Download Eigenvalues and Eigenvectors

These notes closely follow the presentation of the material given in David C. Lay’s
textbook Linear Algebra and its Applications (3rd edition). These notes are intended
primarily for in-class presentation and should not be regarded as a substitute for
thoroughly reading the textbook itself and working through the exercises therein.
Eigenvalues and Eigenvectors
Recall that if A is an n  n (square) matrix, then the mapping x Ax is a linear
transformation from  n into  n . For a given n  n matrix, A, it may or may not be the
case that there are non–zero vectors v   n such that Av is a scalar multiple of v.
Any non–zero vector v   n such that Av is a scalar multiple of v is called an
eigenvector of the matrix A.
Definition If A is an n  n matrix, v is a non–zero vector in  n , and  is a scalar such that
Av v, then v is said to be an eigenvector of the matrix A and  is said to be an
eigenvalue of the matrix A. More specifically, v is said to be an eigenvector of A
associated with the eigenvalue . If  is an eigenvalue of A, then the eigenspace of
, denoted by eig, is defined to be
eig  v   n | Av  v  0 n .
Our use of the term eigenspace is justified by the following theorem.
Theorem If  is an eigenvalue of an n  n matrix, A, then eig is a subspace of  n .
Proof Suppose that  is an eigenvalue of the n  n matrix, A. By definition,
eig  v   n | Av  v .
We will show that eig is closed under addition and scalar multiplication.
Let v 1 and v 2 be vectors in eig. Then
Av 1  v 2   Av 1  Av 2  v 1  v 2  v 1  v 2 .
This shows that v 1  v 2  eig. Therefore, eig is closed under vector
addition.
Let v  eig and let c be a scalar. Then
Acv  cAv  cv  cv.
This shows that cv  eig. Therefore, eig is closed under scalar
multiplication.
We have now proved that eig is a subspace of  n .
Remark If  is an eigenvalue of an n  n matrix, A, then
eig  v   n | Av  v .
It is clear from this definition that 0 n  eig because A0 n  0 n . However, note
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that the eigenvectors of the matrix A associated with the eigenvalue lambda are
the non–zero vectors in eig. Thus, even though 0 n  eig, we do not say that
0 n is an eigenvector of A associated with the eigenvalue .
Example For the matrix
0 1
A
,
1 0
find the eigenvalues of A and the eigenspaces associated with these eigenvalues.
Solution If
x1
v
x2
is an eigenvector of A associated with some eigenvalue , then it must be the case
that v  0 and that
0 1
x1
1 0
x2

x1
x2
.
We can write the above equation as
 x 2  x 1
x 1  x 2
or as
 x 1  x 2  0
x 1  x 2  0
or as
 1
x1

x2
1

0
0
.
Our assumption that v  0 means that the above homogeneous system has a
non–trivial solution. This means that the coefficient matrix of the above system
must not be invertible and hence that
 1
det
1

 0.
However, note that
det
 1
1

 2  1
and that  2  1  0 for any real number . We conclude that the matrix A has no
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eigenvalues (and hence no eigenvectors).
Note that
0 1
A
1 0
is the standard matrix of the linear transformation that rotates all vectors in  2
counterclockwise 90  about the origin. Since all vectors are rotated 90  , it is clear
that there is no non–zero vector whose image (under this linear transformation) is
a scalar multiple of itself. This is a simple explanation of why A has no eigenvalues
or eigenvectors.
Example For the matrix
1 0
A
0 1
,
find the eigenvalues of A and the eigenspaces associated with these eigenvalues.
Solution If
x1
v
x2
is an eigenvector of A associated with some eigenvalue , then it must be the case
that v  0 and that
1 0
x1
0 1
x2
x1

.
x2
We can write the above equation as
 x 1  x 1
x 2  x 2
or as
1  x 1  0
1  x 2  0
or as
1  
0
x1
0
1
x2

0
0
.
Our assumption that v  0 means that the above homogeneous system has a
non–trivial solution. This means that the coefficient matrix of the above system
must not be invertible and hence that
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det
1  
0
0
1
 0.
Since
det
1  
0
0
1
 1  1  ,
we see that if  is an eigenvalue of A, then
1  1    0,
which means that either   1 or   1. Both of these numbers are eigenvalues
of the matrix A.
To find the eigenvectors of A associated with the eigenvalue   1, we must
find the non–trivial solutions of the system
1  1
0
x1
0
1  1
x2
0

0
which is the same as
0 0
x1
0 2
x2
0

.
0
The above system is clearly equivalent to
2x 2  0
and we see that any vector of the form
x1
v
t

x2
0
t
1
0
is an eigenvector of A associated with the eigenvalue   1. Therefore,
eig1  Span
1
.
0
Geometrically, eig1 is the x 1 axis.
To find the eigenvectors of A associated with the eigenvalue   1, we must
find the non–trivial solutions of the system
1  1
0
x1
0
1  1
x2
0

