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The RIP and the NSP

Module by: Mark A. Davenport. E-mail the author

Summary: This module describes the relationship between the restricted isometry property (RIP) and the null space property (NSP). Specifically, it is shown that a matrix which satisfies the RIP will also satisfy the NSP.

Next we will show that if a matrix satisfies the restricted isometry property (RIP), then it also satisfies the null space property (NSP). Thus, the RIP is strictly stronger than the NSP.

Theorem 1

Suppose that ΦΦ satisfies the RIP of order 2K2K with δ2K<2-1δ2K<2-1. Then ΦΦ satisfies the NSP of order 2K2K with constant

C = 2 δ 2 K 1 - ( 1 + 2 ) δ 2 K . C = 2 δ 2 K 1 - ( 1 + 2 ) δ 2 K .
(1)

Proof

The proof of this theorem involves two useful lemmas. The first of these follows directly from standard norm inequality by relating a KK-sparse vector to a vector in RKRK. We include a simple proof for the sake of completeness.

Lemma 1

Suppose uΣKuΣK. Then

u 1 K u 2 K u . u 1 K u 2 K u .
(2)
Proof

For any uu, u1=u, sgn (u)u1=u, sgn (u). By applying the Cauchy-Schwarz inequality we obtain u1u2 sgn (u)2u1u2 sgn (u)2. The lower bound follows since sgn (u) sgn (u) has exactly KK nonzero entries all equal to ±1±1 (since uΣKuΣK) and thus sgn (u)=K sgn (u)=K. The upper bound is obtained by observing that each of the KK nonzero entries of uu can be upper bounded by uu.

Below we state the second key lemma that we will need in order to prove Theorem 1. This result is a general result which holds for arbitrary hh, not just vectors hN(Φ)hN(Φ). It should be clear that when we do have hN(Φ)hN(Φ), the argument could be simplified considerably. However, this lemma will prove immensely useful when we turn to the problem of sparse recovery from noisy measurements later in this course, and thus we establish it now in its full generality. We state the lemma here, which is proven in "11 minimization proof".

Lemma 2

Suppose that ΦΦ satisfies the RIP of order 2K2K, and let hRNhRN, h0h0 be arbitrary. Let Λ0Λ0 be any subset of {1,2,...,N}{1,2,...,N} such that |Λ0|K|Λ0|K. Define Λ1Λ1 as the index set corresponding to the KK entries of hΛ0chΛ0c with largest magnitude, and set Λ=Λ0Λ1Λ=Λ0Λ1. Then

h Λ 2 α h Λ 0 c 1 K + β Φ h Λ , Φ h h Λ 2 , h Λ 2 α h Λ 0 c 1 K + β Φ h Λ , Φ h h Λ 2 ,
(3)

where

α = 2 δ 2 K 1 - δ 2 K , β = 1 1 - δ 2 K . α = 2 δ 2 K 1 - δ 2 K , β = 1 1 - δ 2 K .
(4)

Again, note that Lemma 2 holds for arbitrary hh. In order to prove Theorem 1, we merely need to apply Lemma 2 to the case where hN(Φ)hN(Φ).

Towards this end, suppose that hN(Φ)hN(Φ). It is sufficient to show that

h Λ 2 C h Λ c 1 K h Λ 2 C h Λ c 1 K
(5)

holds for the case where ΛΛ is the index set corresponding to the 2K2K largest entries of hh. Thus, we can take Λ0Λ0 to be the index set corresponding to the KK largest entries of hh and apply Lemma 2.

The second term in Lemma 2 vanishes since Φh=0Φh=0, and thus we have

h Λ 2 α h Λ 0 c 1 K . h Λ 2 α h Λ 0 c 1 K .
(6)

Using Lemma 1,

h Λ 0 c 1 = h Λ 1 1 + h Λ c 1 K h Λ 1 2 + h Λ c 1 h Λ 0 c 1 = h Λ 1 1 + h Λ c 1 K h Λ 1 2 + h Λ c 1
(7)

resulting in

h Λ 2 α h Λ 1 2 + h Λ c 1 K . h Λ 2 α h Λ 1 2 + h Λ c 1 K .
(8)

Since hΛ12hΛ2hΛ12hΛ2, we have that

( 1 - α ) h Λ 2 α h Λ c 1 K . ( 1 - α ) h Λ 2 α h Λ c 1 K .
(9)

The assumption δ2K<2-1δ2K<2-1 ensures that α<1α<1, and thus we may divide by 1-α1-α without changing the direction of the inequality to establish Equation 5 with constant

C = α 1 - α = 2 δ 2 K 1 - ( 1 + 2 ) δ 2 K , C = α 1 - α = 2 δ 2 K 1 - ( 1 + 2 ) δ 2 K ,
(10)

as desired.

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