LMI MODELLING 4. CONVEX LMI MODELLING. Didier HENRION. LAASCNRS 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 LAASCNRS 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 levelsets) 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 liftedlmi 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 liftedlmi representability We say that a convex set X R n is liftedlmi 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 liftedlmi 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 xspace and u are additional, or lifting variables In other words, lifting variables u are not allowed in LMI representations
15 LMI and liftedlmi functions Similarly, a convex function f representable if its epigraph : R n R is LMI (or liftedlmi) {x, t : f(x) t} is an LMI (or liftedlmi) representable set x x 1
16 Conic quadratic forms The Lorentz, or icecream 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 secondorder conic sets are LMI representable In the sequel we give a list of LMI and liftedlmi representable sets (following BenTal, 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 liftedlmi 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 liftedlmi representable
23 Rational power functions Similarly, the decreasing rational power functions g(x) = x p/q, x 0 with rational p/q 0, are liftedlmi representable
24 Rational power functions Example: {x, t : x 0, x 7/3 t} Start from liftedlmi 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 liftedlmi representable Indeed {x, t : x 2p t} {x, y, t : x 2 y} and {x, y, t : y 0, y p t}, both liftedlmi representable Use lifting y and project back onto x, t Similarly, even power polynomials are liftedlmi 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 (finitedimensional) 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 nbyn symmetric matrix X The epigraph {X = X T R n n, t R : S k (X) t} is liftedlmi 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 liftedlmi representable [ ] X T 0 diag t (δ 1 δ n ) q since we know that the latter constraint (hypograph of a concave monomial) is liftedlmi 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 ndimensional 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 LöwnerJohn 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 nbym matrix X Then the epigraph {X R n m, t R : Σ k (X) t} is liftedlmi representable since X T ([ 0 σ i (X) = λ i X 0 ]) and the sum of largest eigenvalues of a symmetric matrix is liftedlmi representable
37 Positive polynomials The set of univariate polynomials that are positive on the real axis is liftedlmi 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 Sumofsquares 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 sumofsquares (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 skewsymmetric
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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