An Introduction to Linear Matrix Inequalities

Raktim Bhattacharya Aerospace Engineering, Texas A&M University Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Linear Matrix Inequalities What are they?

Inequalities involving matrix variables Matrix variables appear linearly Represent convex sets polynomial inequalities Critical tool in post-modern control theory

AERO 632, Instructor: Raktim Bhattacharya 2 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Standard Form

F (x) := F0 + x1F1 + ··· + xnFn > 0 where   x1 x   2 m x :=  .  ,Fi ∈ S m × m symmetric matrix  .  xn Think of F (x): Rn 7→ Sm. Example: 1 x 1 0 0 1 > 0 ⇔ + x > 0. x 1 0 1 1 0

AERO 632, Instructor: Raktim Bhattacharya 3 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Positive Deﬁniteness

Matrix F > 0 represents positive definite matrix F > 0 ⇐⇒ xT F x > 0, ∀x =6 0 F > 0 ⇐⇒ leading principal minors of F are positive Let   F11 F12 F13 ···   F21 F22 F23 ···  F =   F31 F32 F33 ··· ············

n Polynomial Constraints as a Linear Matrix Inequality

F11 F12 F13 F11 F12 F > 0 ⇐⇒ F11 > 0, > 0, F21 F22 F23 > 0, ··· F21 F22 F31 F32 F33

AERO 632, Instructor: Raktim Bhattacharya 4 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Deﬁniteness

Positive Semi-Definite F ≥ 0 ⇐⇒ iff all principal minors are ≥ 0 not just leading Negative Definite F < 0 ⇐⇒ iff every odd leading principal minor is < 0 and even leading principal minor is > 0 they alternate signs, starting with < 0 Negative Semi-Definite F ≤ 0 ⇐⇒ iff every odd principal minor is ≤ 0 and even principal minor is ≥ 0

F > 0 ⇐⇒ −F < 0 F ≥ 0 ⇐⇒ −F ≤ 0

Matrix Analysis, Roger Horn.

AERO 632, Instructor: Raktim Bhattacharya 5 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Example 1

y x y > 0, y − x2 > 0, ⇐⇒ > 0 x 1 y x LMI written as > 0 is in general form. x 1 We can write it in standard form as 0 0 1 0 0 1 + y + x > 0 0 1 0 0 1 0

General form saves notations, may lead to more eﬀicient computation

AERO 632, Instructor: Raktim Bhattacharya 6 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Example 2

  1 0 x1 2 2 ⇐⇒   x1 + x2 < 1 0 1 x2 > 0 x1 x2 1

Leading Minors are

1 > 0 1 0 > 0 0 1

1 x2 1 x1 0 1 1 − 0 + x1 > 0 x2 1 x1 1 x1 x2

Last inequality simplifies to − 2 2 1 (x1 + x2) > 0

AERO 632, Instructor: Raktim Bhattacharya 7 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Eigenvalue Minimization

n Let Ai ∈ S , i = 0, 1, ··· , n.

Let A(x) := A0 + A1x1 + ··· + Anxn.

T Find x := [x1 x2 ··· xn]

that minimizes J(x) := min λmaxA(x). x How to solve this problem?

AERO 632, Instructor: Raktim Bhattacharya 8 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Eigenvalue Minimization (contd.)

Recall for M ∈ Sn λmaxM ≤ t ⇐⇒ M − tI ≤ 0.

Linear algebra result: Matrix Analysis – R.Horn, C.R. Johnson Optimization problem is therefore

min t x,t such that A(x) − tI ≤ 0.

AERO 632, Instructor: Raktim Bhattacharya 9 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Matrix Norm Minimization

n Let Ai ∈ R , i = 0, 1, ··· , n.

Let A(x) := A0 + A1x1 + ··· + Anxn.

T Find x := [x1 x2 ··· xn]

that minimizes J(x) := min kA(x)k2. x How to solve this problem?

AERO 632, Instructor: Raktim Bhattacharya 10 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Matrix Norm Minimization contd.

Recall T kAk2 := λmaxA A.

