Random Matrix Lecture 05

[[lecture-data]]

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Class Notes

Pages 27-30

Important

2. Rectangular Matrices

Geometric method later, not at all, or in the notes

2.5 Application: Compressed Sensing

eigenvalues for Gaussian random matrices

Let G∈Rd×m. The goal of compressed sensing is to recover x∈Rm from measurement(s) / observation(s) y=Gx∈Rd.

Example

Medical imaging (eg, MRI). In this case, we get to choose G.

Question

How many measurements (d) are needed to recover x∈Rm?

Answer

In general, we need G to be injective (ie, d≥m).

BUT if x has some sort of structure, then compressed sensing wants to make d much smaller.

Question

What kind of structure do we need for x?

Answer

We want x to be sparse.

Remark

Note that knowing that I can design G (called the sensing matrix) to recover sparse x is equivalent to knowing that there is a basis A in which x is sparse (thus, WLOG, we assume is sparse)

Demonstration

  • Assume that x is sparse is some known invertibleA such that x=Ax~ is sparse. (it is important that A is known as it encodes our prior assumptions about x)
  • Then we have GAx~=Gx.
  • Assuming I have designed my G to recover (sparse) x, I can then recover x~=A−1x

Suppose I have some G that recovers sparse x. Then I can also recover x~ that is sparse in basis A.

2.5.1 Null Space and Restricted Isometry Properties

Question

  1. When does Gx uniquely determine x for x sparse? (when is G injective?)
  1. How do we actually recover x (quickly)?

Sparsity

We denote the sparsity of a vector x by

||x||0:=#{i:xi≠0}=|{i:xi≠0}|

If ||x||0≤k, we say that x is k-sparse

Note

This is not a [[norm]]; it is not homogeneous.

see [[sparsity]]

Distinguishes k-sparse vectors

G distinguishes k-sparse vectors if

||x||0,||x′||0≤k,Gx=Gx′⟹x=x′

ie G is injective on {x:||x||0≤k}⊂Rn.

see [[distinguishes sparse vectors]]

If such a G exists, then yes, a way to recover x exists! Now, we want to find for a d≪m.

k-nullspace property

G has the k-nullspace property (NSP) if for all ||x||0≤k,

x≠0,Gx≠0

see [[nullspace property]]

Proposition

G distinguishes k-sparse if and only if G has the 2k-NSP

Proof

(⟸) (via contrapositive)

Suppose G does not distinguish k-sparse vectors. Then there is some x≠x′ with ||x||0,||x′||0≤k such that Gx=Gx′. Note that ||x−x′||0≤2k, but

Gx−Gx′=G(x−x′)≠0$$Thus$G$does∗not∗havethe$2k$NSP.
(⟹)

Now, suppose G does distinguish k sparse vectors. Suppose ||x||0≤2k and x≠0. Then there exist x′,x″ k-sparse such that x=x′−x″. Then

0≠Gx′−Gx″=G(x′−x″)=Gx

ie G has the 2k NSP.

◼

see [[distinguishing is equivalent to double nullspace property]]

(k,δ) restricted isometry property

G has the (k,δ) restricted isometry property (RIP) if for all x≠0 with ||x||0≤k we have

(1-\delta)\lvert \lvert x \rvert \rvert { #2} \leq \lvert \lvert Gx \rvert \rvert { #2} \leq (1+\delta) \lvert \lvert x \rvert \rvert { #2}

see [[restricted isometry property]]

Restricted to sparse vectors, G is approximately an isometry.

Theorem

If G has (k,δ) RIP with δ<1, then G has k NSP

This follows immediately from the definition.

See [[restricted isometry implies nullspace property]]

Note

The RIP is more useful than the NSP for algorithms, especially when noise is added to y=Gx⟹y=Gx+ε

2.5.2 Random Sensing Matrices

Theorem

Suppose k≥1 and δ∈(0,1) and d≥64klog⁡mδ2. Let G∼N(0,1d)⊗d×m. Then

P[G has (k,δ)-RIP]≥1−2mk

(We want to reduce the number of measurements required)

Proof

Suppose S⊂[m],|S|=k and x∈Rm. Define

  1. x(S)∈Rk as the restriction to indices in S
  2. G(S)∈Rd×k be the restriction to columns with indices in S

If ||x||0≤k and all non-zero indices are in S, ie supp(x)⊂S, then we have

Gx=G(S)x(S) and ||x||=||x(S)||

In this case, the (k,δ) RIP is equivalent to requiring that
(for all S⊂[m] with |S|=k) we have

(1-\delta)\lvert \lvert x^{(S)} \rvert \rvert { #2} \leq \lvert \lvert G^{(S)}x^{(S)} \rvert \rvert { #2} \leq (1+\delta) \lvert \lvert x^{(S)} \rvert \rvert { #2}

Now, let x^(S):=x(S)||x(S)||. Then

⟺−δ≤x^(S)(G(S)TG(S)−Ik)x^(S)≤δ⟺||G(S)TG(S)−Ik||≤δ

So, applying the union bound, we can see

P[G not (k,δ)-RIP]≤UB∑S⊆[m],|S|=kP[||G(S)TG(S)−I+k||>δ]

Now, Law(G(S))=N(0,1d)⊗d×m. So introduce H∼N(0,1)⊗k×d. Then we have

P[G not (k,δ)-RIP]≤(mk)P[||HHT−dIk||>δd](∗)≤(mk)⋅2exp⁡(−18min{δ2d24d,δd2})

Where we get (∗) from the last time. And continuing the calculation, we see

P[G not (k,δ)-RIP]≤mk⋅2exp⁡(−δ232d)=2exp⁡(klog⁡m−δ232d)≤2exp⁡(−klog⁡m)=2mk◼

see [[dimension requirements for gaussian random matrix to be a restricted isometry with high probability]]

Summary

To recover x∈Rm (information theoretically)


Next Time:
Singular Vectors of G∼N(0,1)⊗d×m ([[singular value decomposition]])

G=UΣVT=∑1dσiuiviT

ui(G),vi(G) with σd≤⋯≤σ1

Only well-defined (up to sign) if the singular values are distinct

Theorem
The singular values of G are almost surely distinct.

Review

#flashcards/math/rmt

Remark

Note that knowing that I can design G (called the sensing matrix) to recover sparse x is equivalent to {1||knowing that I can do so and that there is a basis A in which x is sparse} (thus, WLOG, we assume x is {1||sparse})

Demonstration

TODO

const { dateTime } = await cJS()

return function View() {
	const file = dc.useCurrentFile();
	return <p class="dv-modified">Created {dateTime.getCreated(file)}     ֍     Last Modified {dateTime.getLastMod(file)}</p>
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