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6.1
Eigenvalues and Eigenvectors
¶ permalink
Objectives
Learn the definition of eigenvector and eigenvalue.
Learn to find eigenvectors and eigenvalues geometrically.
Learn to decide if a number is an eigenvalue of a matrix, and if so, how to find an associated eigenvector.
Recipe:
find a basis for the
λ
-eigenspace.
Pictures:
whether or not a vector is an eigenvector, eigenvectors of standard matrix transformations.
Theorem:
the expanded invertible matrix theorem.
Vocabulary:
eigenspace
.
Essential Vocabulary:
eigenvector
,
eigenvalue
.
In this section, we define eigenvalues and eigenvectors. These form the most important facet of the structure theory of square matrices. As such, eigenvalues and eigenvectors tend to play a key role in the real-life applications of linear algebra.
Subsection
6.1.1
Eigenvalues and Eigenvectors
Here is the most important definition in this text.
Definition
Let
A
be an
n
×
n
matrix.
An
eigenvector
of
A
is a
nonzero
vector
v
in
R
n
such that
Av
=
λ
v
,
for some scalar
λ
.
An
eigenvalue
of
A
is a scalar
λ
such that the equation
Av
=
λ
v
has a
nontrivial
solution.
If
Av
=
λ
v
for
v
A
=
0,
we say that
λ
is the
eigenvalue for
v
,
and that
v
is an
eigenvector for
λ
.
The German prefix “eigen” roughly translates to “self” or “own”. An eigenvector of
A
is a vector that is taken to a multiple of itself by the matrix transformation
T
(
x
)=
Ax
,
which perhaps explains the terminology. On the other hand, “eigen” is often translated as “characteristic”; we may think of an eigenvector as describing an intrinsic, or characteristic, property of
A
.
Eigenvectors are
by definition nonzero
. Eigenvalues may be equal to zero.
We do not consider the zero vector to be an eigenvector: since
A
0
=
0
=
λ
0
for
every
scalar
λ
,
the associated eigenvalue would be undefined.
If someone hands you a matrix
A
and a vector
v
,
it is easy to check if
v
is an eigenvector of
A
:
simply multiply
v
by
A
and see if
Av
is a scalar multiple of
v
.
On the other hand, given just the matrix
A
,
it is not obvious at all how to find the eigenvectors. We will learn how to do this in
Section 6.2
.
Example
(Verifying eigenvectors)
Example
(Verifying eigenvectors)
Example
(An eigenvector with eigenvalue
0
)
To say that
Av
=
λ
v
means that
Av
and
λ
v
are
collinear with the origin
. So, an eigenvector of
A
is a nonzero vector
v
such that
Av
and
v
lie on the same line through the origin. In this case,
Av
is a scalar multiple of
v
;
the eigenvalue is the scaling factor.
v
Av
w
Aw
0
v
isaneigenvector
w
isnotaneigenvector
For matrices that arise as the standard matrix of a linear transformation, it is often best to draw a picture, then find the eigenvectors and eigenvalues geometrically by studying which vectors are not moved off of their line. For a transformation that is defined geometrically, it is not necessary even to compute its matrix to find the eigenvectors and eigenvalues.
Example
(Reflection)
Here is an example of this. Let
T
:
R
2
→
R
2
be the linear transformation that reflects over the line
L
defined by
y
=
−
x
,
and let
A
be the matrix for
T
.
We will find the eigenvalues and eigenvectors of
A
without doing any computations.
This transformation is defined geometrically, so we draw a picture.
L
u
Au
0
The vector
u
is not an eigenvector, because
Au
is not collinear with
u
and the origin.
L
z
Az
0
The vector
z
is not an eigenvector either.
L
v
Av
0
The vector
v
is an eigenvector because
Av
is collinear with
v
and the origin. The vector
Av
has the same length as
v
,
but the opposite direction, so the associated eigenvalue is
−
1.
L
w
Aw
0
The vector
w
is an eigenvector because
Aw
is collinear with
w
and the origin: indeed,
Aw
is equal to
w
!
