Chapter 6
Solution of Linear Equations
1. As in Example 6.8, characterize all left inverses of a matrix
A
∈
IR
m
×
n
.
Answer 6.1
A
has a left inverse
⇐⇒
A
T
has a right inverse
R
(
I
n
)
⊆
R
(
A
T
)
A
T
(
A
T
)
+
I
n
=
I
n
rank(
A
T
)=
r
=
n
(since
r
≤
n
)
A
T
is onto (
(
A
T
)
+
is then a right inverse, so
A
+
is a left inverse). Thus, all left inverses of
A
are of the form
L
=
±
(
A
T
)
+
I
n
+(
I
m

(
A
T
)
+
A
T
)
Z
T
²
T
=
±
(
A
T
)
+
I

(
A
T
)
+
A
T
)
Z
T
²
T
=
A
+
+
Z
(
I

AA
+
)
where
Z
∈
n
×
m
is arbitrary.
2. Let
A
∈
m
×
n
,
B
∈
m
×
k
and suppose
A
has an SVD as in Theorem
5.1. Assuming
R
(
B
)
⊆
R
(
A
), characterize all solutions of the matrix
linear equation
AX
=
B
in terms of the SVD of
A
.
X
=
A
+
B
I

A
+
A
)
Y
=
V
1
S
+
U
T
1
B
I

V
1
V
T
1
)
Y
=
V
1
S

1
U
T
1
B
+
V
2
V
T
2
Y
where
Y
∈
n
×
k
is arbitrary.
21
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CHAPTER 6. SOLUTION OF LINEAR EQUATIONS
3. Let
x,y
∈
IR
n
and suppose further that
x
T
y
±
= 1. Show that
(
I

xy
T
)

1
=
I

1
x
T
y

1
xy
T
.
Answer 6.2
By the ShermanMorrisonWoodbury formula,
(
I

xy
T
)

1
=(
I
+
x
(

1)
y
T
)

1
=
I

Ix
(

1+
y
T
)

1
y
T
I
=
I

x
(
y
T
x

1)

1
y
T
=
I

1
x
T
y

1
xy
T
.
4. Let
∈
n
and suppose further that
x
T
y
±
= 1. Show that
±
y
T
1
²

1
=
±
I
+
cxy
T

cx

cy
T
c
²
where
c
=1
/
(1

x
T
y
).
Answer 6.3
By formula 7 in Section 6.4,
±
y
T
1
²

1
=
±
I
+
Ixcy
T
I

Ixc

cy
T
Ic
²
=
±
I
+
cxy
T

cx

cy
T
c
²
where
c
=
1
1

y
T
=
1
1

x
T
y
.
5. Let
A
∈
n
×
n
n
and let
A

1
have columns
c
1
,...,c
n
and individual
elements
γ
ij
. Assuming that
γ
ji
±
= 0, show that the matrix
B
=
A

1
γ
e
i
e
T
j
(i.e.,
A
with
1
γ
subtracted from its
ij
th element) is
singular.
Hint:
Show that
c
i
∈
N
(
B
).
Answer 6.4
If we show that
N
(
B
)
±
=0
, then this will mean that
B
is
singular since it will not have full rank. Now
Bc
i
=
Ac
i

1
γ
e
i
e
T
j
c
i
=
e
i

1
γ
e
i
γ
=
e
i

e
i
. Hence,
c
i
∈
N
(
B
)
. But
c
i
±
since
A
is
nonsingular. Therefore,
B
is singular.
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 Spring '11
 LAUB
 inner product, Inner product space, orthogonal projection

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