31
CHALLENGE 5.20.
(a) Since
U
Σ
V
∗
x
=
b
,wehave
x
=
V
Σ
−
1
U
∗
b
.
If we let
c
=
U
∗
b
, then
α
1
=
c
1
/σ
1
,and
α
2
=
c
2
/σ
2
.
(b) Here is one way to look at it. This system is very illconditioned. The condition
number is the ratio of the largest singular value to the smallest, so this must be
large. In other words,
σ
2
is quite small compared to
σ
1
.
For the perturbed problems,
Ax
(
i
)
=
b
−
E
(
i
)
x
(
i
)
,
so it is as if we solve the linear system with a slightly perturbed righthand side.
So, letting
f
(
i
)
=
U
∗
E
(
i
)
x
(
i
)
, the computed solution is
x
(
i
)
=
α
(
i
)
1
v
1
+
α
(
i
)
2
v
2
,
with
α
(
i
)
1
=(
c
1
+
f
(
i
)
1
)
/σ
1
,α
(
i
)
2
c
2
+
f
(
i
)
2
)
/σ
2
.
From the ±gure, we know that
f
(
i
)
must be small, so
α
(
i
)
1
≈
α
1
. But because
σ
2
is
close to zero,
α
(
i
)
2
can be quite di²erent from
α
2
,so
the solutions lie almost on a
straight line in the direction
v
2
.
CHALLENGE 5.21.
Some possibilities:
•
Any right eigenvector
u
of
A
corresponding to a zero eigenvalue satis±es
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
 Fall '11
 Dr.Robin

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