Chapter 8
SSM:
Linear Algebra
17. If
A
is the
n
×
n
matrix with all 1’s, then the eigenvalues of
A
are 0 (with multiplicity
n

1) and
n
.
Now
B
=
qA
+ (
p

q
)
I
n
, so that the eigenvalues of
B
are
p

q
(with
multiplicity
n

1) and
qn
+
p

q
. Thus det(
B
) = (
p

q
)
n

1
(
qn
+
p

q
).
19. Let
L
(
~x
) =
A~x
.
Then
A
T
A
is symmetric, since (
A
T
A
)
T
=
A
T
(
A
T
)
T
=
A
T
A
, so that
there is an orthonormal eigenbasis
~v
1
, . . . ,~v
m
for
A
T
A
. Then the vectors
A~v
1
, . . . , A~v
m
are orthogonal, since
A~v
i
·
A~v
j
= (
A~v
i
)
T
A~v
j
=
~v
T
i
A
T
A~v
j
=
~v
i
·
(
A
T
A~v
j
) =
~v
i
·
(
λ
j
~v
j
) =
λ
j
(
~v
i
·
~v
j
) = 0 if
i
6
=
j
.
21. For each eigenvalue there are two unit eigenvectors:
±
~v
1
,
±
~v
2
, and
±
~v
3
.
We have 6
choices for the first column of
S
, 4 choices remaining for the second column, and 2 for
the third.
Answer: 6
·
4
·
2 = 48.
23. The eigenvalues are real (by Fact 8.1.3), so that the only possible eigenvalues are
±
1 (by
Fact 7.1.2).
Since
A
is symmetric,
E
1
and
E

1
are orthogonal complements.
Thus
A
represents a
reflection
about
E
1
.
25. Note that
A
is symmetric an orthogonal, so that the eigenvalues of
A
are 1 and

1.
E
1
= span
1
0
0
0
1
,
0
1
0
1
0
,
0
0
1
0
0
,
E

1
= span
1
0
0
0

1
,
0
1
0

1
0
The columns of
S
must form an eigenbasis for
A
:
S
=
1
√
2
1
0
0
1
0
0
1
0
0
1
0
0
√
2
0
0
0
1
0
0

1
1
0
0

1
0
is one
possible choice.
27. If
n
is even, then this matrix is
J
n
+
I
n
, for the
J
n
introduced in Exercise 26, so that the
eigenvalues are 0 and 2, with multiplicity
n
2
each.
E
2
is the span of all
~
e
i
+
~e
n
+1

i
, for
i
= 1
, . . . ,
n
2
, and
E
0
is spanned by all
~
e
i

~
e
n
+1

i
. If
n
is odd, then
E
2
is spanned by all
~e
i
+
~
e
n
+1

i
, for
i
= 1
, . . . ,
n

1
2
;
E
0
is spanned by all
~
e
i

~
e
n
+1

i
, for
i
= 1
, . . . ,
n

1
2
, and
E
1