LECTURE 25: EIGENVALUE METHOD AND MULTIPLE EIGENVALUE
MINGFENG ZHAO
November 03, 2014
Theorem 1. Let A be a 2 × 2 matrix, ~y (t) and ~z(t) be two linearly independent solutions to ~x0 = A~x, then the general
solution to ~x0 = A~x is
~x(t) = C1 ~y (t) + C2 ~z(t).
~
In this case, the 2 × 2 matrix X(t)
= [~y (t) ~z(t)] is called a fundamental matrix of the system ~x0 = A~x. It’s easy to
~ 0 (t) = AX(t).
~
see that X
Theorem 2 (Eigenvalue Method). Let λ be an eigenvalue of A and ~v be an eigenvector corresponding to λ, then
~x(t) = eλt~v is a solution to ~x0 = A~x.
Eigenvalue method with distinct eigenvalues
Theorem 3. Let A be a 2×2 matrix with two distinct eigenvalues λ1 and λ2 , and ~v1 and ~v2 are eigenvectors corresponding
to λ1 and λ2 , respectively.
I. If λ1 and λ2 are two different real eigenvalues, then the general solution to ~x0 = A~x is:
~x = C1 eλ1 t~v1 + C2 eλ2 t~v2 .
II. If λ1 and λ2 are two different complex eigenvalues, then λ2 = λ1 , ~v2 = ~v1 , and the general solution to ~x0 = A~x
is:
~x = C1 Re eλ1 t~v1 + C2 Im eλ1 t~v1 .

Example 1. Solve ~x0 = 
2
2
1
3

 ~x.
1
2
MINGFENG ZHAO

First, let’s find eigenvalues of A := 
2
2
1
3

det (λI2 − A) = det 

, we need to solve

λ−2
−2
−1
λ−3
 = (λ − 2)(λ − 3) − 2 = λ2 − 5λ + 4 = 0.
Then
λ1 = 1,
and λ2 = 4.
For λ1 = 1, let’s find the eigenvalue corresponding to λ1 = 1, we need to solve A~x = ~x, that is,


 

−1 −2
x1
0


=
.
−1 −2
x2
0

Then 
x1


 = x2 
x2
−2

. That is,
1

−2


 is an eigenvalue corresponding to λ1 = 1.
1
For λ2 = 4, let’s find the eigenvalue corresponding to λ2 = 4, we need to solve A~x = 4~x, that is,


 

2 −2
x1
0


=
.
−1 1
x2
0

Then 
x1
x2


 = x1 
1

, that is,
1

1


 is an eigenvalue corresponding to λ2 = 4.
1

0
Therefore, the general solution to ~x = 
2
2
1
3

 ~x is:

~x(t) = C1 et 
−2
1


 + C2 e4t 
1
1

.
LECTURE 25: EIGENVALUE METHOD AND MULTIPLE EIGENVALUE

2
2
−4

6
Example 2. Solve the initial value problem ~x0 = 

First, let’s find the eigenvalues of A = 

det (λI2 − A) = det 
2
2
−4
6


1
 ~x and ~x(0) = 
3

.
1
, then
λ−2
−2
4
λ−6

 = (λ − 2)(λ − 6) + 8 = λ2 − 8λ + 20 = 0.
Then
λ1 = 4 + 2i,
and λ2 = 4 − 2i.
For λ = 4 + 2i, let’s solve A~v = λ~v , that is,

−2
2 + 2i

−2 + 2i
4

x1



=
x2
0

.
0
So we have

x1



 = x1 
x2

That is, 
1

1
.
1+i


 is an eigenvalue corresponding to 4 + 2i, then e(4+2i)t 
i+i
Euler’s identity, we have
1

 is a solution to ~x0 = A~x. By
1+i

e(4+2i)t 
1


1+i


e4t cos(2t) + ie4t sin(2t)
=

= 
(1 + i)e(4+2i)t
(1 + i) e4t cos(2t) + ie4t sin(2t)




e4t cos(2t)
e4t sin(2t)
 + i
.
= 
e4t cos(2t) − e4t sin(2t)
e4t cos(2t) + e4t sin(2t)
e(4+2i)t