Implies min t2 t,x A(x)T A(x) − t2I ≤ 0. or Optimization problem is therefore tI A(x) min t2 subject to ≥ 0. t,x A(x)T tI

AERO 632, Instructor: Raktim Bhattacharya 11 / 39 Important Inequalities Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Generalized Square Inequalities

Lemma For arbitrary scalar x, y, and δ > 0, we have √ y 2 1 δx − √ = δx2 + y2 − 2xy ≥ 0. δ δ Implies 1 2xy ≤ δx2 + y2. δ

AERO 632, Instructor: Raktim Bhattacharya 13 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Generalized Square Inequalities Restriction-Free Inequalities

Lemma Let X,Y ∈ Rm×n,F ∈ Sm, F > 0, and δ > 0 be a scalar, then

XT FY + Y T FX ≤ δXT FX + δ−1Y T FY .

When X = x and Y = y

2xT F y ≤ δxT F x + δ−1yT F y.

Proof: Using completion of squares. √ √ T √ √ δX − δ−1Y F δX − δ−1Y ≥ 0.

AERO 632, Instructor: Raktim Bhattacharya 14 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Generalized Square Inequalities Inequalities with Restrictions

Let F = {F | F ∈ Rn×n,F T F ≤ I}.

Lemma Let X ∈ Rm×n,Y ∈ Rn×m, then for arbitrary δ > 0

XFY + Y T F T XT ≤ δXXT + δ−1Y T Y , ∀F ∈ F.

Proof: Approach 1: Using completion of squares. Start with √ √ T √ √ δXT − δ−1FY δXT − δ−1FY ) ≥ 0.

AERO 632, Instructor: Raktim Bhattacharya 15 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Generalized Square Inequalities Inequalities with Restrictions

Let F = {F | F ∈ Rn×n,F T F ≤ I}. Lemma Let X ∈ Rm×n,Y ∈ Rn×m, then for arbitrary δ > 0

XFY + Y T F T XT ≤ δXXT + δ−1Y T Y , ∀F ∈ F.

Proof: Approach 2: Use following Lemma. (To do)

AERO 632, Instructor: Raktim Bhattacharya 16 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Schur Complements Very useful for identifying convex sets

Let Q(x) ∈ Sm1 ,R(x) ∈ Sm2 Q(x),R(x),S(x) are aﬀine functions of x Q(x) S(x) Q(x) > 0 > 0 ⇐⇒ ST (x) R(x) R(x) − ST (x)Q(x)−1S(x) > 0 Generalizing,

Q(x) ≥ 0 Q(x) S(x) ≥ 0 ⇐⇒ ST (x) I − Q(x)Q†(x) = 0 ST (x) R(x) R(x) − ST (x)Q(x)†S(x) ≥ 0

Q(x)† is the pseudo-inverse This generalization is used when Q(x) is positive semidefinite but singular

AERO 632, Instructor: Raktim Bhattacharya 17 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Schur Complement Lemma

Let A A A := 11 12 . A21 A22 Deﬁne

− −1 Sch(A11) := A22 A21A11 A12 − −1 Sch(A22) := A11 A12A22 A21

For symmetric A,

A > 0 ⇐⇒ A11 > 0,Sch(A11) > 0 ⇐⇒ A22 > 0,Sch(A22) > 0

AERO 632, Instructor: Raktim Bhattacharya 18 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Example 1

I x x2 + x2 < 1 ⇐⇒ 1 − xT x > 0 ⇐⇒ > 0 1 2 xT 1 Here R(x) = 1, Q(x) = I > 0.

AERO 632, Instructor: Raktim Bhattacharya 19 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Example 2

P −1 x kxk < 1 ⇐⇒ 1 − xT P x > 0 ⇐⇒ > 0 P xT 1 or √ √ √ I ( P x) 1 − xT P x = 1 − ( P x)T ( P x) > 0 ⇐⇒ √ > 0 ( P x)T 1 √ where P is matrix square root.

AERO 632, Instructor: Raktim Bhattacharya 20 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs LMIs are not unique

If F is positive definite then congruence transformation of F is also positive definite

F > 0 ⇐⇒ xT F x, ∀x =6 0 ⇐⇒ yT M T F My > 0, ∀y =6 0 and nonsingular M ⇐⇒ M T F M > 0

Implies, rearrangement of matrix elements does not change the feasible set QS 0 I QS 0 I RST > 0 ⇐⇒ > 0 ⇐⇒ > 0 ST R I 0 ST R I 0 SQ

AERO 632, Instructor: Raktim Bhattacharya 21 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Variable Elimination Lemma

Lemma: For arbitrary nonzero vectors x, y ∈ Rn, there holds

max (xT F y)2 = (xT x)(yT y). F ∈F:F T F ≤I

Proof: From Schwarz inequality, √ p |xT F y| ≤ xT x yT F T F y √ p ≤ xT x yT y.

Therefore for arbitrary x, y we have

(xT F y)2 ≤ (xT x)(yT y).

Next show equality.

AERO 632, Instructor: Raktim Bhattacharya 22 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Variable Elimination Lemma contd.

Let xyT F0 = √ p . xT x yT y Therefore, yxT xyT yyT F T F = = . 0 0 (xT x)(yT y) yT y We can show that T T σmax(F0 F0) = σmax(F0F0 ) = 1.

⇒ T ≤ ∈ F = F0 F0 1, thus F0 .

AERO 632, Instructor: Raktim Bhattacharya 23 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Variable Elimination Lemma contd.

Therefore, ! 2 T T 2 T xy T T (x F0y) = x √ p y = (x x)(y y). xT x yT y

AERO 632, Instructor: Raktim Bhattacharya 24 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Variable Elimination Lemma contd.

Lemma: Let X ∈ Rm×n,Y ∈ Rn×m, and Q ∈ Rm×m. Then

Q + XFY + Y T F T XT < 0, ∀F ∈ F,

iff ∃ δ > 0 such that 1 Q + δXXT + Y T Y < 0. δ Proof: Suﬀiciency

T T T T 1 T Q + XFY + Y F X ≤ Q + δXX + Y Y from previous Lemma δ < 0.

AERO 632, Instructor: Raktim Bhattacharya 25 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Variable Elimination Lemma contd.

Proof: Necessity Suppose Q + XFY + Y T F T XT < 0, ∀F ∈ F is true. Then for arbitrary nonzero x xT (Q + XFY + Y T F T XT )x < 0, or xT Qx + 2xT XF Y x < 0. Using previous lemma result q max(xT XF Y x) = (xT XXT x)(xT Y T Y x), F ∈F q =⇒ xT Qx + 2 (xT XXT x)(xT Y T Y x) < 0.

AERO 632, Instructor: Raktim Bhattacharya 26 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Variable Elimination Lemma contd.

q xT Qx + 2 (xT XXT x)(xT Y T Y x) < 0

q =⇒ xT Qx − 2 (xT XXT x)(xT Y T Y x) < 0, and xT Qx < 0.

Therefore, T 2 − T T T T |(x {zQx)} 4 (|x XX{z x}) |(x Y{z Y x}) > 0. b2 a c or b2 − 4ac > 0.

AERO 632, Instructor: Raktim Bhattacharya 27 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Variable Elimination Lemma contd.

Or the quadratic equation aδ2 + bδ + c = 0 has real-roots √ −b b2 − 4ac . 2a Recall,

a := (xT XXT x) > 0, b := (xT Qx) < 0, c := (xT Y T Y x) > 0.

Implies b − > 0, 2a or at least one positive root.

AERO 632, Instructor: Raktim Bhattacharya 28 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Variable Elimination Lemma contd.

Therefore, ∃δ > 0 such that aδ2 + bδ + c < 0. Dividing by δ we get c aδ + b + < 0, δ or 1 xT Qx + δxT XXT x + xT Y T Y x < 0, δ or 1 xT (Q + δXXT + Y T Y )x < 0, δ or 1 Q + δXXT + Y T Y < 0. δ

AERO 632, Instructor: Raktim Bhattacharya 29 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Elimination of Variables In a Partitioned Matrix

Lemma: Let Z11 Z12 ∈ Rn×n Z = T ,Z11 , Z12 Z22 be symmetric. Then ∃ X = XT such that   Z11 − XZ12 X  T  ⇐⇒ Z12 Z22 0 < 0 Z < 0. X 0 −X

Proof: Apply Schur complement lemma.   Z11 − XZ12 X  T  ⇐⇒ − − Z12 Z22 0 < 0 X < 0,Sch( X) < 0. X 0 −X

AERO 632, Instructor: Raktim Bhattacharya 30 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Elimination of Variables In a Partitioned Matrix (contd.)

0 > S (−X), ch Z − XZ X − = 11 12 − (−X) 1 X 0 , ZT Z 0 12 22 Z − XZ X 0 = 11 12 + , ZT Z 0 0 12 22 Z11 Z12 = T . Z12 Z22

AERO 632, Instructor: Raktim Bhattacharya 31 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Elimination of Variables In a Partitioned Matrix (contd.)