This means that
w
is an eigenvector with eigenvalue
1.
It appears that all eigenvectors lie either on
L
,
or on the line perpendicular to
L
.
The vectors on
L
have eigenvalue
1,
and the vectors perpendicular to
L
have eigenvalue
−
1.
Figure
8
An eigenvector of
A
is a vector
x
such that
Ax
is collinear with
x
and the origin. Click and drag the head of
x
to convince yourself that all such vectors lie either on
L
,
or on the line perpendicular to
L
.
We will now give five more examples of this nature
Example
(Projection)
Example
(Identity)
Example
(Dilation)
Example
(Shear)
Example
(Rotation)
Here we mention one basic fact about eigenvectors.
Fact
(Eigenvectors with distinct eigenvalues are linearly independent)
Let
v
1
,
v
2
,...,
v
k
be eigenvectors of a matrix
A
,
and suppose that the corresponding eigenvalues
λ
1
,
λ
2
,...,
λ
k
are
distinct
(all different from each other). Then
{
v
1
,
v
2
,...,
v
k
}
is linearly independent.
When
k
=
2,
this says that if
v
1
,
v
2
are eigenvectors with eigenvalues
λ
1
A
=
λ
2
,
then
v
2
is not a multiple of
v
1
.
In fact, any nonzero multiple
cv
1
of
v
1
is also an eigenvector with eigenvalue
λ
1
:
A
(
cv
1
)=
cAv
1
=
c
(
λ
1
v
1
)=
λ
1
(
cv
1
)
.
As a consequence of the above
fact
, we have the following.
An
n
×
n
matrix
A
has at most
n
eigenvalues.
Subsection
6.1.2
Eigenspaces
Suppose that
A
is a square matrix. We already know how to check if a given vector is an eigenvector of
A
and in that case to find the eigenvalue. Our next goal is to check if a given real number is an eigenvalue of
A
and in that case to find all of the corresponding eigenvectors. Again this will be straightforward, but more involved. The only missing piece, then, will be to find the eigenvalues of
A
;
this is the main content of
Section 6.2
.
Let
A
be an
n
×
n
matrix, and let
λ
be a scalar. The eigenvectors with eigenvalue
λ
,
if any, are the nonzero solutions of the equation
Av
=
λ
v
.
We can rewrite this equation as follows:
Av
=
λ
v
⇐⇒
Av
−
λ
v
=
0
⇐⇒
Av
−
λ
I
n
v
=
0
⇐⇒
(
A
−
λ
I
n
)
v
=
0.
Therefore, the eigenvectors of
A
with eigenvalue
λ
,
if any, are the nontrivial solutions of the matrix equation
(
A
−
λ
I
n
)
v
=
0,
i.e., the nonzero vectors in
Nul
(
A
−
λ
I
n
)
.
If this equation has no nontrivial solutions, then
λ
is not an eigenvalue of
A
.
The above observation is important because it says that
finding the eigenvectors for a given eigenvalue means solving a homogeneous system of equations.
For instance, if
A
=
C
713
−
32
−
3
−
3
−
2
−
1
D
,
then an eigenvector with eigenvalue
λ
is a nontrivial solution of the matrix equation
C
713
−
32
−
3
−
3
−
2
−
1
DC
x
y
z
D
=
λ
C
x
y
z
D
.
This translates to the system of equations
E
7
x
+
y
+
3
z
=
λ
x
−
3
x
+
2
y
−
3
z
=
λ
y
−
3
x
−
2
y
−
z
=
λ
z
−−−→
E
(
7
−
λ
)
x
+
y
+
3
z
=
0
−
3
x
+(
2
−
λ
)
y
−
3
z
=
0
−
3
x
−
2
y
+(
−
1
−
λ
)
z
=
0.
This is the same as the homogeneous matrix equation
C
7
−
λ
13
−
32
−
λ
−
3
−
3
−
2
−
1
−
λ
DC
x
y
z
D
=
0,
i.e.,
(
A
−
λ
I
3
)
v
=
0.
