Semi-definite representibility. For fun and profit

Transcription

1 Semi-definite representation For fun and profit March 9, 2010

2 Introduction Any convex quadratic constraint x T C T Cx a + b T x Can be recast as the linear matrix inequality (LMI): ( ) In Cx (Cx) T a + b T 0 x If you introduce extra variables, you can express a great variety of polynomial constraints.

3 For example: {(x, y) R 2 : x 2 + y 4 1} Is the projection onto x, y space of {(x, y, w) R 3 : x 2 + w 2 1, y 2 w} Which is expressible as In other words: 1 x 1 w x w 1 ( 1 y y w 1 x 1 w x w 1 0 ) 1 y y w 0 0

4 LMI-representation This raises the question of which convex sets S R n can be expressed as an LMI, for some A 0,, A n symmetric nxn matrices S = {x R n : A 0 + A 1 x A n x n 0} Without loss of generality, 0 S 0, so A 0 0, and we can take A 0 = I.

5 LMIr: necessary conditions Consider f (x) = det(a 0 + A 1 x A n x n ). The zero-set of f divides R n into connected components. Claim: S 0 is the connected component containing 0. Since f (x) > 0 on S 0 and 0 S 0, S 0 must at least contain the connected component and cannot contain more. Thus S has to be semi-algebraic: S = {x R n : g 1 (x) 0,, g m (x) 0}, and the problem is (basically) equivalent to finding A 0,, A n such that f (x) = g 1 (x) g m (x). 1 1 More properly you want f to be in I (δs), the ideal of polynomials vanishing on the boundary.

6 LMIr: rigid convexity condition Let x R n and consider f (tx) = det(i + t(a 1 x 1 + A n x n )) = t n det(i /t + (A 1 x A n x n )) But A 1 x 1 + A n x n is symmetric so we have its characteristic polynomial at 1/t, which can only have real roots. This property of f (x) is called rigid convexity. It is necessary for LMI representibility (Helton and Vinnikov) and sufficient for n = 2. The problem is still open for n > 2. For example, f (x, y) = 2 x 4 y 4 has (i, i) as a root so is not rigidly convex, so does not have a determinantal representation, and the region x 4 + y 4 2 is not LMI representible. If you drop the requirement that the A 0,, A n are symmetric, then all polynomials have a determinantal representation (Helton, McCullough, Vinnikov) that can be explicitly constructed (Quarez).

7 Lifted LMI, Semidefinite representation This leads to the concept of a lifted representation, adding extra variables to make the problem LMI representible. Say S is Semidefinite representible (SDr) if there exist symmetric matrices A, B 1,, B n, C 1,, C M if x S y R M such that A + B i x i + C i y i 0 Example, the region x 4 + y 4 2. x 4 + y 4 2 u 2 + v 2 2, u 2 x, v 2 y 1 u 1 v u v 2 1 u u x 1 v v y 0

8 Which sets are SDr? Conjecture: all convex semi-algebraic sets (convex regions cut out by polynomial inequalities) are SDr. This is known (Helton Nie) in the case where you have either positive curvature or sos-convexity.

9 Positive curvature This is different from the Hessian being positive definite. Definition from differential geometry. For nondegenerate defining functions g i, we want the Hessian to be negative definite on the tangent space: v T 2 g i (x)v > 0 for all 0 v g i (x)

10 Positive curvature - example S = {x R n + : g(x) = x 1 x 2 x n 1 0. S is convex. g(x) g(x) + 1 = ( 1 x 1 1 x x n ) T 2 g(x) g(x) + 1 = 0 1 x 1 x x 1 x n 1 x 1 x x n 1 x n. 1 x 1 x n... 1 x n 1 x n 0 The Hessian has trace 0 so is not positive definite. However: 2 g(x) + g(x) g(x) T = (g(x) + 1) diag( 1 x1 2,, 1 xn 2 ) 0 The result follows.

