-
6.2 Definitions and examples
DEF
INITION6.1.1
(Eigenvalue,eigenvector)
Let
A
be a
complex square matrix. Then if
?
is a complex number and
X
a
non
–
zero complex column
vector satisfying
AX
?
?
X
,
we call
X
an eigenvector
of
A
, while
?
is called an
eigenvalue of
A
. We also say
that
.
X
is an eigenvector corresponding to the
eigenvalue ?
So in the above example
P
1
and
P
2
are
eigenvectors corresponding to
?
1
and
?
2
,
respectively. We shall give an
algorithm which starts from the
eigenvalues of
?
a
h
?
and constructs
a rotation matrix
A
such that
P
t
AP
is diagonal.
A
?
?
?
?
h
b
?
As noted
above, if
?
is an eigenvalue
of an
n
?
n
matrix
A
, with
corresponding
eigenvector
X
, then
(
< br>A
?
?
I
n
)
X
?
0
, with
X
?
0
,
so
det(
A
?
< br>?
I
n
)
?
0
and there are
at
most n distinct eigenvalues of
A
.
Conversely if
det(
A
?
?
I
n
)
?<
/p>
0
, then
(
A
?
?
I
n
)
X
?
0<
/p>
has a
non
–
trivial solution
X
and
so
,
is
an eigenvalue of
A
with
X
a corresponding eigenvector.
DEFINITION6.1.2
(Characteristicpolynomi
al,equation)
The polynomial
det(
A
?
?
I
n
)
is called the characteristic polynomial
of
A
and is often
denoted by
ch
A
(
?
)
. The
equation
det(
A
?
p>
?
I
n
)
?
0
is
called the characteristic equation of
A. Hence the eigenvalues of
A
are the roots
of the characteristic polynomial of
A
.
For a
2
?
2
matrix
A
?
< br>?
?
a
b
?
, it is easily verified that the
characteristic polynomial is
?
?
c
d
?
?
2
?
(
tr
aceA
)
?
?
det
A
, where
trac
eA
?
a
?
d
is the sum of the diagonal
elements of
A
.
?<
/p>
2
1
?
EXAM
PLE6.2.1
Find the
eigenvalues of
A
?
?
?
and find all eigen-
vectors.
1
2
?
?
Solution.
The characteristic equation of
A
is
?<
/p>
2
?
4
?
?
3
?
0
, or
(
?
?
1)(
?
?
3)
p>
?
0
.
Hence
?
?
1
or 3. The eigenvector equation
< br>(
A
?
?
I
n
)
X
?
0
reduces to
1
?
?
x
?
?
0
?
< br>?
2
?
?
?
?
?
,
?
1
?
?
p>
?
2
?
?
?
?
y
?
?
0
?
?
< br>or
(
2
?
< br>?
)
x
?
y
?
0
.
x
?
(
2
?<
/p>
?
)
y
?
0
Taking
?
?
1
gives
x
?
y
?
0
x
?
y
?
0
,
which has solution
x
< br>?
?
y
,
y
arbitrary.
Consequently the eigenvectors corresponding to
?
?
1
are the vectors
?
?
,
with
y
?
0
. <
/p>
?
y
?
Taki
ng
?
?
3
gives
?
?
y
?
?
x
?
y
?
0
,
x
?
y
?
p>
0
which has solution
x
?
?
y
,
y
arbitrary.
Consequently the eigenvectors corre-
sponding to
?
?
3
are the vectors
?
?
, with
y
?
0
.
Our next result has wide applicability:
THEOREM6.2.1
Let
A
be a
2
?
2
matrix having distinct eigenvalues
?
1
and
?
2
and
corresponding eigenvectors
X
1
and
X
2
. Let
P
be the matrix
whose columns are
X
1
and
?
y
?
?
p>
y
?
?
?
1
0
?
X
2
, respectively. Then
P
is
non
–
singular and
P
AP
?
?<
/p>
?
0
?
?
2
?
?
1
Proof. Suppose
AX
1
?
?
1
X
1
and
AX
2
?
?
2
X
2
. We show
that the system of homogeneous
equations
xX
1
?
yX
2
?
0
has only the trivial
solution. Then by theorem 2.5.10 the matrix
P
?
?
x
< br>1
x
2
?
is
non
–
singular.
So
assume
xX
1
?<
/p>
yX
2
?
0
p>
.
(6.3)
Then
A
(
xX
1
?
yX
2
)
?
< br>A
0
?
0
, so
x
(
AX
1
)
?
y
(
AX
2
)
?
0
. Hence
x
?
p>
1
X
1
?
y
?
2
X
2
?
0
.
(6.4)
Multiplying equation
6.3 by
?
1
and subtracting from equation 6.4 gives
(
?
2
?
p>
?
1
)
yX
2
?
0
.
Hence
y
?
0
,
as
(
?
2
?
?
1
)
?<
/p>
0
and
X
2
?
0
.
Then from equation 6.3,
xX
p>
1
?
0
and hence
x
?
0
.
Then the equations
AX
1
?
?
1
X
1
and
< br>AX
2
?
?
2
X
2
give
AP
?
A
?
x
1
x
2
?
?
?<
/p>
Ax
1
Ax
2<
/p>
?
?
?
?
x
1
?
x
2
?
?
?
0
?
?
?
1
0
?
?
?
x
1
x<
/p>
2
?
?
1
?
P
?
?
?
0
?
?
?
0
?
2
?
?
2
?
EXAMPLE6.2.2 Let
A
?
?
?
2
1
?
be the matrix of example 6.2.1. Then <
/p>
?
?
1
2
?
?
?
1
?
?
1
?
X
1
?
?
?
and
X
< br>2
?
?
?
are eigenvectors corresponding to
eigenvalues
?
1
?
p>
?
1
?
1 and
3, respectively. Hence if
P
?<
/p>
?
?
1
0
?
?
1
?
1
?
?
1
, we
have
There are two
P
AP
?
?
?
?
p>
?
0
3
?
?
1
1
?
immediate applications of
theorem first is to the calculation of
A
n
: If
P
?
1
AP
?
diag
(
?
1
,
?
2
)
, then
A
?
< br>Pdiag
(
?
1
,
?
2
)
< br>P
?
?
1
n
?
?
1
0
?
?
1
p>
n
?
?
1
0
?
?
1
n
and
A
?
(
P
?
?
< br>P
)
?
P
?
0
?
?
P
?
P
?
0
p>
?
?
2
?
?
2
?
?
0
n
0
?
< br>?
1
?
P
.
?
1
n
?
The second application is to solving a
system of linear differential equations
dx
dy
?
ax
?
by
?
p>
xc
?
dy
p>
dt
dt
where
A
?
?
?
a
b
?
is a matrix of
real or complex numbers and
x
and
y
?
?
p>
c
d
?
?
are functions of t. The system can be
written in matrix form as
X
?
AX
,
?
x
?
whe
re
X
?
?
?
and
?
y
?
X
?
?
?<
/p>
?
?
dx
?
p>
?
x
?
?
dt
?
?
?
?
?
?
?
?
dy
?
< br>y
?
?
?
?
?
?
?
d
t
?
?
?
x<
/p>
1
?
We make the
substitution
X
?
PY
,
where
Y
?
?
< br>?
. Then
x
1
and
y
1
are
also functions
?
x
2
?
of
t
and
X
?
PY
?
AX
?
A
(
PY
)
, so
Y
?
(
P
?
1
AP
)
Y
?
?
?
?
p>
?
?
?
1
0
?
Y
.
?
?
p>
0
?
2
?
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