Definition
Let
A
be an
n
×
n
matrix, and let
λ
be an eigenvalue of
A
.
The
λ
-eigenspace
of
A
is the solution set of
(
A
−
λ
I
n
)
v
=
0,
i.e., the subspace
Nul
(
A
−
λ
I
n
)
.
The
λ
-eigenspace
is a subspace because it is the null space of a matrix, namely, the matrix
A
−
λ
I
n
.
This subspace consists of the zero vector and all eigenvectors of
A
with eigenvalue
λ
.
Example
(Computing eigenspaces)
Example
(Computing eigenspaces)
Example
(Reflection)
Recipes: Eigenspaces
Let
A
be an
n
×
n
matrix and let
λ
be a number.
λ
is an eigenvalue of
A
if and only if
(
A
−
λ
I
n
)
v
=
0
has a nontrivial solution, if and only if
Nul
(
A
−
λ
I
n
)
A
=
{
0
}
.
In this case, finding a basis for the
λ
-eigenspace
of
A
means finding a basis for
Nul
(
A
−
λ
I
n
)
,
which can be done by finding the parametric vector form of the solutions of the homogeneous system of equations
(
A
−
λ
I
n
)
v
=
0.
The dimension of the
λ
-eigenspace
of
A
is equal to the number of free variables in the system of equations
(
A
−
λ
I
n
)
v
=
0,
which is the number of columns of
A
−
λ
I
n
without pivots.
The eigenvectors with eigenvalue
λ
are the nonzero vectors in
Nul
(
A
−
λ
I
n
)
,
or equivalently, the nontrivial solutions of
(
A
−
λ
I
n
)
v
=
0.
We conclude with an observation about the
0
-eigenspace
of a matrix.
Fact
Let
A
be an
n
×
n
matrix.
The number
0
is an eigenvalue of
A
if and only if
A
is not invertible.
In this case, the
0
-eigenspace
of
A
is
Nul
(
A
)
.
Proof
We know that
0
is an eigenvalue of
A
if and only if
Nul
(
A
−
0
I
n
)=
Nul
(
A
)
is nonzero, which is equivalent to the noninvertibility of
A
by the
invertible matrix theorem in Section 4.6
. In this case, the
0
-eigenspace
is by definition
Nul
(
A
−
0
I
n
)=
Nul
(
A
)
.
Concretely, an eigenvector with eigenvalue
0
is a nonzero vector
v
such that
Av
=
0
v
,
i.e., such that
Av
=
0.
These are exactly the nonzero vectors in the null space of
A
.
Subsection
6.1.3
The Invertible Matrix Theorem: Addenda
We now have two new ways of saying that a matrix is invertible, so we add them to the
invertible matrix theorem
.
Invertible Matrix Theorem
Let
A
be an
n
×
n
matrix, and let
T
:
R
n
→
R
n
be the matrix transformation
T
(
x
)=
Ax
.
The following statements are equivalent:
A
is invertible.
A
has
n
pivots.
Nul
(
A
)=
{
0
}
.
The columns of
A
are linearly independent.
The columns of
A
span
R
n
.
Ax
=
b
has a unique solution for each
b
in
R
n
.
T
is invertible.
T
is one-to-one.
T
is onto.
det
(
A
)
A
=
0.