0
which is the same as
2 0
x1
0 0
x2

0
0
.
The above system is clearly equivalent to
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 2x 1  0
and we see that any vector of the form
v
x1

x2
0
t
t
0
1
is an eigenvector of A associated with the eigenvalue   1. Therefore
0
eig1  Span
.
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Geometrically, eig1 is the x 2 axis.
Now let us interpret our findings in terms of the linear transformation for
which
A
1 0
0 1
is the standard matrix. This linear transformation reflects all vectors in  2
through the x 2 axis. Therefore, every vector, v, that lies on the x 1 axis is mapped
onto its opposite (that is, Av   1v), and every vector, v, that lies on the x 2 axis is
mapped onto itself (that is, Av 1v). No other vector in  2 (other than those that
lie on the x 1 and x 2 axes) are mapped onto scalar multiples of themselves.
Before proceeding to develop a general procedure for finding eigenvalues,
eigenvectors, and eigenspaces, we look at two more examples:
Example Find the eigenvalues and associated eigenspaces of the matrix
A
2
0
15 3
.
Solution If
x1
v
x2
is an eigenvector of A associated with some eigenvalue , then it must be the case
that v  0 and that
2
0
x1
15 3
x2

x1
x2
.
We can write the above equation as
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2x 1  x 1
15x 1  3x 2  x 2
or as
2  x 1  0
15x 1  3  x 2  0
or as
2
0
x1
15
3  
x2
0

.
0
Our assumption that v  0 means that the above homogeneous system has a
non–trivial solution. This means that the coefficient matrix of the above system
must not be invertible and hence that
det
2
0
15
3  
 0.
Since
det
2
0
15
3  
 2  3  ,
we see that if  is an eigenvalue of A, then
2  3    0,
which means that either   2 or   3. Both of these numbers are eigenvalues
of the matrix A.
To find the eigenvectors of A associated with the eigenvalue   2, we must
find the non–trivial solutions of the system
2  2
0
x1
15
3  2
x2
0

0
which is the same as
0
0
x1
15 5
x2

0
0
.
or as
15x 1  5x 2  0.
We can now see that any vector of the form
v
x1
x2

t
3t
t
1
3
is an eigenvector of A associated with the eigenvalue   2. Therefore,
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1
eig2  Span
.
3
Geometrically, eig2 is the line x 2  3x 1 .
To find the eigenvectors of A associated with the eigenvalue   3, we must
find the non–trivial solutions of the system
2  3
0
x1
15
3  3
x2
0

0
which is the same as
5 0
x1
15 0
x2

0
.
0
The above system is clearly equivalent to
x1  0
and we see that any vector of the form
v
x1
x2

0
t
t
0
1
is an eigenvector of A associated with the eigenvalue   3. Therefore
eig3  Span
0
1
.
Geometrically, eig3 is the x 2 axis.
The eigenspaces, eig2 and eig3, are pictured below.
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Example Find the eigenvalues and associated eigenspaces of the matrix
A
2 0
0 2
.
Solution It is easily seen that if v is any vector in  2 , then
Av 2I 2 v 2I 2 v  2v.
Therefore, the only eigenvalue of A is   2, and eig2   2 .
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A General Procedure for Finding Eigenvalues,
Eigenvectors, and Eigenspaces
Guided by the examples that we have studied, we now develop a general
procedure for finding eigenvalues, eigenvectors, and eigenspaces.
To find the eigenvalues of an n  n matrix, A, and the eigenvectors associated with
these eigenvalues, we must study the equation
Av v.
This equation has two unknowns:  (which is a scalar) and v (which is a vector).
Of course, one obvious solution of the above equation is  any scalar and v  0 n .
However, we are interested only in identifying cases that there exist solutions in which
v  0 n . Such solutions, if they exist at all, will only exist for certain values of  (which
are what we call the eigenvalues).
The first step in our solution process is to write the above equation as
Av I n v
and to observe that this is the same as
Av I n v
which is the same as
Av I n v  0 n
which is the same as
A  I n v  0 n .
The above equation is a homogeneous equation with coefficient matrix A  I n . We
are seeking non–trivial solutions, v, of this homogeneous system. From the theory
that we have been studying in this course, we know that the above system will have
non–trivial solutions if and only if its coefficient matrix, A  I n , is not invertible. This
will be true if and only if
detA  I n   0.
We conclude that the eigenvalues, , of the matrix A are precisely those values of
 that satisfy the above equation, which is called the characteristic equation of the
matrix A. Once we have found the eigenvalues of A, we must then proceed (for each
eigenvalue ) to solve the equation A  I n v  0 n in the usual way. The solution set
of this equation (which will definitely contain non–trivial solutions) is the eigenspace of
.
Example Find the eigenvalues and associated eigenspaces of the matrix
A
1 2
3 4
.
Solution Since
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1 2
A  I 2 
 0

3 4
0 

1
2
3
4
,
the eigenvalues of A are the solutions of
det
1
2
3
4
0
which can be written as
1  4    6  0
or, in expanded form as
 2  5  2  0.
Using the quadratic formula, we obtain the solutions
5  25  412
5  33

.
2
2
Therefore, the eigenvalues of A are
5  33

 0. 372
2
and
5  33

 5. 372.
2

To find eig
5 33
2
1
, we must solve the system
5 33
2
2
0

x2
5 33
2
4
3
x1
.
0
Since
1
5 33
2
2
4
3
~
5 33
2
1
3 33
6
0
0
,
we see that all solutions of the above homogeneous system have the form
x1
x2
eig
To find eig
5 33
2

5  33
2
3  33 t
6t
 Span
3  33
t
6
3  33
6
.
, we must solve the system
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1
5 33
2
2
3
5 33
2
4
x1
0

x2
.
0
Since
1
5 33
2
2
3
5 33
2
4
~
1
3 33
6
0
0
,
we see that all solutions of the above homogeneous system have the form
x1
x2
eig

5  33
2
3  33 t
6t
 Span
3  33
t
6
3  33
6
.
The figure below shows both of the eigenspaces of A.
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