LECTURE 26: EIGENVALUE METHOD
MINGFENG ZHAO
November 16, 2015
2
Theorem 1.
to
~x = A~x,
0
Let
A be a 2 2 matrix, and ~y(t) = 4
then the general solution to
~x = A~x
0
y1 (t)
y2 (t)
3
5
2
and
~z(t) = 4
z1 (t)
z2 (t)
3
5
be two linearly independent solutions
is
~x(t) = C1 ~y(t) + C2~z(t):
22
In this case, the
see that
~ (t) = AX
~ (t),
X
0
matrix
~ (t) = [~y(t) ~z(t)]
X
where
2
~ (t) = 4
X
0
Theorem 2 (Eigenvalue Method).
~x(t) = et~v
is a solution to
Let
is called a fundamental matrix of the system
y1 (t) z1 (t)
y2 (t) z2 (t)
30
5
2
:= 4
be an eigenvalue of
A
0
0
and
~v
It's easy to
5:
y2 (t) z2 (t)
0
0
3
y1 (t) z1 (t)
0
~x = A~x.
be an eigenvector corresponding to
,
then
~x = A~x.
0
Eigenvalue method with distinct real eigenvalues
Theorem 3.
Let
corresponding to
A
1
be a
and
22
2 ,
matrix with two distinct real eigenvalues
respectively, then the general solution to
1
~x = A~x
0
and
2 ,
and
~v1
and
~v2
are eigenvectors
~v2
are eigenvectors
is:
~x = C1 e1 t~v1 + C2 e2 t~v2 :
Eigenvalue method with distinct complex eigenvalues
Theorem 4.
Let
corresponding to
A
1
be a
and
22
2 ,
matrix with two distinct complex eigenvalues
respectively, then
2 = 1 , ~v2 = ~v1 ,
1
2 ,
and
and the general solution to
~x = C1 Re e1 t~v1 + C2 Im e1 t~v1 :
1
and
~v1
and
~x = A~x
0
is:
2
MINGFENG ZHAO
Recall the Euler's identity:
ea+ib = ea cos(b) + iea sin(b):
2
Example 1. Find the general solution to the system ~x = 4
0
2
1
First, let's nd the eigenvalues of A = 4
1
1 1
3
5,
1
5~
x.
1 1
then
2
det (A
3
1
I2 ) = det
4
1
1
)2 + 1
= 2
2 + 2
5
1
= (1
3
1
= 0:
Then
1 = 1 + i;
and 2 = 1
i:
For = 1 + i, let's solve A~v = ~v , that is,
2
4
So we get
1
1
1 1
2
4
So we have
32
54
i
1
1
i
2
4
2
That is, 4
1
i
identity, we have
x1
x2
x1
x2
32
54
3
2
5
= (1 + i) 4
x1
x2
3
2
5
=4
3
2
5
= x1 4
1
i
0
0
x1
x2
5:
3
5:
2
5
is an eigenvalue corresponding to 1 + i, then e(1+i)t 4
e(1+i)t 4
1
i
3
2
5
=4
e(1+i)t
ie(1+i)t
3
2
5
=4
5:
3
3
2
3
et cos(t) + iet sin(t)
i [et cos(t) + iet sin(t)]
3
2
5
=4
1
i
3
5
is a solution to ~x0 = A~x. By Euler's
et cos(t)
et sin(t)
3
2
5 + i4
et sin(t)
et cos(t)
3
5:
LECTURE 26: EIGENVALUE METHOD
3
Therefore, the general solution to ~x0 = A~x is:
2
3
et cos(t)
~x(t) = C1 4
et sin(t)
2
5 + C2 4
2
2
Example 2. Find the general solution to the system ~x = 4
0
2
2
First, let's nd the eigenvalues of A = 4
2
4 6
5,
:
5~
x.