Lemma:   Z11 Z12 Z Z Z T < 0 11 12 13 Z Z22  T T  ⇐⇒ 12 Z12 Z22 Z23 + X < 0 T T Z Z Z Z + XZ33 11 13 13 23 T < 0, Z13 Z33 with T −1 − T X = Z13Z11 Z12 Z23. Proof: Necessity ⇒ Apply rules for negative definiteness.

Suﬀiciency ⇐ Following are true from Schur complement lemma.

Z11 < 0 − T −1 − T −1 Z22 Z12Z11 Z12 < 0 Z33 Z13Z11 Z13 < 0

AERO 632, Instructor: Raktim Bhattacharya 32 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Elimination of Variables In a Partitioned Matrix (contd.)

Look at Schur complement of   Z11 Z12 Z13  T T  Z12 Z22 Z23 + X . T T Z13 Z23 + XZ33

Z Z + XT ZT 22 23 − 12 Z−1 Z Z ZT + XZ ZT 11 12 13 23 33 13 − T −1 T − T −1 Z22 Z12Z11 Z12 Z23 + X Z12Z11 Z13 = T − T −1 − T −1 Z23 + X Z13Z11 Z12 Z33 Z13Z11 Z13 < 0.

Also Z!1 < 0.

AERO 632, Instructor: Raktim Bhattacharya 33 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Elimination of Variables Projection Lemma

m×n Deﬁnition Let A ∈ R . Then Ma is left orthogonal complement of A if it satisfies

MaA = 0, rank(Ma) = m − rank(A).

m×n Deﬁnition Let A ∈ R . Then Na is right orthogonal complement of A if it satisfies

ANa = 0, rank(Na) = n − rank(A).

AERO 632, Instructor: Raktim Bhattacharya 34 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Elimination of Variables Projection Lemma (contd.)

T Lemma: Let P,Q, and H = H be matrices of appropriate dimensions. Let Np, Nq be right orthogonal complements of P,Q respectively.

Then ∃ X such that

T T T ⇐⇒ T T H + P X Q + Q XP < 0 Np HNp < 0 and Nq HNq < 0.

Proof: Necessity ⇒: Multiply by Np or Nq. Suﬀiciency ⇐: Little more involved – Use base kernel of P,Q, followed by Schur complement lemma.

AERO 632, Instructor: Raktim Bhattacharya 35 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Elimination of Variables Reciprocal Projection Lemma

Lemma: Let P be any given positive definite matrix. The following statements are equivalent: 1. Ψ + S + ST < 0. 2. The LMI problem Ψ + P − (W + W T ) ST + W T < 0, S + W −P

is feasible with respect to W . Proof: Apply projection lemma w.r.t general variable W . Let Ψ + PST X = , Y = −I 0 , Z = I −I . S −P n n n

AERO 632, Instructor: Raktim Bhattacharya 36 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Elimination of Variables Reciprocal Projection Lemma (contd.)

Let Ψ + PST X = , Y = −I 0 , Z = I −I . S −P n n n

Right orthogonal complements of Y,Z are 0 In Ny = −1 , Nz = . −P In

Verify that YNy = 0 and ZNz = 0. We can show

T − −1 T T Ny XNy = P , Nz XNz = Ψ + S + S.

Apply projection lemma.

AERO 632, Instructor: Raktim Bhattacharya 37 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Elimination of Variables Reciprocal Projection Lemma (contd.)

T − −1 T T Ny XNy = P , Nz XNz = Ψ + S + S.

The expression Ψ + P − (W + W T ) ST + W T X + Y T W T Z + ZT WY = . S + W −P

Therefore, if

T Ny XNy < 0 Ψ + P − (W + W T ) ST + W T =⇒ < 0. T S + W −P Nz XNz < 0

AERO 632, Instructor: Raktim Bhattacharya 38 / 39 Introduction to LMIs Generalized Square Inequalities Schur Complement Lemma Variable Elimination Lemma Trace of LMIs Trace of Matrices in LMIs

Lemma Let A(x) ∈ Sm be a matrix function in Rn, and γ ∈ R > 0. The following statements are equivalent: 1. ∃ x ∈ Rn such that trA(x) < γ, 2. ∃ x ∈ Rn,Z ∈ Sm such that

A(x) < Z, trZ < γ.

Proof: Homework problem.

AERO 632, Instructor: Raktim Bhattacharya 39 / 39