Sparse solutions of underdetermined systems

Transcription

1 Sparse solutions of underdetermined systems I-Liang Chern September 22, / 16

2 Outline Sparsity and Compressibility: the concept for measuring sparsity and compressibility of data Minimum measurements to recover sparse data: Find a measurement A so that Ax = y is solvable For general sparse data For a specific sparse data x This is a note from S. Foucart and H. Rauhut, A Mathematical Introduction to Compressive Sensing, Springer / 16

3 Sparsity Notations: [N] := {1,..., N}, S := [N]\S; A vector x C N, its support is Supp (x) := {j x j 0}. For S [N], x C N, x S is defined by { x i if i S x S,i = 0 otherwise. s-sparse vectors Σs = {x C N, x is s-sparse.} Definitions: x 0 := # Supp (x); ( N ) 1/p, x p := j=1 x j p 0 < p < ; x p x 0 as p 0. 3 / 16

4 Sparsity Non increasing rearrangement of x is defined by x 1 x 2 x N 0 where x j = x π(j) and π is a permutation on [N]. Given x, we can find its s-sparse projection x s, the part of component of x which contains s largest absolute values of x. Such x s may not be unique. A vector x is s-sparse if x = x s Best s-term approximation: An x s of x. 4 / 16

5 Compressibility Best s-term approximation error in p-norm (p 1) σ s (x) p := inf{ x z p z Σ s } A vector is called compressible if σ s (x) p decays fast in s. (e.g. σ s (x) p = O(s r ) for some p 1 and some r > 1.) Many data are compressible. The following theorem says that σ s (x) q can be controlled by x p with 0 < p < q. 5 / 16

6 Compressibility of a vector Theorem The following inequality holds where 0 < p < q and c p,q := σ s (x) q [ (p q c p,q s 1/p 1/q x p, ) p/q ( 1 p q ) 1 p/q ] 1/p 1. 6 / 16

7 Remarks 1. For q = 2, p 1, we have σ s (x) 2 1 s 1/p 1/2 x p. This suggests that the unit ball in l p quasi-norm for p 1 are good models for compressible vectors. 2. In particular, p = 1, σ s (x) s x 1. The vectors with x 1 1 can also serve as compressible vectors. 7 / 16

8 Proof. 1. Set α j = (x j) p / x p p, where x is a non-increasing rearrangement of x, then the theorem is equivalent to: implies (α 1 α N, α α N 1) α q/p s α q/p N cq p,q s q/p 1. This is a constrained optimization problem: let r = q/p > 1, we want to maximize over a convex polygon f(α 1,..., α N ) := α r s α r N C := {(α 1,..., α N ) α 1... α N 0, α α N 1} 8 / 16

9 2 The maximum occurs at corners. If α1 = = α N = 0, then f(α) = 0. If α1 + + α N = 1, α 1 = = α k > α k+1 = = α N = 0, 1 k s, then f(α) = 0. If α1 + + α N = 1, α 1 = = α k > α k+1 = = α N = 0, s + 1 k N, then α i = 1/k for 1 i k and f(α) = (k s)/k r. It follows that max f(α) = max k s α C s+1 k N k r 3 Taking k as a continuous variable, we obtain maximum at k = (r/(r 1))s. Hence max α C f(α) 1 r ( 1 1 r ) r 1 1 s = r 1 cq p,q 1 s q/p 1 9 / 16

10 Minimal Measurements for Reconstructing Sparse Vectors Linear measurement: y = Ax. A is m N. Question: minimal number of linear measurements needed to reconstruct s-sparse vectors from these measurements, regardless of the practicality of the reconstruction scheme. Two meanings: Reconstruction of all s-sparse vectors Given a specific s-sparse x, construct a measurement A such that it can recover x exactly. Minimal measurements: For the first, m = 2s. For the second, m = s / 16

11 Remarks The meaning for the second is in the measure sense, which means that the Lebesgue measure of the set of matrices which cannot recover a given specific s-sparse vector x is zero. For the recovery of all vectors, however, the reconstruction may not be stable, in the sense that a small perturbation of y causes large change of x. If we add the stability requirement, then the minimal measurements become m Cs ln(n/s), where C depends on the stability criterion. 11 / 16

12 Recover all s-sparse vectors Theorem Given A C m N, the following properties are equivalent: (a) Every s-sparse vector x C N is the unique s-sparse solution of Az = Ax, that is, if Ax = Az and x, z Σ s, then x = z. (b) The null space kera does not contain any 2s-sparse vector other than the zero vector,that is, kera Σ 2s = {0}. (c) For every S [N] with S 2s, the submatrix A S is injective as a map from C S to C m. (d) Every set of 2s columns of A is linearly independent. 12 / 16

13 Proof 1. (b) (a): If Az = Ax, then z x kera. Since both z and x are s-sparse, we get z x is 2s-sparse. From (b), z x = (a) (b): For any v kera Σ 2s, we can split v = z x such that supp z supp x = and both x and z are s-sparse. By (a), z = x. Thus, v = (b) (c) (d): For any S [N], noting that Av = A S v S. Use S = dim Dom(A S ) = dim kera S + dim RanA S, we get that kera = {0} if and only if dim RanA S = S. 13 / 16

14 Recover all s-sparse vectors We look for a measurement matrix with 2s N which can recover all s-sparse vectors. Answer 1: Given any 0 < t 1 < < t N, define t 1 t 2 t N A = t 2 1 t 2 2 t 2 N t 2s 1 1 t 2s 1 2 t 2s 1 N Let S [N] with card(s) = 2s. Any column sub matrix A S is a Vandermonde matrix, which is invertible. 14 / 16

15 Recover all s-sparse vectors Answer 2: Fourier sub matrix, we choose ω j = e 2πi(j 1)/N A = ω 1 ω 2 ω N ω1 2 ω2 2 ωn ω1 2s 1 ω2 2s 1 ω 2s 1 N 15 / 16

16 Recover a specific s-sparse vector Given an s-sparse vector, find an (s + 1) N matrix A so that the measurement y = Ax can recover x exactly via (P0). Answer: The set of all (s + 1) N matrices which cannot do so has measure zero. 16 / 16