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 deFnition of the ±robenius 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 X
i =1
± Ua
i
±
2
2
=
n X
i =1
± a
i
±
2
2
= ± A ±
2
F
.
Similarly, letting b a
T
i
be the i th row of A ,
± AV ±
2
F
=
m X
i =1
± b a
T
i
V ±
2
2
=
n X
i =1
± b a
i
±
2
2
= ± A ±
2
F
,
and the conclusion follows.
(c) ±rom the constraint
£
K + Eg + r
¤
·
f
− 1
¸
= 0 ,
we see that
˜
U
T
£
K + Eg + 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
¸
. ±rom 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 deFcient, so that the constraint can be satisFed
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
˜