' Minimization: Resolution/ Bandwidth # Samples

`1 minimization
We will now focus on underdetermined systems of equations:
# samples resolution/
= bandwidth
data
acquisition
system
unknown
signal/image
Suppose we observe y = Φx0, and given y we attempt to estimate

x0 by applying the pseudo-inverse of Φ to y — we are implicitly
solving the program
min kxk2 subject to Φx = y.

x
Of course, we will recover x0 exactly only under very special cir-

cumstances, namely that x0 is in Range(ΦT); if x0 = ΦTα for some
α ∈ RM , then
ΦT(ΦΦT)−1y = ΦT(ΦΦT)−1Φx0
= ΦT(ΦΦT)−1ΦΦTα
= ΦTα
= x0 .
So if x0 lies in a particular M -dimensional subspace of RN , it can

be recovered from observations through the M × N matrix Φ.
1
Notes by J. Romberg – October 16, 2013 – 21:17
Yesterday, we hinted that a different variational framework, one
based on `1 minimization instead of `2 minimization, would allow
us to recover sparse vectors. Given y, we solve
min kxk1 subject to Φx = y. (1)

x
It is our goal today to formalize the conditions under which this

recovery procedure is effective.
Necessary and sufficient conditions for `1 recovery
Let x0 be a vector supported on a set Γ ⊂ {1, 2, . . . , N }, and let

y = Φx0. It is clear that x0 is a solution to (1) if and only if
kx0 + hk1 ≥ kx0k1 ∀h with Φh = 0. (2)
It is always true that

X X
kx0 + hk1 − kx0k1 = (|x0[γ] + h[γ]| − |x0[γ]|) + |h[γ]|
γ∈Γ γ∈Γc
X X
≥ sgn(x0[γ])h[γ] + |h[γ]|
γ∈Γ γ∈Γc
since
|a + b| − |a| ≥ sgn(a)b for all a, b ∈ R.
Thus (2) holds when
X X
− sgn(x0[γ])h[γ] ≤ |h[γ]| for all h ∈ Null(Φ). (3)
γ∈Γ γ∈Γc
This condition is also necessary, since if there is an h ∈ Null(Φ)

with X X
|h[γ]| < − sgn(x0[γ])h[γ],
γ∈Γc γ∈Γ
2
then the same is true for h for an > 0. Since h ∈ Null(Φ),
Φ(x0 + h) = y and
X X
kx0 + hk1 = |x0[γ] + h[γ]| + |h[γ]|
γ∈Γ γ∈Γc
X
< |x0[γ] + h[γ]| − sgn(x0[γ])h[γ]
γ∈Γ
X
≤ |x0[γ]| for some small enough > 0
γ∈Γ
= kx0k1.
So this would imply that there is another vector (namely x0 + h)
that has smaller `1 norm and is also feasible.
It is interesting to note that given the observation matrix Φ, our

ability to recover a vector x0 is determined only by
1. the set Γ on which x0 is supported (the locations of its “active
elements”), and
2. the signs of the elements on this set.
The magnitudes of the entries are not involved at all. Geometri-
!"#$L
facet of the1`1!%&'(
cally, the support set coupled with the sign sequence specifies the
ball on which x0 lives:
),)$L1 (%-,.*%+
(01&(2(.$(%-,.*%+$*3
random orientation
dimension N-M
3
Duality and optimality
We can get a more workable sufficient condition for exact recovery
of x0 by looking at the dual of the convex program
min kxk1 subject to Φx = y. (4)

x
Let’s start by considering a general optimization program with

linear equality constraints:
min f (x) subject to Φx = y. (5)

x
The Lagrangian for this problem is
L(x, v) = f (x) + v T(Φx − y)
If f is differentiable, then x is a solution to (5) if it is feasible,

Φx = y, and there exists a v ∈ RM such that
∇xL(f, v) = ∇xf (x) + ΦTv = 0.
We are interested in the particular functional f (x) = kxk1, which

is not differentiable but is convex. Fortunately, there is an easy
modification to the condition above in this case — x is a solution
to (5) if it is feasible and there exists a v ∈ RM such that
ΦTv is a “subgradient” of f at x.
A vector u ∈ RN is a subgradient of a function f at x0 if it is

normal to a “supporting hyperplane”; that is if
f (x) ≥ f (x0) + uT(x − x0).
If f is differentiable at x0, then ∇f (x0) is the only subgradient.
4
Subgradients
Definition
To make this concept more concrete, consider the (very relevant)
u is a subgradient of f at x0 if for all x
one dimensional example f (t) = |t|.
f (x) f (x0 ) + u · (x x0 )
f f is differentiable at x0 , the only subgradient is rf (x0 )
Subgradients of f (t) = |t|, t 2 R

