Differential Equations Solutions 74 - f nonzero and the...

84 Chapter 14. Solutions: Case Study: Blind Deconvolution: Errors
its eigenvalues). We can use the fact that trace( AB ) = trace( BA ). (See Challenge
12.3.)
It can also be proven just from the definition of the Frobenius norm and the
fact that Ux
2
= x
2
for all vectors x and orthogonal matrices U . Using this
fact, and letting a
i
be the i th column of A , we see that
UA
2
F
=
n
i =1
Ua
i
2
2
=
n
i =1
a
i
2
2
= A
2
F
.
Similarly, letting a
T
i
be the i th row of A ,
AV
2
F
=
m
i =1
a
T
i
V
2
2
=
n
i =1
a
i
2
2
= A
2
F
,
and the conclusion follows.
(c) From the constraint
K + E g + r
f
− 1
= 0 ,
we see that
˜
U
T
K + E g + r
˜
V
˜
V
T
f
− 1
= 0 ,
so
(
˜
Σ +
˜
E )
˜
f = 0 ,
where
˜
E =
˜
U
T
[ E , r ]
˜
V and
˜
f =
˜
V
T
f
− 1
. From part (b) we know that minimizing
[ E , r ]
F
is the same as minimizing
˜
E
F
.
Therefore, to solve our problem, we want to make the smallest change to
˜
Σ
that makes the matrix
˜
Σ +
˜
E rank deficient, so that the constraint can be satisfied
by a nonzero
˜
f . Changing the ( n + 1 , n + 1) element of
˜
Σ from ˜ σ
n +1
to 0 certainly
makes the constraint feasible (by setting the last component of
˜
f