0
is not an eigenvalue of
A
. |
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# [Interactive Linear AlgebraUBC edition](https://personal.math.ubc.ca/~tbjw/ila/index.html)
Dan Margalit, Joseph Rabinoff, Ben Williams
## [Front Matter](https://personal.math.ubc.ca/~tbjw/ila/index2.html)
- [Colophon](https://personal.math.ubc.ca/~tbjw/ila/colophon-1.html)
- [Contributors to this textbook](https://personal.math.ubc.ca/~tbjw/ila/preface-1.html)
- [History of the UBC edition](https://personal.math.ubc.ca/~tbjw/ila/preface-2.html)
- [Overview](https://personal.math.ubc.ca/~tbjw/ila/overview.html)
## [1Vectors: Algebra and Geometry](https://personal.math.ubc.ca/~tbjw/ila/chap-vectors.html)
- [Vectors](https://personal.math.ubc.ca/~tbjw/ila/vectors.html)
- [Vector Equations and Spans](https://personal.math.ubc.ca/~tbjw/ila/spans.html)
## [2Systems of Linear Equations: Algebra](https://personal.math.ubc.ca/~tbjw/ila/chap-algebra.html)
- [Systems of Linear Equations](https://personal.math.ubc.ca/~tbjw/ila/systems-of-linear-equations.html)
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- [Parametric Form](https://personal.math.ubc.ca/~tbjw/ila/parametric-form.html)
- [Matrix Equations](https://personal.math.ubc.ca/~tbjw/ila/matrix-equations.html)
## [3Solution Sets and Subspaces](https://personal.math.ubc.ca/~tbjw/ila/chap-geometry.html)
- [Solution Sets](https://personal.math.ubc.ca/~tbjw/ila/solution-sets.html)
- [Linear Independence](https://personal.math.ubc.ca/~tbjw/ila/linear-independence.html)
- [Subspaces](https://personal.math.ubc.ca/~tbjw/ila/subspaces.html)
- [Basis and Dimension](https://personal.math.ubc.ca/~tbjw/ila/dimension.html)
- [Bases as Coordinate Systems](https://personal.math.ubc.ca/~tbjw/ila/bases-as-coord-systems.html)
- [The Rank Theorem](https://personal.math.ubc.ca/~tbjw/ila/rank-thm.html)
## [4Linear Transformations and Matrix Algebra](https://personal.math.ubc.ca/~tbjw/ila/chap-matrices.html)
- [Matrix Transformations](https://personal.math.ubc.ca/~tbjw/ila/matrix-transformations.html)
- [One-to-one and Onto Transformations](https://personal.math.ubc.ca/~tbjw/ila/one-to-one-onto.html)
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- [Matrix Inverses](https://personal.math.ubc.ca/~tbjw/ila/matrix-inverses.html)
- [The Invertible Matrix Theorem](https://personal.math.ubc.ca/~tbjw/ila/invertible-matrix-thm.html)
## [5Determinants](https://personal.math.ubc.ca/~tbjw/ila/chap-determinant.html)
- [Determinants: Definition](https://personal.math.ubc.ca/~tbjw/ila/determinants-definitions-properties.html)
- [Cofactor Expansions](https://personal.math.ubc.ca/~tbjw/ila/determinants-cofactors.html)
- [Determinants and Volumes](https://personal.math.ubc.ca/~tbjw/ila/determinants-volumes.html)
## [6Eigenvalues and Eigenvectors](https://personal.math.ubc.ca/~tbjw/ila/chap-eigenvalues.html)
- [Eigenvalues and Eigenvectors](https://personal.math.ubc.ca/~tbjw/ila/eigenvectors.html)
- [The Characteristic Polynomial](https://personal.math.ubc.ca/~tbjw/ila/characteristic-polynomial.html)
- [Similarity](https://personal.math.ubc.ca/~tbjw/ila/similarity.html)
- [Diagonalization](https://personal.math.ubc.ca/~tbjw/ila/diagonalization.html)
- [Complex Eigenvalues](https://personal.math.ubc.ca/~tbjw/ila/complex-eigenvalues.html)
- [Discrete Dynamical Systems](https://personal.math.ubc.ca/~tbjw/ila/dds.html)
- [Stochastic Matrices and the Steady State](https://personal.math.ubc.ca/~tbjw/ila/stochastic2.html)
## [7Orthogonality](https://personal.math.ubc.ca/~tbjw/ila/chap-orthogonality.html)
- [Dot Products and Orthogonality](https://personal.math.ubc.ca/~tbjw/ila/dot-product.html)
- [Orthogonal Complements](https://personal.math.ubc.ca/~tbjw/ila/orthogonal-complements.html)
- [Orthogonal Projection](https://personal.math.ubc.ca/~tbjw/ila/projections.html)
- [Orthogonal Sets](https://personal.math.ubc.ca/~tbjw/ila/orthogonal-sets.html)
- [The Method of Least Squares](https://personal.math.ubc.ca/~tbjw/ila/least-squares.html)
## [Back Matter](https://personal.math.ubc.ca/~tbjw/ila/backmatter.html)
- [Complex Numbers](https://personal.math.ubc.ca/~tbjw/ila/complex-numbers.html)
- [Notation](https://personal.math.ubc.ca/~tbjw/ila/appendix-2.html)
- [GNU Free Documentation License](https://personal.math.ubc.ca/~tbjw/ila/appendix-gfdl.html)
- [Index](https://personal.math.ubc.ca/~tbjw/ila/index-1.html)
- [Colophon](https://personal.math.ubc.ca/~tbjw/ila/colophon-2.html)
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# Section6\.1Eigenvalues and Eigenvectors[¶ permalink](https://personal.math.ubc.ca/~tbjw/ila/eigenvectors.html)
