Math 51 Exam 2 Solutions — May 22, 2007
1. (12 points)
(a) Find the eigenvalues of the following matrix.
A
=
1

1
1
1

1
1

1
1
1
(6 points)
To find the eigenvalues, we must solve the equation det(
A

λI
) = 0 for
λ
.
det(
A

λI
) =
1

λ

1
1
1

1

λ
1

1
1
1

λ
To evaluate the determinant, we expand it along a row or column.
Here, we show it
expanded along the top row:
(1

λ
)

1

λ
1
1
1

λ

(

1)
1
1

1

1

λ
+
1

1

λ

1
1
= (1

λ
)(
λ
2

2) + (2

λ
)

λ
=

λ
3
+
λ
2
+ 2
λ

2 + 2

λ

λ
=

λ
3
+
λ
2
Setting this equal to 0 and factoring yields
λ
2
(1

λ
) = 0 from which we conclude that
λ
= 0 or 1. So the eigenvalues are 0 or 1.
(b) Compute the eigenspace of the largest eigenvalue you found in part (a).
(3 points)
The largest eigenvalue is
λ
= 1. The eigenspace
E
1
is precisely the nullspace
of the matrix
A

I
. Let’s compute
N
(
A

I
) by rowreducing:
A

I
=
0

1
1
1

2
1

1
1
0
;
1

2
1
0

1
1

1
1
0
;
1

2
1
0

1
1
0

1
1
;
1

2
1
0

1
1
0
0
0
;
1

2
1
0
1

1
0
0
0
;
1
0

1
0
1

1
0
0
0
which is rowreduced. There are two pivots, and their equations are
x
1

x
3
= 0
⇒
x
1
=
x
3
and
x
2

x
3
= 0
⇒
x
2
=
x
3
So
N
(
A

I
) =
→
x
=
x
1
x
2
x
3
=
x
3
x
3
x
3
= span
1
1
1
.
(c) Is there a nonzero vector
→
v
∈
R
3
such that
A
→
v
=
→
v
? Briefly explain.
(3 points)
Yes. There is a nonzero vector
→
v
such that
A
→
v
=
→
v
exactly when and only
when 1 is an eigenvalue of
A
. In part a), we found that 1 is an eigenvalue of
A
.
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Math 51, Spring 2007
Exam 2 Solutions — May 22, 2007
Page 2 of 9
2. (12 points)
Let
A
=
1
3

3
4

2
16
.
(a) Find a basis for the orthogonal complement of the null space
N
(
A
).
(5 points)
Since
N
(
A
)
⊥
=
C
(
A
T
), we will find a basis for
C
(
A
T
).
A
T
=
1
4
3

2

3
16
By definition, the two columns of
A
T
span
C
(
A
T
). These vectors are linearly independent,
since they are clearly not scalar multiples of each other. Thus a basis for
N
(
A
)
⊥
=
C
(
A
T
)
is
1
3

3
,
4

2
16
.
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