(
{subgradients} = {sgn(t)} t 6= 0
{subgradients} = [ 1, 1] t=0
In this case, the set of subgradients is simply the derivative (±1)

for t away from the origin, and all slopes between −1 and +1 at
the origin: (
sgn(t) t 6= 0
{subgradients} = .
[−1, 1] t = 0
In RN , for f (x) = kxk1, the subgradients of f at x0 are the

collection of vectors u such that
u[γ] = sgn(x0[γ]), γ ∈ Γ
|u[γ]| ≤ 1, γ ∈ Γc,
where Γ is the support set of x0.
5
Given an M × N matrix Φ and a vector x0 ∈ RN supported
on Γ, set y = Φx0. Then x0 is a solution to (4) if there exits a
v ∈ RM such that
(ΦTv)[γ] = sgn(x0[γ]), γ ∈ Γ
|(ΦTv)[γ]| ≤ 1, γ ∈ Γc.
In addition, if Φ is injective on the set of all vectors supported on

Γ and we can find a v such that the second condition is
|(ΦTv)[γ]| < 1, γ ∈ Γc
then x0 is the unique solution.
Choosing a particular dual vector
Our recovery conditions boxed above simply ask that we be able to

find one such v (there could be many). Many of the results in the
fields of sparse recovery and compressed sensing have narrowed
down the condition down by simply testing a prescribed vector.
Let
ΦΓ = the M × |Γ| submatrix containing the columns of Φ indexed by Γ
and let
z0 ∈ R|Γ| contain the signs of x0 on Γ.
We set
u0 = ΦTΦΓ(ΦTΓΦΓ)−1z0. (6)
Now sufficient conditions for x0 to be the unique minimizer of (4)
are
6
1. ΦTΓΦΓ is invertible. If this is true, then the expression above
for u0 is well-behaved, and b construction we will have u0[γ] =
sgn(x0[γ]);
2. If 1) holds, then we need |u0[γ]| < 1.
Note on complex vectors: Almost everything we have said

so far about `1 minimization extends in a straightforward manner
to complex-valued vectors. First, it is worth mentioning that if
x ∈ CN , then
N
X N q
X 2 2
kxk1 = |x[n]| = Re {x[n]} + Im {x[n]} .
n=1 n=1
In this case, the `1 minimization program can no longer be re-

cast as a linear program, but rather is what is called a “sum of
norms” program (which is a particular type of “second order cone
program”). This type of problem, however, is not too much more
difficult to solve from a practical perspective.
The sufficient conditions for recovery are the same, but now we
take sgn(z) to be the phase of the complex number z. That is, if
z = Ae jθ , then sgn(z) = θ.
7
Example: Spikes and Sines Dictionary
Let us return to the Φ created by concatenating the “spikes” or-

thobasis and the Fourier orthobasis:

Φ= I F .
Here the matrix is n × 2n. Suppose that
α0 ∈ C2n is supported on a set Γ = T ∪ Ω
where as before T denotes the “spike” support and Ω the “sine”

support.
Yesterday, in establishing the Donoho-Starck uncertainty principle,
we saw that we could write
|T | |Ω|
ΦHΓΦΓ = I + G, where kGk ≤ .
n
Thus, if
|T | |Ω| ≤ δn,
for some 0 < δ < 1, we have
1
k(ΦHΓΦΓ)−1k ≤ .
1−δ
So our first sufficient condition holds when |T | |Ω| < n, which is
implied by √
|Γ| = |T | + |Ω| < 2 n.
For the second condition, construct u0 as in (6), where now z0 are

the phases of α0 on Γ. Then for a particular entry γ ∈ Γc,
u0[γ] = ΦHγ ΦΓ(ΦHΓΦΓ)−1z0

= hwγ , z0i
8
where Φγ is the column of Φ corresponding to γ, and wγ is the
vector
wγ = (ΦHΓΦΓ)−1ΦHΓΦγ .
Using Cauchy-Schwarz,
|u0[γ]| ≤ kwγ k kz0k

q
≤ k(ΦHΓΦΓ)−1k · kΦHΓΦγ k2 · |T | + |Ω|.
The entries in√the vector ΦHΓΦγ are either equal to zero or have
magnitude 1/ n, and so
s
1 |T | + |Ω| q
|u0[γ]| ≤ · · |T | + |Ω|.
1−δ n
When can we guarantee that the expression on the right less than
1? We know that we will at least need
√
|Γ| = |T | + |Ω| ≤ n.
In this case, |T | · |Ω| ≤ n/4, so we can take δ = 1/4, and so if

