Fall 2003 Math 308/501–502
9 Matrix Methods for Linear Systems
9.5 Homogeneous Linear Systems
with Constant Coefficients
Fri, 14/Nov
c 2003, Art Belmonte
Summary
Today we’ll take a broad view of how to solve a linear system
x
0
=
Ax
with constant coefficients. That is, the
n
×
n
coefficient
matrix
A
has constant elements.
Definitions
An
eigenvalue
of
A
is a number
r
such that
Av
=
r
v
for some
nonzero vector
v
, which is called an
eigenvector
associated with
the eigenvalue
r
. Collectively,
r
and
v
are called an
eigenpair
.
Equivalently, we have
(
A

r
I
)
v
=
0
, where
I
is the
n
×
n
identity
matrix and
0
the
n
×
1 column vector of zeros. Thus
v
is a
nontrivial vector in nullspace of
A

r
I
, a.k.a, the
eigenspace
of
r
,
a subspace of
R
n
or
C
n
. Observe that any nonzero multiple of an
eigenvector
v
is also an eigenvector associated with eigenvalue
r
.
The
characteristic polynomial
of
A
is
p
(
r
)
=
det
(
A

r
I
)
. The
characteristic equation
of
A
is
p
(
r
)
=
0. Its roots (i.e., the
values for which the polynomial is zero) are precisely the
eigenvalues of
A
. These are also known as
characteristic roots
.
Theorem
If
r
is an eigenvalue of a matrix
A
with assoicated eigenvector
v
,
then the function
x
(
t
)
=
e
rt
v
is a solution of the system
x
0
=
Ax
satisfying the vector initial condition
x
(
0
)
=
v
.
Roots various
1. If the eigenvalues
r
k
,
k
=
1
, . . . ,
n
,
of a matrix
A
are
distinct
real roots
, then a fundamental solution set for
x
0
=
Ax
is
given by
x
k
(
t
)
=
e
r
k
t
v
k
,
k
=
1
, . . . ,
n
,
where
v
k
is an eigenvector associated with
r
k
.
2. If some of the eigenvalues are complex numbers, exponential
solutions given by the preceding theorem are complex
valued. We’ll learn how to extract realvalued solutions from
these in Section 9.6.
3. If the eigenvalues (be they real or complex) are
not
distinct
(i.e., there are repeated roots), then this
may
present
complications. The ultimate resolution to this difficulty is
found in the notion of
generalized eigenvectors
. But that will
come later in the chapter
. . .
Hand Examples
541/2
Find the eigenvalues and eigenvectors of
A
=
6

3
2
1
.
Solution
•
To find the characteristic polynomial
p
(
r
)
, we compute the
determinant of the matrix
A

r
I
, where
I
is the identity
matrix having the same size as
A
.
p
(
r
)
=
det
6

r

3
2
1

r
=
6

7
r
+
r
2
+
6
=
r
2

7
r
+
12
•
The eigenvalues of
A
are roots of the characteristic equation,
0
=
p
(
r
)
=
r
2

7
r
+
12
=
(
r

3
)(
r

4
)
, whence
r
=
3
,
4.
•
For
r
=
3
,
compute the reduced row echelon form of
A

r
I
.
3

3
2

2
rref
→
1

1
0
0
In order for
(
A

r
I
)
v
=
0
, we must have
v
1

v
2
=
0 or
v
1
=
v
2
. Hence 3
↔
1
1
is an eigenpair.
•
Similarly, for
r
=
4, we have
A

r
I
=
2

3
2

3
rref
→
1

3
2
0
0
,
whence 4
↔
3
2
is an eigenpair.
•
We thus have a full set of linearly independent eigenvectors.
541/6
Find the eigenvalues and eigenvectors of
A
=
0
1
1
1
0
1
1
1
0
.
Solution
•
Solve 0
=
p
(
r
)
=
det
(
A

r
I
)
=
det

r
1
1
1

r
1
1
1

r
= 
r
(
r
2

1
)

(

r

1
)
+
(
1
+
r
)
=
(
r
+
1
)(
2
+
r

r
2
)
=
(
1
+
r
)
2
(
2

r
)
to obtain eigenvalues
r
= 
1
,

1
,
2.