then
2
det (A
5
et cos(t)
3
2
4 6
3
3
et sin(t)
I2 ) = det
4
2
3
2
4
= (2
)(6
= 2
8 + 20
5
6
) + 8
= 0
Then
1 = 4 + 2i;
and 2 = 4
2i:
For = 4 + 2i, let's solve A~v = ~v , that is,
2
2i
2
4
4
So we have
2
4
2
1
32
2
2i
2
x1
x2
54
3
2
5
= x1 4
x1
x2
3
2
5
=4
0
0
3
5:
3
1
1+i
5:
3
2
1
5 is an eigenvalue corresponding to 4 + 2i, then e(4+2i)t 4
i+i
1+i
Euler's identity, we have
That is, 4
2
e(4+2i)t 4
1
1+i
3
2
5
= 4
2
= 4
e(4+2i)t
(1 + i)e(4+2i)t
3
2
5
=4
5
is a solution to ~x0 = A~x. By
e4t cos(2t) + ie4t sin(2t)
3
(1 + i) e4t cos(2t) + ie4t sin(2t)
e4t cos(2t)
e4t cos(2t)
3
e4t sin(2t)
3
2
5 + i4
e4t sin(2t)
e4t cos(2t) + e4t sin(2t)
5
3
5:
4
MINGFENG ZHAO
Hence, the general solution to ~x0 = A~x is:
2
~x(t) = C1 4
3
e4t cos(2t)
e4t cos(2t)
e4t sin(2t)
2
5 + C2 4
e4t sin(2t)
e4t cos(2t) + e4t sin(2t)
3
5:
Eigenvalue method with the same real eigenvalue
Denition 1. Let A be a 2 2 matrix, f (t) = t2 (a + d)t + (ad bc) be the characteristic polynomial of A, and is
an eigenvalue of A, that is, f () = 0, then
I. The multiplicity of as a solution to f (t) is called the algebraic multiplicity of as an eigenvalue of A.
II. The dimension of eigenvector space corresponding to is called the geometric multiplicity of .
III. If the algebraic multiplicity and the geometric multiplicity are equal, we say that is complete
IV. If the algebraic multiplicity is 2, but the geometric multiplicity is 1, we say that is a defective eigenvalue with
defect 1.
Let A be a 2 2 matrix with the same real eigenvalue and ~v be the eigenvector corresponding to , then ~x(t) = et~v
is a solution to ~v 0 = A~x.
If I2
2
A = 0, then 4
1
3
2
3
0
are two linearly independent eigenvectors corresponding to , that is, 0
1
is a complete eigenvalue of A. Hence the general solution to ~x0 = A~x is
5
and 4
5
2
~x(t) = et 4
If I2
1
0
3
2
5 + et 4
0
1
3
5:
A 6= 0, that is, is a defective eigenvalue of A, then another linearly independent solution ~y to ~x = A~x
0
has the form ~y (t) = et (t~v + w
~ ) for some vector w
~ to be determined. That is, ~y(t) = t~x(t) + et w
~ . So we get
~y (t) = ~x(t) + t~x (t) + et w
~
0
0
= ~x(t) + tA~x(t) + et w
~
= A~y (t)
= tA~x(t) + et Aw:
~
Then ~x(t) + et w
~ = et Aw
~ , that is, et~v + et w
~ = et Aw
~ , so w
~ is a solution to ~v + w
~ = Aw
~ , that is,
(A
I2 )w
~ = ~v :
LECTURE 26: EIGENVALUE METHOD
In this case, we have (A
I2 )2 w
~ = (A
5
I2 )~v = 0, we call w
~ a generalized eigenvector of A.
Problems you can do:
Lebl's Book [2]: All Exercise on Page 108, Page 109, Page 127 and Page 128.
Braun's Book [1]: All exercises on Page 344, Page 345, Page 352, Page 353, Page 354 and Page 355. Read all
materials in Section 3.9 and Section 3.10.
References
[1] Martin Braun.
[2] Jiri Lebl.
Dierential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
Notes on Diy Qs: Dierential Equations for Engineers. Createspace, 2014.
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