LMI MODELLING 4. CONVEX LMI MODELLING. Didier HENRION. LAAS-CNRS Toulouse, FR Czech Tech Univ Prague, CZ. Universidad de Valladolid, SP March 2009

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1 LMI MODELLING 4. CONVEX LMI MODELLING Didier HENRION LAAS-CNRS Toulouse, FR Czech Tech Univ Prague, CZ Universidad de Valladolid, SP March 2009

2 Minors A minor of a matrix F is the determinant of a submatrix of F, say with row index I and column index J If I = J this is a principal minor If I = J = 1... k this is a leading principal minor A symmetric matrix F is positive definite iff all its leading principal minors are positive A symmetric matrix F is positive semidefinite iff all its principal minors are nonnegative

3 Positive semidefiniteness If F R m m we have 2 m 1 principal minors A simpler and equivalent criterion follows from the fact that a univariate polynomial t f(t) = k f m k t k = k(t t k ) which has only real roots satisfies t k 0 iff f k 0 Apply to characteristic polynomial t f(t) = det(ti m + F ) = m k=0 f m k (F )t k A symmetric matrix F is positive semidefinite iff f i (F ) 0, i Only m polynomials to be checked, they are (signed) sums of principal minors

4 Geometry of LMI sets Given symmetric matrices F i we want to characterize the shape in R n of the LMI set F = {x R n : F (x) = F 0 + n i=1 x i F i 0} Build characteristic polynomial t f(t, x) = det(ti m + F (x)) = which is monic, i.e. f 0 (x) = 1 m k=0 f m k (x)t k Matrix F (x) is PSD iff f i (x) 0 for all i = 1,..., m

5 Semialgebraic description Diagonal minors are multivariate polynomials of the x i So the LMI set can be described as F = {x R n : f i (x) 0, i = 1, 2,...} which is a basic semialgebraic set (basic = intersection of polynomial level-sets) Moreover, it is a convex set

6 Example of 2D LMI feasible set F (x) = 1 x 1 x 1 + x 2 x 1 x 1 + x 2 2 x 2 0 x x 2 0 System of 3 polynomial inequalities f i (x) 0 1st order minors: f 1 (x) = 4 x 1 0 LMI set = intersection of an infinite number of halfspaces {x : y T F (x)y 0} for all y R 3

7 2nd order: f 2 (x) = 5 3x 1 + x 2 2x 2 1 2x 1x 2 2x x x 1

8 3rd order: f 3 (x) = 2 2x 1 +x 2 3x 2 1 3x 1x 2 2x 2 2 x 1x 2 2 x x x 1

9 LMI feasible set = intersection of sets {x : f i (x) 0}, i = 1, 2, x x 1 Boundary of LMI region shaped by determinant Other polynomials only isolate convex connected component

10 Example of 3D LMI feasible set F = {x R 3 : 1 x 1 x 2 x 1 1 x 3 x 2 x 3 1 } {{ } F (x) 0} A smoothened tetrahedron.. vertices correspond to points x for which rank F (x) = 1

11 Semialgebraic formulation F = {x R 3 : 1+2x 1 x 2 x 3 (x 2 1 +x2 2 +x2 3 ) 0, 3 x2 1 x2 2 x2 3 0}

12 Intersection of LMI sets Intersection of LMI feasible sets is also LMI F (x) 0 x 1 2 2x 1 + x 2 0

13 LMI sets LMI sets are convex basic semialgebraic sets.. but are all convex basic semialgebraic sets LMI? Let us make a fundamental distinction between LMI representable sets lifted-lmi representable sets We say that a convex set X R n is LMI representable if there exists an affine mapping F (x) such that x X F (x) 0

14 LMI and lifted-lmi representability We say that a convex set X R n is lifted-lmi representable if there exists an affine mapping F (x, u) such that x X u R m : F (x, u) 0 A set X is lifted-lmi representable when x X u : F (x, u) 0 i.e. when it is the projection of the solution set of the LMI F (x, u) 0 onto the x-space and u are additional, or lifting variables In other words, lifting variables u are not allowed in LMI representations

15 LMI and lifted-lmi functions Similarly, a convex function f representable if its epigraph : R n R is LMI (or lifted-lmi) {x, t : f(x) t} is an LMI (or lifted-lmi) representable set x x 1

16 Conic quadratic forms The Lorentz, or ice-cream cone is LMI representable as {x, t R n R : x 2 t} { [ tin x ] } x, t R n R : x T t 0 As a result, all second-order conic sets are LMI representable In the sequel we give a list of LMI and lifted-lmi representable sets (following Ben-Tal, Nemirovski, Nesterov)

17 Quadratic forms The Euclidean norm {x, t R n R : x 2 t} is LMI representable (see previous slide) The squared Euclidean norm {x, t R n R : x T x t} is also LMI representable as [ t x T x I n ] 0

18 Quadratic forms (2) The convex quadratic set {x R n A = A T 0 is LMI representable as : x T Ax + b T x + c 0} with [ b T x c x T D T ] Dx I n 0 where D is the Cholesky factor of A = D T D

19 Hyperbola The branch of hyperbola {x, y R 2 : x 0, xy 1} is LMI representable as [ x 1 1 y ] 0

20 Geometric mean of two variables The hypograph of the geometric mean of 2 variables {x 1, x 2, t R 3 : x 1, x 2 0, x 1 x 2 t} is LMI representable as [ x1 t t x 2 ] 0

21 Geometric mean of several variables The hypograph of the geometric mean of 2 k variables {x 1,..., x 2 k, t R 2k +1 : x i 0, (x 1 x 2 k) 1/2k t} is lifted-lmi representable Proof: iterate the previous construction by introducing lifting variables Example with k = 3: x1 x 2 x 11 x3 x 4 x 12 x5 x 6 x 13 x7 x 8 x 14 (x 1 x 2 x 8 ) 1/8 t } x11 x 12 x 21 x21 x x13 x 14 x 22 t 22 Useful idea in other LMI representability problems

22 Rational power functions Following the same ideas, the increasing rational power functions f(x) = x p/q, x 0 with rational p/q 1, are lifted-lmi representable

23 Rational power functions Similarly, the decreasing rational power functions g(x) = x p/q, x 0 with rational p/q 0, are lifted-lmi representable

24 Rational power functions Example: {x, t : x 0, x 7/3 t} Start from lifted-lmi representable ˆt (ˆx 1 ˆx 8 ) 1/8 and replace to get ˆt = ˆx 1 = x 0 ˆx 2 = ˆx 3 = ˆx 4 = t 0 ˆx 5 = ˆx 6 = ˆx 7 = ˆx 8 = 1 x x 1/8 t 3/8 x 7/8 t 3/8 x 7/3 t Same idea works for any rational p/q 1 lift = use additional variables, and project in the space of original variables

25 Even power monomial The epigraph of even power monomial F = {x, t : x 2p t} where p is a positive integer is lifted-lmi representable Indeed {x, t : x 2p t} {x, y, t : x 2 y} and {x, y, t : y 0, y p t}, both lifted-lmi representable Use lifting y and project back onto x, t Similarly, even power polynomials are lifted-lmi representable (several monomials)

26 Quartic level set Model quartic level set F = {x, t : x 4 t} as LMI in x, t and y F = {x, t : y : y x 2, t y 2, y 0} [ 1 x x y ] 0 [ 1 y y t ] 0 It can be shown that it is impossible to remove the lifting variable y while keeping a (finite-dimensional) LMI formulation

27 Quartic level set: from 3D to 2D F = {x, t : x 4 t}

28 Largest eigenvalue Function largest eigenvalue of a symmetric matrix {X = X T R n n, t R : λ max (X) t} is LMI representable as X ti n Eigenvalues of matrix [ ] 1 x1 x 1 x 2

29 Sums of largest eigenvalues Let S k (X) = k i=1 λ i (X), k = 1,..., n denote the sum of the k largest eigenvalues of an n-by-n symmetric matrix X The epigraph {X = X T R n n, t R : S k (X) t} is lifted-lmi representable as t ks trace Z 0 Z 0 Z X + si n 0 where Z and s are liftings

30 The determinant Determinant of a PSD matrix det(x) = n λ i (X) i=1 is not a convex function of X, but the function is convex when q [0, 1/n] is rational f q (X) = det q (X), X = X T 0 The epigraph {X = X T R n n, t R : f q (X) t} is lifted-lmi representable [ ] X T 0 diag t (δ 1 δ n ) q since we know that the latter constraint (hypograph of a concave monomial) is lifted-lmi representable Here is a lower triangular matrix of liftings with diagonal entries δ i

31 Application: extremal ellipsoids A little excursion in the world of ellipsoids and polytopes.. Various representations of an ellipsoid in R n E = {x R n : x T P x + 2x T q + r 0} = {x R n : (x x c ) T P (x x c ) 1} = {x = Qy + x c R n : y T y 1} = {x R n : Rx x c 1} where Q = R 1 = P 1/2 0

32 Ellipsoid volume Volume of ellipsoid E = {Qy + x c : y T y 1} vol E = k n det Q where k n is volume of n-dimensional unit ball k n = 2 (n+1)/2 π (n 1)/2 n(n 2)!! 2π n/2 n(n/2 1)! for n odd for n even n k n Unit ball has maximum volume for n = 5

33 Outer and inner ellipsoidal approximations Let S R n be a solid = a closed bounded convex set with nonempty interior the largest volume ellipsoid E in contained in S is unique and satisfies E in S ne in the smallest volume ellipsoid E out containing S is unique and satisfies E out /n S E out These are Löwner-John ellipsoids Factor n reduces to n if S is symmetric How can these ellipsoids be computed?

