Xk such that for j = 1, . . . , k: 

 

−Xj 

0 

0 

(Aj + BjF )Xj + Xj(Aj + BjF ) ⊤ 

 

≺ 0 not an LMI!! 

Sufficient condition: X = X1 = . . . = Xk, K = F X, yields 

 

−X 0 

 

≺ 0 an LMI!! 

0 AjX + XA ⊤ j + BjK + K ⊤ B ⊤ j 

Set feedback F = KX −1 . 



Example 3: Eigenvalue problem 

Given F : V → S affine, minimize over all x 

f(x) = λmax (F (x)) . 

Observe that, with γ > 0, and using Schur complement: 

λmax(F ⊤ (x)F (x)) < γ 2 ⇔ 1 

γ F ⊤ (x)F (x)−γI ≺ 0 ⇔ 

We can define 

y := 

 

x 

; G(y) := 

γ 

−γI F ⊤ (x) 

F (x) −γI 

 

−γI F ⊤ (x) 

; g(y) := γ 

F (x) −γI 

then G affine in y and minx f(x) = min G(y)≺0 g(y). 

 

≺ 0 


Truss topology design 

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Trusses 

Trusses consist of straight members (‘bars’) connected at joints. 

One distinguishes free and fixed joints. 

Connections at the joints can rotate. 

The loads (or the weights) are assumed to be applied at the free 

joints. 

This implies that all internal forces are directed along the members, 

(so no bending forces occur). 

Construction reacts based on principle of statics: the sum of the 

forces in any direction, or the moments of the forces about any joint, 

are zero. 

This results in a displacement of the joints and a new tension 

distribution in the truss. 

Many applications (roofs, cranes, bridges, space structures, . . . ) !! 

Design your own bridge 



Problem features: 

Goal: 

Connect nodes by N bars of length ℓ = col(ℓ1, . . . , ℓN) (fixed) and 

cross sections s = col(s1, . . . , sN) (to be designed) 

Impose bounds on cross sections ak ≤ sk ≤ bk and total volume 

ℓ ⊤ s ≤ v (and hence an upperbound on total weight of the truss). 

Let a = col(a1, . . . , aN) and b = col(b1, . . . , bN). 

Distinguish fixed and free nodes. 

Apply external forces f = col(f1, . . . fM) to some free nodes. These 

result in a node displacements d = col(d1, . . . , dM). 

Mechanical model defines relation A(s)d = f where A(s) 0 is the 

stiffness matrix which depends linearly on s. 

Maximize stiffness or, equivalently, minimize elastic energy f ⊤ d 



Problem 

Find s ∈ R N which minimizes elastic energy f ⊤ d subject to the constraints 

A(s) ≻ 0, A(s)d = f, a ≤ s ≤ b, ℓ ⊤ s ≤ v 

Data: Total volume v > 0, node forces f, bounds a, b, lengths ℓ and 

symmetric matrices A1, . . . , AN that define the linear stiffness matrix 

A(s) = s1A1 + . . . + sNAN. 

Decision variables: Cross sections s and displacements d (both 

vectors). 

Cost function: stored elastic energy d ↦→ f ⊤ d. 

Constraints: 

Semi-definite constraint: A(s) ≻ 0 

Non-linear equality constraint: A(s)d = f 

Linear inequality constraints: a ≤ s ≤ b and ℓ ⊤ s ≤ v. 


From truss topology design to LMI’s 

First eliminate affine equality constraint A(s)d = f: 

minimize f ⊤ (A(s)) −1 f 

subject to A(s) ≻ 0, ℓ ⊤ s ≤ v, a ≤ s ≤ b 

Push objective to constraints with auxiliary variable γ: 

minimize γ 

subject to γ > f ⊤ (A(s)) −1 f, A(s) ≻ 0, ℓ ⊤ s ≤ v, a ≤ s ≤ b 

Apply Schur lemma to linearize 


 

γ f ⊤ 

subject to 

≻ 0, ℓ 

f A(s) 

⊤s ≤ v, a ≤ s ≤ b 

Note that the latter is an LMI optimization problem as all constraints on s 

are formulated as LMI’s!! 











 

γ f ⊤ 

subject to 

≻ 0, ℓ 

f A(s) 

⊤s ≤ v, a ≤ s ≤ b 













 

γ f ⊤ 

subject to 

≻ 0, ℓ 

f A(s) 

⊤s ≤ v, a ≤ s ≤ b 




Yalmip coding for LMI optimization problem 

Equivalent LMI optimization problem: 


 

γ f ⊤ 

subject to 

≻ 0, ℓ 

f A(s) 

⊤s ≤ v, a ≤ s ≤ b 

The following YALMIP code solves this problem: 

gamma=sdpvar(1,1); x=sdpvar(N,1,’full’); 

lmi=set([gamma f’; f A*diag(x)*A’]); 

lmi=lmi+set(l’*x

Result: optimal truss 

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200 

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Useful software: 

General purpose MATLAB interface Yalmip 

Free code developed by J. Löfberg accessible here 

Get Yalmip now 

Run yalmipdemo.m for a comprehensive introduction. 

Run yalmiptest.m to test settings. 

Yalmip uses the usual Matlab syntax to define optimization problems. 

Basic commands sdpvar, set, sdpsettings and solvesdp. 

Truely easy to use!!! 

Yalmip needs to be connected to solver for semi-definite programming. 

There exist many solvers: 

SeDuMi PENOPT OOQP 

DSDP CSDP MOSEK 

Alternative Matlab’s LMI toolbox for dedicated control applications. 


Gallery 

Joseph-Louis Lagrange (1736) Aleksandr Mikhailovich Lyapunov (1857) 

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Linear Matrix Inequalities in Control

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