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Eigenvalues and
Eigenvectors
6
...
Eigenvector
that L(V)
Eigenvalue
The pair A
...
=
A non-zero vector v E
JR11 such
A
...
is called an eigenvalue of L
...
The pairing of eigenvalues and eigenvectors is not one-to-one
...
2
...
In Chapter 9 we will consider the case where we allow eigenvalues and
eigenvectors to be complex
...
It is
natural because L(O)
esting to consider
=
0 for every
linear transformation, so it is completely uninter
0 as an eigenvector
...
In particular, we will
see that we often want to look for a basis of JR11 that contains eigenvectors of a linear
transformation
...
Determine which of the following vectors are eigenvectors of Land give
the corresponding eigenvalues:
Solution:
So,
[]
To test whether
[ �]
[
!
]
[
rn =;J
...
i, so
[ � ] 2[=�]
=
[�]
17
-15
r
r
[ J J J rJ
l
20
-18
=
3
-28
l
-34 f
...
[17 -15] [3] [ -9]
L(3 4)=
...
L(l ,3) =
[!]
1)
is also an eigenvector of L with eigenvalue 2
...
Since
So,
...
Eigenvectors and Eigenvalues of Projections and Reflections in JR3
1
...
If vis orthogonal to it, then proj,lV) =
0
= Ov, so vis an eigen
vector of proji1 with corresponding eigenvalue 0
...
For an arbitrary vector a
is a multiple of rt, so that a is definitely
a multiple of it or orthogonal to it
...
On the other hand, perp,z(it)
(continued)
eigenvalue 0
...
3
...
For
vorthogonal to ii, refl11(V)
=
1 v
...
EXAMPLE3
Eigenvectors and Eigenvalues of Rotations in JR2
...
der the rotation Re
integer multiple of
such that Re(V)
=
1r
...
cose
...
sme
[
cose
l
...
This linear transformation has no real
eigenvalues or real eigenvectors
...
[
EXERCISE 1
Let Re : IR3
cose
--
- sine
IR3 denote the rotation in IR3 with matrix sine
cose
0
0
H
Derermine
any real eigenvectors of Re and the corresponding eigenvalues
...
However, in many applications of these ideas, it
is a matrix A that is given
...
Definition
Suppose that A is an n x n matrix
...
In Example 1, we saw that
/l is called
[ �] [!]
and
an eigenvalue of
A
...
Finding Eigenvectors and Eigenvalues
If eigenvectors and eigenvalues are going to be of any use, we need a systematic
method for finding them
...
A
is given; then a non-zero
This condition can be rewritten
Av-M=o
(A
- ,i)v
=
0,
but this would be incorrect because A is
a matrix and /l is a number, so their difference is not defined
...
Then the
eigenvector condition can be rewritten
(A - ,U)v=
0
The eigenvector vis thus any non-trivial solution (since it cannot be the zero vector)
of the homogeneous system of linear equations with coefficient matrix (A -tl/)
...
Hence, for tl to be an eigenvalue, we must have det(A-tl/) = 0
...
Theorem 1
Suppose that A is an
n x n matrix
...
Observe that the set of all eigenvectors corresponding to an eigenvalue tl is just
the nullspace of A -tl/, excluding the zero vector
...
We make the
following definition
...
n x n matrix
A
...
Hence, the dimension of the eigenspace
must be at least 1
...
17-tl
-15
20
-18 -,{
]
(You should set up your calculations like this: you will need A - tll later when you
...
) Then
det(A -tl/)
=
1
17-,{
20
-15
_
18
_
tl
I
= (17 - tl)(-18 - tl) - (-15)20
= tl
...
These are all of the eigenvalues of A
...
3
[ i4]
-3
3
{[ i4]}
...