34 Ellipsoid in polytope Let the intersection of hyperplanes S = {x R n : a T i x b i, i = 1,..., m} describe a polytope = bounded nonempty polyhedron The largest volume ellipsoid contained in S is E = {Qy + x c : y T y 1} where Q, x c are optimal solutions of the LMI problem max det 1/n Q Q 0 Qa i 2 b i a T i x c

35 Polytope in ellipsoid Let the convex hull of vertices S = conv {x 1,..., x m } describe a polytope The smallest volume ellipsoid containing S is E = {x : (x x c ) T P (x x c ) 1} where P, x c = P 1 q are optimal solutions of the LMI problem max t t[ det 1/n ] P P q q T 0 r x T i P x i + 2x T i q + r 1

36 Let Sums of largest singular values Σ k (X) = k i=1 σ i (X), k = 1,..., n denote the sum of the k largest singular values of an n-by-m matrix X Then the epigraph {X R n m, t R : Σ k (X) t} is lifted-lmi representable since X T ([ 0 σ i (X) = λ i X 0 ]) and the sum of largest eigenvalues of a symmetric matrix is lifted-lmi representable

37 Positive polynomials The set of univariate polynomials that are positive on the real axis is lifted-lmi representable in the coefficient space Can be proved with cone duality (Nesterov) or with theory of moments (Lasserre) - more on that later The even polynomial p(s) = p 0 + p 1 s + + p 2n s 2n satisfies p(s) 0 for all s R if and only if p k = i+j=k X ij, k = 0, 1,..., 2n = trace H k X for some lifting matrix X = X T 0

38 Sum-of-squares decomposition The expression of p k with Hankel matrices H k comes from p(s) = [1 s s n ]X[1 s s n ] hence X 0 naturally implies p(s) 0 Conversely, existence of X for any polynomial p(s) 0 follows from the existence of a sum-of-squares (SOS) decomposition (with at most two elements) of p(s) = k q 2 k (s) 0 Matrix X has entries X ij = k q ki q kj Seeking the lifting matrix amounts to seeking an SOS decomposition

39 Primal and dual formulations Global minimization of polynomial p(s) = n k=0 p k s k Global optimum p : maximum value of ˆp such that p(s) ˆp 0 Primal LMI problem Dual LMI problem max ˆp = p 0 trace H 0 X s.t. trace H k X = p k, k = 1,..., n X 0 min p 0 + n k=1 p k y k s.t. H 0 + n k=1 H k y k 0 with Hankel structure (moment matrix)

40 Positive polynomials and LMIs Example: Global minimization of the polynomial p(s) = 48 92s + 56s 2 13s 3 + s 4 Solving the dual LMI problem yields p = p(5.25) = min 48 92y y 2 13y 3 + y 4 s.t. 1 y 1 y 2 y 1 y 2 y 3 0 y 2 y 3 y 4

41 Complex LMIs The complex valued LMI F (x) = A(x) + jb(x) 0 is equivalent to the real valued LMI [ A(x) B(x) B(x) A(x) ] 0 If there is a complex solution to the LMI then there is a real solution to the same LMI Note that matrix A(x) = A T (x) is symmetric whereas B(x) = B T (x) is skew-symmetric

42 INTERLUDE

43 LMI set or not? x x 1 x 1 x 2 1 and x 1 0

44 LMI x 1 x 2 1 and x 1 0 [ x1 1 1 x 2 ] 0

45 LMI set or not? x x 1 x 2 x 2 1

46 LMI x 2 x 2 1 [ 1 x1 x 1 x 2 ] 0

47 LMI set or not? x x 1 x x2 2 1

48 LMI x x2 2 1 [ 1 + x1 x 2 x 2 1 x 1 ] 0

49 LMI set or not? x x 1 {x R 2 : t 4 + 2x 1 t 2 + x 2 0, t R}

50 NOT LMI: not basic semialgebraic x x 1 x 2 x 2 1 or x 1, x 2 0

51 LMI set or not? x x 1 1 2x 1 x 2 1 x x3 1 0

52 NOT LMI: not connected x x 1 x 2 1 x x3 1 0 x 1

53 LMI set or not? x x 1 1 2x 1 x 2 1 x x3 1 0 and x 1 1 2

54 LMI 1 2x 1 x 2 1 x x3 1 0 and x x 1 0 x 1 1 x 2 0 x 2 1 2x 1 0

55 LMI set or not? x x 1 x x4 2 1

56 NOT LMI but projection of an LMI 1 + u 1 u 2 u 2 1 u 1 1 x 1 x 1 u 1 1 x 2 0 x 2 u 2 with two liftings u 1 and u 2

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