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CHAPTER 8
Linear Algebra: Matrix Eigenvalue Problems
Prerequisite for this chapter is some familiarity with the notion of a matrix and with the
two algebraic operations for matrices. Otherwise the chapter is independent of Chap. 7,
so that it can be used for teaching eigenvalue problems and their applications, without
first going through the material in Chap. 7.
SECTION 8.1. Eigenvalues, Eigenvectors, page 334
Purpose.
To familiarize the student with the determination of eigenvalues and eigenvectors
of real matrices and to give a first impression of what one can expect (multiple eigenvalues,
complex eigenvalues, etc.).
Main Content, Important Concepts
Eigenvalue, eigenvector
Determination of eigenvalues from the characteristic equation
Determination of eigenvectors
Algebraic and geometric multiplicity, defect
Comments on Content
To maintain undivided attention on the basic concepts and techniques, all the examples in
this section are formal, and typical applications are put into a separate section (Sec. 8.2).
The distinction between the algebraic and geometric multiplicity is mentioned in this
early section, and the idea of a
basis of eigenvectors
(“eigenbasis”)
could perhaps be
mentioned briefly in class, whereas a thorough discussion of this in a later section
(Sec. 8.4) will profit from the increased experience with eigenvalue problems, which the
student will have gained at that later time.
The possibility of
normalizing
any eigenvector is mentioned in connection with Theorem
2, but this will be of greater interest to us only in connection with orthonormal or unitary
systems (Secs. 8.4 and 8.5).
In our present work we find eigen
values
first and are then left with the much simpler task
of determining corresponding eigen
vectors.
Numeric work (Secs. 20.6
2
20.9) may proceed
in the opposite order, but to mention this here would perhaps just confuse the student.
SOLUTIONS TO PROBLEM SET 8.1, page 338
2.
The characteristic equation is
D
(
l
)
5
(
a
2
)(
c
2
)
5
0.
Hence
1
5
a
,
2
5
c
. Components of eigenvectors can now be determined for
1
5
a
from
0
x
1
1
bx
2
5
0,
say,
x
1
5
1,
x
2
5
0
so that an eigenvector is [1
0]
T
, and for
2
5
c
from
(
a
2
c
)
x
1
1
bx
2
5
0,
say,
x
1
5
1,
x
2
5
(
c
2
a
)/
b
,
so that an eigenvector is
[
1
]
T
.
c
2
a
}
b
161
im08.qxd
9/21/05
12:12 PM
Page 161

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*Sign up*Here we must assume that
b
Þ
0. If
b
5
0, we have a diagonal matrix with the same
eigenvalues as before and eigenvectors [1
0]
T
and [0
1]
T
.
4.
This zero matrix, like any square zero matrix, has the eigenvalue 0, whose algebraic
multiplicity and geometric multiplicity are both equal to 2, and we can choose any
basis, for instance [1
0]
T
and [0
1]
T
.
6.
The characteristic equation is
(
a
2
l
)
2
1
b
2
5
0.
Solutions are
5
a
6
ib
.
Eigenvectors are obtained from
(
a
2
)
x
1
1
bx
2
57
ibx
1
1
bx
2
5
0.
Hence we can take
x
1
5
1 and
x
2
56
i
. Note that
b
has dropped out, and the
eigenvectors are the same as in Example 4 of the text.

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