Math 2270, Fall 2014
Instructor: Thomas Goller
11 November 2014
Test #4 Solutions
1
Definition of Eigenvectors (10 points)
(a) Write down the equation that defines what it means for ~x to be an eigenvector of a
matrix A with eigenvalue . (2 points)
Solution: A~x = ~x.
(b) If ~x is an eigenvector of A with eigenvalue , show that 5~x is an eigenvector of A2 . What
is its eigenvalue? (3 points)
Solution: Since
A2 (5~x) = 5A(A~x) = 5A( ~x) = 5 (A~x) = 5 ( ~x) =
5~x is an eigenvector of A2 with eigenvalue
2
2
(5~x),
.
(c) If ~x is an eigenvector of A with eigenvalue 1 and also an eigenvector of B with eigenvalue
x is an eigenvector of A + 5B. What is its eigenvalue? (3 points)
2 , show that ~
Solution: We compute
(A + 5B)~x = A~x + 5(B~x) =
x
1~
so ~x is an eigenvector of A + 5B with eigenvalue
+ 5 2~x = (
1
1
+ 5 2 )~x,
+ 5 2.
(d) The eigenspace V0 of all eigenvectors for A with eigenvalue 0 coincides with one of the
four fundamental subspaces of A. Which subspace is it? (2 points)
Solution: The nullspace N (A).
Math 2270, Fall 2014
2
Instructor: Thomas Goller
11 November 2014
Inventions (10 points)
(a) Invent a 2 ⇥ 2 matrix A that has no real eigenvalues. (2 points)
Possible solution: A =

0
1
1
.
0
(b) Invent a vector ~x that is not an eigenvector of any matrix. (2 points)
Possible solution: ~x =

0
.
0
(c) Invent a 4 ⇥ 4 matrix with no zero entries that is diagonalizable. (2 points)
2
1
61
Possible solution: 6
41
1
1
1
1
1
1
1
1
1
3
1
17
7. (Every symmetric matrix is diagonalizable!)
15
1
(d) Invent a symmetric 2 ⇥ 2 matrix with all positive entries that is not positive definite. (2
points)
Possible solution:

11 10
10 1
(e) Invent a diagonalizable matrix that has an eigenvalue of multiplicity two. (2 points)
Possible solution:

1
0
0
1
Math 2270, Fall 2014
3
Instructor: Thomas Goller
11 November 2014
Di↵erential Equations (10 points)
Consider the following population model. Let h(t) denote the population of horses and g(t)
the population of dragons at time t. Initially, there are h(0) = 42 horses and g(0) = 10
dragons. The growth rates of the two populations are
dh
= 7h
dt
5g
dg
= h + g.
dt
and
(a) Write these growth rates as a di↵erential equation of the form
Solution: Using ~u =

7
5
1
we compute
✓
7
N
N
✓
1
7
1
1,
5
1 6
2
5
1 2
2
2
=
6
1
= A~u. (2 points)
h
, we can write the equations as
g



d~u
dh/dt
7
5 h
=
=
.
dg/dt
1 1
g
dt
(b) Compute the eigenvalues
Solution:
d~
u
dt
◆
◆
and corresponding eigenvectors ~x1 , ~x2 of A. (5 points)
8 + 12 = (
=N
=N
✓
✓
1
1
5
5
5
1
5
1
6)(
◆
◆
= span
= span
2), so
✓
✓
5
1
1
1
◆
◆
1
,
,
= 6 and
so ~x1 =
so ~x2 =
(c) Compute the scalars C1 , C2 that give the unique solution ~u(t) = C1 e
the di↵erential equation. (3 points)


42
5
Solution: We need to solve
= ~u(0) = C1
10
1




C1
1
1 42
32
solution is
= 14
= 14
=
C2
1 5
10
8

5
1
di↵erential equation is ~u(t) = 8e6t
+ 2e2t
.
1
1

1t
2
= 2. Then

5
;
1

1
.
1
~x1 + C2 e
2t
~x2 to


1
5 1 C1
=
. The unique
1
1 1 C2
+ C2

8
. Thus the unique solution to the
2
Math 2270, Fall 2014
4
Instructor: Thomas Goller
11 November 2014
Diagonalization (10 points)
For this problem, let A =

2
4
4
.
8
(a) How can you tell just by glancing at A that A is diagonalizable? (1 point)
Solution: A is symmetric! (Every symmetric matrix is diagonalizable.)
(b) Compute a factorization A = Q⇤QT , where Q is an orthogonal matrix of eigenvectors
and ⇤ is the eigenvalue matrix. (7 points)
4
= 2 10 = (
10), so the eigenvalues of A are 1 = 0 and
4 8
2 = 10. Now we compute unit eigenvectors:
✓
◆
✓ ◆

2
4
2
2
1
N
= span
,
so ~x1 = p5
;
1
4 8
1
✓
◆
✓
◆
✓
◆

2 10
4
8
4
1
1
1
N
=N
= span
,
so ~x2 = p5
.
4
8 10
4
2
2
2

2
1
1
The orthogonal eigenvector matrix is Q = p5
and the eigenvalue matrix is
1
2

0 0
⇤=
, so the decomposition A = Q⇤QT is
0 10




2
4
2
1
0
0
2 1
1
1
p
= p5
.
4 8
1 2
0 10 5
1 2
Solution:
2
(c) Use your answer in (b) to compute the matrix exponential eA . (2 points)
T
Solution: eA = eQ⇤Q = Qe⇤ QT , which equals






10
2
1
1
0
2
1
2
e
2
1
4 + e10 2 2e10
1
1
1
1
p
p
=
=
.
5 1 2e10
5 2
5 1
2
0 e10 5
1 2
1 2
2e10 1 + 4e10
Math 2270, Fall 2014
5
Instructor: Thomas Goller
11 November 2014
Multiplicity of Eigenvalues (10 points)
2
3 0
4
Consider the matrix A = 1 3
0 0
3
1
0 5.
1
(a) Compute the eigenvalues 1 , 2 , 3 of A. You should find that
multiplicities of the distinct eigenvalues 1 and 3 ? (3 points)
1
=
2.
What are the
Solution: Expanding by minors along the third row, we get
3
0
1
0
1
0
3
0
=( 1
)
1
so the eigenvalues of A are 1 = 3,
multiplicity 2 and the eigenvalue 3 =
2
3
0
1
3
=
( + 1)(
= 3, 3 = 1. The eigenvalue
1 has multiplicity 1.
3)2 ,
1
= 3 has
(b) What can you conclude about the dimensions of the two eigenspaces V1 = N (A
and V3 = N (A
3 I) from the multiplicities of 1 and 3 ? (2 points)
1 I)
Solution: 1  dim V1  2 and 1  dim V3  1, so dim V3 = 1.
(c) Now compute the eigenspaces V1 and V3 . What are their dimensions? (3 points)
Solution:
02
31
02
31
02 31
1
0 0 1
0
@
4
5
A
@
4
5
A
@
4
1
3 3
0
1 0 0
15A ;
V1 = N
=N
= span
0
0
1 3
0 0
4
0
02
31
02
31
02 1 3 1
3 ( 1)
0
1
4 0 1
4
1 5A
@
4
5
A
@
4
1
3 ( 1)
0
1 4 05A = span @4 16
V3 = N
=N
.
0
0
1 ( 1)
0 0 0
1
Both eigenspaces have dimension 1.
3
3
0
(d) How many independent eigenvectors does A have? (1 point)
Solution: Two.
(e) Is A diagonalizable? (1 point)
Solution: No! (Since A is 3 ⇥ 3, it would need 3 independent eigenvalues to be diagonalizable.)