90
Chapter 15.
Solutions: Case Study: Blind Deconvolution: A Matter of Norm
∂F
(
e
,
f
)
∂f ∂f
q
=
m
i
=1
(
k
i
+
e
i
)(
k
iq
+
e
iq
) = ((
K
+
E
)
T
(
K
+
E
))
q
,
where outofrange entries in summations are assumed to be zero and
R
is a matrix
whose nonzero entries are components of
r
. So
g
=
∇
F
(
e
,
f
) =
D
2
e
−
F
T
r
(
K
+
E
)
T
r
,
H
(
e
,
f
) =
D
2
+
F
T
F
R
T
+
F
T
(
K
+
E
)
R
+ (
K
+
E
)
T
F
(
K
+
E
)
T
(
K
+
E
)
.
The Newton direction is the solution to
H
(
e
,
f
)
p
=
−
g
.
CHALLENGE 15.3.
The least squares problem is of the form
min
x
Ax
−
b
2
,
where
x
=
Δ
e
Δ
f
and
A
and
b
are the given matrix and vector. So to minimize
Ax
−
b
2
= (
Ax
−
b
)
T
(
Ax
−
b
), we set the derivative equal to zero, obtaining
A
T
Ax
−
A
T
b
=
0
.
The solution to this equation is a minimizer if the second derivative
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 Fall '11
 Dr.Robin
 Derivative, Vector Space, ax, min Ax, Blind deconvolution

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