√
3 n
|Γ| = |T | + |Ω| ≤
4
then
maxc |u0[γ]| < 1
γ∈Γ
and α0 will be the minimizer to (4).
9
Example: Dictionaries with small coherence
Now suppose that Φ is an M ×N matrix with normalized columns,

kΦγ k2 = 1, γ = 1, 2, . . . , N
and coherence
µ= max |hΦγ1 , Φγ2 i|.
1≤γ1 ,γ2 ≤N
γ1 6=γ2
The quantity µ is essentially a measure of how closely aligned any

two columns of Φ are.
Let Γ be a fixed subset of {1, 2, . . . , N }, and x0 be a vector sup-

ported on Γ with sign sequence z0. For the first optimality condi-
tion note that we can write ΦTΓΦΓ as
ΦTΓΦΓ = I + G
where each entry of the |Γ| × |Γ| matrix G is less than or equal to
µ. We have
kGk ≤ kGkF
sX
= |Gj,k |2
j,k
≤ µ|Γ|,
and so we can guarantee ΦTΓΦΓ is invertible when
1
|Γ| ≤ .
µ
To check the second condition, we have for γ ∈ Γc
|u0[γ]| ≤ k(ΦTΓΦΓ)−1k kΦTΓΦγ k2 kz0k2
1 q
T
≤ kΦ Φγ k2 |Γ|.
1 − µ|Γ| Γ
10
Since γ 6∈ Γ, all the entries
p of ΦTΓΦγ will have magnitude less than
µ, and so kΦTΓΦγ k2 ≤ µ |Γ|, and so
µ|Γ|
|u0[γ]| ≤ .
1 − µ|Γ|
Thus
1
|Γ| <
2µ
ensures that x0 will be recoverable from observations y = Φx0.
11

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`1 minimization

We will now focus on underdetermined systems of equations:

Suppose we observe y = Φx0, and given y we attempt to estimate

min kxk2 subject to Φx = y.

Of course, we will recover x0 exactly only under very special cir-

So if x0 lies in a particular M -dimensional subspace of RN , it can

min kxk1 subject to Φx = y. (1)

It is our goal today to formalize the conditions under which this

Necessary and sufficient conditions for `1 recovery

Let x0 be a vector supported on a set Γ ⊂ {1, 2, . . . , N }, and let

kx0 + hk1 ≥ kx0k1 ∀h with Φh = 0. (2)

It is always true that

This condition is also necessary, since if there is an h ∈ Null(Φ)

It is interesting to note that given the observation matrix Φ, our

min kxk1 subject to Φx = y. (4)

Let’s start by considering a general optimization program with

min f (x) subject to Φx = y. (5)

The Lagrangian for this problem is

L(x, v) = f (x) + v T(Φx − y)

If f is differentiable, then x is a solution to (5) if it is feasible,

∇xL(f, v) = ∇xf (x) + ΦTv = 0.

We are interested in the particular functional f (x) = kxk1, which

A vector u ∈ RN is a subgradient of a function f at x0 if it is

f (x) ≥ f (x0) + uT(x − x0).

If f is differentiable at x0, then ∇f (x0) is the only subgradient.

f f is differentiable at x0 , the only subgradient is rf (x0 )

Subgradients of f (t) = |t|, t 2 R

In this case, the set of subgradients is simply the derivative (±1)

In RN , for f (x) = kxk1, the subgradients of f at x0 are the

where Γ is the support set of x0.

In addition, if Φ is injective on the set of all vectors supported on

then x0 is the unique solution.

Choosing a particular dual vector

Our recovery conditions boxed above simply ask that we be able to

ΦΓ = the M × |Γ| submatrix containing the columns of Φ indexed by Γ

Note on complex vectors: Almost everything we have said

In this case, the `1 minimization program can no longer be re-

Let us return to the Φ created by concatenating the “spikes” or-

Here the matrix is n × 2n. Suppose that

α0 ∈ C2n is supported on a set Γ = T ∪ Ω

where as before T denotes the “spike” support and Ω the “sine”

For the second condition, construct u0 as in (6), where now z0 are

u0[γ] = ΦHγ ΦΓ(ΦHΓΦΓ)−1z0

|u0[γ]| ≤ kwγ k kz0k

In this case, |T | · |Ω| ≤ n/4, so we can take δ = 1/4, and so if

and α0 will be the minimizer to (4).

Now suppose that Φ is an M ×N matrix with normalized columns,

The quantity µ is essentially a measure of how closely aligned any

Let Γ be a fixed subset of {1, 2, . . . , N }, and x0 be a vector sup-

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