##### Objectives
1. Learn the definition of eigenvector and eigenvalue.
2. Learn to find eigenvectors and eigenvalues geometrically.
3. Learn to decide if a number is an eigenvalue of a matrix, and if so, how to find an associated eigenvector.
4. *Recipe:* find a basis for the
λ
\-eigenspace.
5. *Pictures:* whether or not a vector is an eigenvector, eigenvectors of standard matrix transformations.
6. *Theorem:* the expanded invertible matrix theorem.
7. *Vocabulary:* *eigenspace*.
8. *Essential Vocabulary:* *eigenvector*, *eigenvalue*.
In this section, we define eigenvalues and eigenvectors. These form the most important facet of the structure theory of square matrices. As such, eigenvalues and eigenvectors tend to play a key role in the real-life applications of linear algebra.
# Subsection6\.1.1Eigenvalues and Eigenvectors
Here is the most important definition in this text.
##### Definition
Let A be an n × n matrix.
1. An *eigenvector* of
A
is a *nonzero* vector
v
in
R
n
such that
Av
\=
λ
v
,
for some scalar
λ
.
2. An *eigenvalue* of
A
is a scalar
λ
such that the equation
Av
\=
λ
v
has a *nontrivial* solution.
If Av \= λ v for v A \= 0, we say that λ is the *eigenvalue for* v , and that v is an *eigenvector for* λ .
The German prefix “eigen” roughly translates to “self” or “own”. An eigenvector of A is a vector that is taken to a multiple of itself by the matrix transformation T ( x )= Ax , which perhaps explains the terminology. On the other hand, “eigen” is often translated as “characteristic”; we may think of an eigenvector as describing an intrinsic, or characteristic, property of A .
##### Note
Eigenvalues and eigenvectors are only for square matrices.
Eigenvectors are *by definition nonzero*. Eigenvalues may be equal to zero.
We do not consider the zero vector to be an eigenvector: since A 0 \= 0 \= λ 0 for *every* scalar λ , the associated eigenvalue would be undefined.
If someone hands you a matrix A and a vector v , it is easy to check if v is an eigenvector of A : simply multiply v by A and see if Av is a scalar multiple of v . On the other hand, given just the matrix A , it is not obvious at all how to find the eigenvectors. We will learn how to do this in [Section 6.2](https://personal.math.ubc.ca/~tbjw/ila/characteristic-polynomial.html "Section 6.2 The Characteristic Polynomial").
[Example(Verifying eigenvectors)]("Example 6.1.4 Verifying eigenvectors")
[Example(Verifying eigenvectors)]("Example 6.1.5 Verifying eigenvectors")
[Example(An eigenvector with eigenvalue 0 )]("Example 6.1.6 An eigenvector with eigenvalue 0")
To say that Av \= λ v means that Av and λ v are *collinear with the origin*. So, an eigenvector of A is a nonzero vector v such that Av and v lie on the same line through the origin. In this case, Av is a scalar multiple of v ; the eigenvalue is the scaling factor.
v
Av
w
Aw
0
v
isaneigenvector
w
isnotaneigenvector
For matrices that arise as the standard matrix of a linear transformation, it is often best to draw a picture, then find the eigenvectors and eigenvalues geometrically by studying which vectors are not moved off of their line. For a transformation that is defined geometrically, it is not necessary even to compute its matrix to find the eigenvectors and eigenvalues.