11 Convexity How do you tell if a function is convex? Very hard (since even determining if a polynomial is always positive is hard). There is a more manageable positivity condition: sum of squares. We say that a function f (x) is sos-convex if there is sum polynomial matrix M(x) such that H f (x) = M T (x)m(x) If f (x) is sos-convex you can find M(x) by solving an SDP. There are convex functions that are not sos-convex (Ahmadi and Parrilo): p(x) = 32x x 6 x x 6 1 x x 4 1 x x 4 1 x 2 2 x x 4 1 x x 2 1 x 4 2 x x 2 1 x 2 2 x x 2 1 x x x 6 2 x x 4 2 x x 2 2 x x 8 3

12 LMI Construction. Natural SDP relaxation (Lasserre and Parrilo) Suppose S is convex given in the form S = {x R n : g 1 (x) 0,, g m (x) 0} Make this linear by setting y α = x α for each multi-index α.then the inequality: g 1 (x) 0. = g 0 + x i g i + x α A α g m (x) Becomes 0 g 0 + x i g i + y α A α = g(x, y)

13 LMI Construction cont d Now we need a constraint that requires y α to basically be x α. Let m d (x) be the vector of monomials of degree up to d Then m d (x) = ( 1 x 1 x 2... x 2 1 x 1 x 2... x d n 0 m d (x)m d (x) T = A 0 + x i A i + x α A α And since this is rank 1 one may expect the given region to be small. Let M d (x, y) = A 0 + x i A i + y α A α Our SDP relaxation is now R = {x : ys.t.g(x, y) 0, M d (x, y) 0} In many cases you can show that R = S for sufficiently large d. Usually sufficient to take 2d to be the highest degree of the g i. ) T

14 SDP relaxation: example Consider the set satisfying g(x) 0 where g(x) = 1 (x1 4 + x 2 4 x 1 2x 2 2 ). Direct computation: ( ) ( ) ( ) 2 x x1 g(x) = x x 2 ( ) 12 4 And 0. Hence g is sos-concave. The SDP 4 12 constructed is 1 y 40 y 04 + y x 1 x 2 y 20 y 11 y 02 x 1 y 20 y 11 y 30 y 21 y 12 x 2 y 11 y 03 y 21 y 12 y 03 y 20 y 30 y 21 y 40 y 31 y 22 0 y 11 y 21 y 12 y 31 y 22 y 13 y 02 y 12 y 03 y 22 y 13 y 04 The number of auxiliary variables grows exponentially in m, but you can utilize sparseness to cut this down a bit (Helton, Nie).

15 Some useful lemmas Given R 1,, R n SDr Intersection. R 1 R n is SDr Minkowski sum. R R n = {r r n : r i R i for all i} is SDr Union. conv(r 1 R n ) = {a 1 r a n r n : r i R i, a i 0, a a n = 1} is SDr. Consequence: it is enough to just have SDr locally. Theorem: R is SDr if and only if each x R has a neighbourhood that is SDr. By compactness only finitely many neighbourhoods are necessary to cover R, then apply the union result to this finite number of local SDr s. You also have flexibility in picking the defining polynomials

16 Proof of relaxation Basic idea of the proofs: R contains S. If R contains some other point a then we can seperate a from S by a linear functional l T x. Let l be the minimum of l T x on S, attained at some u S. Note l T a < l There exist Lagrange multipliers λ 1,, λ m 0 such that l = i λ i g i (u). Define f l (x) := l T x l λ i g i (x) Then f l (x) is convex and nonnegative polynomial such that f l (u) = 0 and f l (u) = 0. Use properties of the g i to express f l in a nice form.

17 R = S for sos-concave case If the g i are sos-concave then integrating the Hessian twice, you get that f l (x) is SOS, ie for some symmetric matrix W 0 Or with the y α : l T x l = λ i g i (x) + [x dg ] T W [x dg ] l T x l = λ i L gi (y) + Tr(W M dg (y)) 0 Which contradicts the separation of a. We get a similar case in general, but we need some bounded degree positivstellensatz to tell us that l T x l is in the quadratic module generated by g i, with bounded degree.

18 Sources Helton and Nie Semidefinite representation of convex sets Ahmadi and Parrilo A convex polynomial that is not sos-convex Lasserre Representation of nonnegative convex polynomials Brändén Obstructions to determinantal representability Helton and Nie Structured Semidenite Representation of Some Convex Sets Quarez Symmetric Determinantal Representations of Polynomials Helton and Vinnikov Linear matrix inequality representation of sets