Hence, the general solution to ~x0 = A~x is:

~x(t) = C1 
e4t cos(2t)
4t
4t
e cos(2t) − e sin(2t)


 + C2 
e4t sin(2t)
4t
4t
e cos(2t) + e sin(2t)

.
4
MINGFENG ZHAO

On the other hand, since ~x(0) = 
1

, then
1

C1 
1


0
 + C2 
1

Then C1 = 0 and C2 = 0. Therefore, the solution to ~x0 = 


1
=
1

~x(t) = 

.
1
2
2
−4
6


 ~x and ~x(0) = 
1

 is:
1

e4t cos(2t)
e4t cos(2t) − e4t sin(2t)
.
Eigenvalue method with the same real eigenvalue
Let A be a 2 × 2 matrix with the same real eigenvalue λ and ~v be the eigenvector corresponding to λ, then ~x(t) = eλt~v
is a solution to ~v 0 = A~x. If λI2 − A = 0, then the general solution to ~x0 = A~x is


 
1
0
 + eλt   .
~x(t) = eλt 
0
1
If λI2 − A 6= 0, then another linearly independent solution ~y to ~x0 = A~x has the form ~y (t) = eλt (t~v + w)
~ for some
vector w
~ to be determined. That is, ~y (t) = t~x(t) + eλt w.
~ So we get
~y 0 (t) = ~x(t) + t~x0 (t) + λeλt w
~ = ~x(t) + tA~x(t) + λeλt w
~ = A~y (t) = tA~x(t) + eλt Aw.
~
Then ~x(t) + λeλt w
~ = eλt Aw,
~ that is, eλt~v + λeλt w
~ = eλt Aw,
~ so w
~ is a solution to ~v + λw
~ = Aw,
~ that is,
(λI2 − A)w
~ = −~v .

 x0 = 3x1 + x2
1
.
Example 3. Find the general solution to
 x0 = 3x
2
2


 x0 = 3x1 + x2
3
1
The coefficient matrix of the system
is A = 
 x0 = 3x
0
2
2

det (λI2 − A) = det 
λ−3
−1
0
λ−3
1

. First, let’s find eigenvalues of A, then
3

 = (λ − 3)2 = 0.
LECTURE 25: EIGENVALUE METHOD AND MULTIPLE EIGENVALUE
5
Then λ1 = λ2 = 3. Now let’s solve


3
1
0
3
0
−1

x1




x1
 = 3
x2
.
x2
Then


So 
x1


 = x1 
x2
1
x1

0


0


=
x2

0
.
0


. Hence we know that λ = 3 is the only one eigenvalue for A, and ~v = 
0
3t
1

 is the eigenvector.
0
0
Then ~y (t) = e ~v is a solution to ~x = A~x.
Let ~x = e3t (t~v + w)
~ = t~y (t) + e3t w
~ be another linearly independent solution to ~x0 = A~x, then
~x0 = ~y (t) + t~y 0 (t) + 3e3t w
~ = A(t~y (t) + e3t w)
~ = tA~y (t) + e3t Aw.
~
Since ~y (t) is a solution to ~x0 = A~x, then
~y (t) + 3e3t w
~ = e3t Aw.
~
Since ~y (t) = e3t~v , then e3t~v + 3e3t w
~ = e3t Aw,
~ that is, ~v + 3w
~ = Aw.
~ Hence, (3I2 − A)w
~ = ~v , that is,


 

0 −1
w
1

 1  = 

0 0
w2
0

So we can take w1 = 0 and w2 = −1, that is, w
~ =
 
e3t (t~v + w)
~ = e3t t 
0
−1
1


. So another solution is

+
0
0
−1


 = e3t 
t
−1

.
Therefore, the genera solution to ~x0 = A~x is:

3t
~x = C1 e 
1
0


3t
 + C2 e 
t
−1

.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca