In conclusion: The kernel of
L
consists of all symmetric matrices, and the image consists
of all skew-symmetric matrices.
61. Note that the first three matrices of the given basis
B
are symmetric, so that
L
(
A
) =
A
−
A
T
= 0, and the coordinate vector [
L
(
A
)]
B
is 0 for all three of them. The last matrix
of the basis is skew-symmetric, so that
L
(
A
) = 2
A
, and [
L
(
A
)]
B
= 2
e
4
. Using Fact 4.3.2,
we find that the
B
-matrix of
L
is
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
.
63. By Exercise 2.4.62b, the given LDU factorization of
A
is unique.
By Fact 5.3.9a,
A
=
A
T
= (
LDU
)
T
=
U
T
D
T
L
T
=
U
T
DL
T
is another way to write the
LDU factorization of
A
(since
U
T
is lower triangular and
L
T
is upper triangular). By the
uniqueness of the LDU factorization, we have
U
=
L
T
(and
L
=
U
T
), as claimed.
65. Write 10
A
=
a
b
c
d
; it is required that
a, b, c
and
d
be integers. Now
A
=
a
10
b
10
c
10
d
10
must be an orthogonal matrix, implying that (
a
10
)
2
+(
c
10
)
2
= 1, or
a
2
+
c
2
= 100. Checking
135

Chapter 5
SSM:
Linear Algebra
the squares of all integers from 1 to 9, we see that there are only two ways to write 100 as
a sum of two positive perfect squares: 100 = 36+64 = 64+36. Since
a
and
c
are required
to be positive, we have either
a
= 6 and
c
= 8 or
a
= 8 and
c
= 6. In each case we have
two options for the second column of
A
, namely, the two unit vectors perpendicular to
the first column vector. Thus we end up with four solutions:
A
=
.
6
−
.
8
.
8
.
6
,
.
6
.
8
.
8
−
.
6
,
.
8
−
.
6
.
6
.
8
or
.
8
.
6
.
6
−
.
8
.
67. a. We need to show that
A
T
Ac
=
A
T
x
, or, equivalently, that
A
T
(
x
−
Ac
) = 0.
But
A
T
(
x
−
Ac
) =
A
T
(
x
−
c
1
v
1
− · · · −
c
m
v
m
) is the vector whose
i
th
component is
(
v
i
)
T
(
x
−
c
1
v
1
− · · · −
c
m
v
m
) =
v
i
·
(
x
−
c
1
v
1
− · · · −
c
m
v
m
), which we know to be zero.
b. The system
A
T
Ac
=
A
T
x
has a unique solution
c
for a given
x
, since
c
is the coordinate
vector of proj
V
x
with respect to the basis
v
1
, . . . , v
m
. Thus the coeﬃcient matrix
A
T
A
must be invertible, so that we can solve for
c
and write
c
= (
A
T
A
)
−
1
A
T
x
.
Then
proj
V
x
=
c
1
v
1
+
· · ·
+
c
m
v
m
=
Ac
=
A
(
A
T
A
)
−
1
A
T
x
.
5.4
1. A basis of ker(
A
T
) is
−
3
2
. (See Figure 5.8.)
3. We will first show that the vectors
v
1
, . . . , v
p
, w
1
, . . . , w
q
span
R
n
. Any vector
v
in
R
n
can
be written as
v
=
v
+
v
⊥
, where
v
is in
V
and
v
⊥
is in
V
⊥
(by definition of a projection,
Fact 5.1.4).
Now
v
is a linear combination of
v
1
, . . . , v
p
, and
v
⊥
is a linear combination of
w
1
, . . . , w
q
,
showing that the vectors
v
1
, . . . , v
p
, w
1
, . . . , w
q
span
R
n
.
Note that
p
+
q
=
n
, by Fact 5.1.8c; therefore, the vectors
v
1
, . . . , v
p
, w
1
, . . . , w
q
form a
basis of
R
n
, by Fact 3.3.4d.
5.
V
= ker(
A
), where
A
=
1
1
1
1
1
2
5
4
.
Then
V
⊥
= (ker
A
)
⊥
= im(
A
T
), by Exercise 4.
136