W riting A - ill and row reducing gives
A
_
so that the general solution of (A-ill)v
of A corresponding to il
eigenspace for il
=
-3
=
-15
-15
20
20
=
are v
is v
=
=
t E R Thus, all eigenvectors
t
=
for any non-zero value oft, and the
is Span
A_ 21
The general solution of (A - ill)V
{[ � ]}
0
t
We repeat the process for the eigenvalue il
Span
1
�
=
=
[
15
20
0 is v
- 15
-20
t
=
=
2:
] [
1
�
0
-1
0
]
[ �],
t E JR, so the eigenspace for il
...
This moti
vates the following definition
...
Then C(il)
Characteristic Polynomial
polynomial of A
...
Note that the term of highest degree il" has
coefficient (-1)'1; some other books prefer to work with the polynomial det(ill - A)
so that the coefficient of tl
...
In our notation, the constant term in the char
acteristic polynomial is det A (see Problem 6
...
D7)
...
(2) The total number of roots (real and complex, counting repetitions) is n
...
If n is odd, there must be at least one real root
...
EXAMPLES
[b � l
Find the eigenvalues and eigenvectors of A=
Solution: The characteristic polynomial is
So, A
...
- 1) appears as a factor of C(A
...
= 1
is the only eigenvalue of A
...
= 1, we have
A-A
...
b
]
rn
lit Thus, the eigenspace for A = 1 is
t E
·
Find the eigenvalues and eigenvectors of A=
[� �]
...
Solution: We have
-3-A
...
I)=
-7
-7
5
-5
9-A
...
Also, we will end up with a degree 3 polynomial, which may not be easy
to factor
...
Since adding a multiple of one row to another does not change the
determinant, we get by subtracting row 2 from row 3,
C(A
...
5
-7
9- ,i
-2+,1
0
-5
-5
2-,1
Expanding this along the bottom row gives
C(A
...
)(-1)((-3
-
,1)(-5)- (-5)(-7))
+( 2-,1)((-3-,1)(9- ,1)- 5(-7))
( 2- ,i)(( 5A
...
2 - 6,i- 27 + 35))
-(,1- 2)(,12 - ,i - 12)= -(,1 - 2)(A
...
+ 3)
[-5 5 -51 �[
-5
ol
[n
EXAMPLE6
Hence, the eigenvalues of Aare
...
t2 = 4, and A3 = -3
...
t1/= -7
7
-7
7
-5
0
Hence, the general solution of (A - ;1 J)V =
eigenspace of;, is
For
...
t2/ =
7
-7
Hence, the general solution of (A - A2J)il =
{[11}
For A3 = -3,
0
0
is V =
�[
1
0
t
�
6
0
0
is V =
t
Thus, a basis for the
1
-7
7
0
Hence, the general solution of (A - ,!3l)V =
l{[; }
Thus, a basis for the
0
[ � � =�1�[� � =�1
m·
A-
...
t
C(
...
t)((l
2
2
= -,l(-,l + ,t - 2A) = -
...
t - 3)
1
-
...
t)(l -
...
t1 = 0 (which occurs twice) and
...
For
...
t1/=
1 1 l
1 1 1
1 1 1
[
�
l
1
l
0
0
0
0
0
0
EXAMPLE 7
(continued)
Hence, a basis forthe eigenspace of 0 is{[ 1l [ �]}
For
�
n -2 -2�i [0 0� =0�i
Thus, a basis for the eigenspace of is{[;]}·
These examples motivate the following definitions
...
The of is the
of is the dimension of the eigenspace of
A'
/l
...
n x n
/l
/l
multiplicity
algebraic multiplicity
/l
...
IInn Exampl
e
each
ei
g
enval
u
e
has
al
g
ebrai
c
mul
t
i
p
l
i
c
i
t
y
and
geomet
r
i
c
mul
t
i
p
l
i
c
i
t
y
0
has
al
g
ebrai
c
Exampl
e
t
h
e
ei
g
enval
u
e
and
geomet
r
i
c
mul
t
i
p
l
i
c
i
t
y
2,
and
t
h
e
eigenvalue has algebraic and geometric multiplicity
5,
(/l
1)(/1
...
6,
7,
3
EXERCISE 3
/l
1
...
Show that and -2 are both eigenvalues of and
determine the algebraic and geometric multiplicity of both of these eigenvalues
...
Let be an eigenvalue of an matrix Then
geometric multiplicity algebraic multiplicity
A
5
=
-3
-4
/l
/l
...
2
A
...
However, if
distinct eigenvalues A1,
•
•
•
A is an n x n matrix
with
, Ab which all have the property that their geometric multi
plicity equals their algebraic multiplicity, then the sum of the geometric multiplicities
of all eigenvalues equals the sum of the algebraic multiplicities, which equals n (since
an n-th degree polynomials has exactly n roots)
...
I!
...
Theorem 3
Suppose that Ai,
...
, vb respectively
...
, vk} is linearly
independent
...
If k
1, then the result is trivial,
since by definition of an eigenvector, v1 * 0
...
1
...
, vb Vk+ i} is linearly independent, we consider
=
(6
...
1) by A
-
AJ)Vi
=
0 and
Ak+I I gives
{v1,
vk} is linearly independent; thus, all the coeffi
AJ; hence, we must have c1
ck
0
...
1)
By our induction hypothesis,
cients must be 0
...
0
=
0
since it is an eigenvector; hence, Ck+l
0, and the set is linearly
•
Remark
In this book, most eigenvalues tum out to be integers
...
Effective computer
methods for finding eigenvalues depend on the theory of eigenvectors and eigenvalues
...
1
Practice Problems
Al Let A
=
[= ��2 �3 �]
...
If they are,
determine the corresponding eigenvalues
...
(e)
of the following matrices
...
(e)
[� -�J
(b)
22 22 221
[2 2 2
(d)
(f)
[� -�J
3
[-2-� =� 6]
-7
1
10
=
[-� -� -�1
...
If they are,
determine the corresponding eigenvalues
...
32
[ 53 1
1
(e) 0
0
0
7
1
4
6
6
[-2 23 1
-
(f)
7
[; � 3�i
Homework Problems
Bl Let A
--6]3
3[6
A3 For each of the following matrices, determine the
(a)
A2 Find the eigenvalues and corresponding eigenspaces
[! � ]
(f)
4
-1
1
B3 For each of the following matrices, determine the
algebraic multiplicity of each eigenvalue and deter
mine the geometric multiplicity of each eigenvalue
[[-; =�J
by writing a basis for its eigenspace
...
(a)
(c)
[ �2
[=�
!]
_;]
(c)
(e)
[;
[�
�]
�]
[� � � ]
(b)
(d)
(t)
4
��
0
0
-� �
1
[=� =; j]
Computer Problems
Cl
2
...
70562
1
...
(a)
(b)
(c)
9
[�
[l -i]
-
-
-5
-2
-5
-2
-2
2
2
4
4
1
3
2
2
0
2
2
-1
...
76414
-0
...
42918
-0
...
34270
[[ ] [ ] [ ]
C2 Let A
1
...
34 ,
0
...
31
2
...
21
...
85
,
and
are
0
...
10
(approximately) eigenvectors of A
...
3
-4
4
4
Conceptual Problems
Dl Suppose that vis an eigenvector of both the matrix
A and the matrix B, with corresponding eigenvalue
D4 (a) Let A be an n x n matrix with rank(A) = r < n
...
,1 for A and corresponding eigenvalueµ for B
...
De
(b) Give an example of a 3 x 3 matrix with
rank(A) = r < n such that the algebraic mul
termine the corresponding eigenvalues
...
in is an eigenvalue of A11• How are the cor
responding eigenvectors related?
(b) Give an example of a 2 x 2 matrix A such that
geometric multiplicity
...
How are
the corresponding eigenvalues related?
[�
sum of the entries in each row is the same
...
(Hint: See Problem 3
...
D4
...
Show that v =
is
1
an eigenvector of A
...
)