##### Example(Reflection)
Here is an example of this. Let T : R 2 → R 2 be the linear transformation that reflects over the line L defined by y \= − x , and let A be the matrix for T . We will find the eigenvalues and eigenvectors of A without doing any computations.
This transformation is defined geometrically, so we draw a picture.
L
u
Au
0
The vector u is not an eigenvector, because Au is not collinear with u and the origin.
L
z
Az
0
The vector z is not an eigenvector either.
L
v
Av
0
The vector v is an eigenvector because Av is collinear with v and the origin. The vector Av has the same length as v , but the opposite direction, so the associated eigenvalue is − 1\.
L
w
Aw
0
The vector w is an eigenvector because Aw is collinear with w and the origin: indeed, Aw is equal to w \! This means that w is an eigenvector with eigenvalue 1\.
It appears that all eigenvectors lie either on L , or on the line perpendicular to L . The vectors on L have eigenvalue 1, and the vectors perpendicular to L have eigenvalue − 1\.
Figure8An eigenvector of
A
is a vector
x
such that
Ax
is collinear with
x
and the origin. Click and drag the head of
x
to convince yourself that all such vectors lie either on
L
,
or on the line perpendicular to
L
.
We will now give five more examples of this nature
[Example(Projection)]("Example 6.1.9 Projection")
[Example(Identity)]("Example 6.1.11 Identity")
[Example(Dilation)]("Example 6.1.13 Dilation")
[Example(Shear)]("Example 6.1.15 Shear")
[Example(Rotation)]("Example 6.1.17 Rotation")
Here we mention one basic fact about eigenvectors.
##### Fact(Eigenvectors with distinct eigenvalues are linearly independent)
Let v 1 , v 2 ,..., v k be eigenvectors of a matrix A , and suppose that the corresponding eigenvalues λ 1 , λ 2 ,..., λ k are *distinct* (all different from each other). Then { v 1 , v 2 ,..., v k } is linearly independent.
[Proof]()
Suppose that { v 1 , v 2 ,..., v k } were linearly *de*pendent. According to the [increasing span criterion in Section 3.2]("Theorem 3.2.14 Increasing Span Criterion"), this means that for some j , the vector v j is in Span { v 1 , v 2 ,..., v j − 1 } . If we choose the *first* such j , then { v 1 , v 2 ,..., v j − 1 } is linearly independent. Note that j \> 1 since v 1 A \= 0\.
Since v j is in Span { v 1 , v 2 ,..., v j − 1 } ,, we can write
v
j
\=
c
1
v
1
\+
c
2
v
2
\+
···
\+
c
j
−
1
v
j
−
1
for some scalars c 1 , c 2 ,..., c j − 1 . Multiplying both sides of the above equation by A gives
λ
j
v
j
\=
Av
j
\=
A
A
c
1
v
1
\+
c
2
v
2
\+
···
\+
c
j
−
1
v
j
−
1
B
\=
c
1
Av
1
\+
c
2
Av
2
\+
···
\+
c
j
−
1
Av
j
−
1
\=
c
1
λ
1
v
1
\+
c
2
λ
2
v
2
\+
···
\+
c
j
−
1
λ
j
−
1
v
j
−
1
.
Subtracting λ j times the first equation from the second gives
0
\=
λ
j
v
j
−
λ
j
v
j
\=
c
1
(
λ
1
−
λ
j
)
v
1
\+
c
2
(
λ
2
−
λ
j
)
v
2
\+
···
\+
c
j
−
1
(
λ
j
−
1
−
λ
j
)
v
j
−
1
.
Since λ i A \= λ j for i \< j , this is an equation of linear dependence among v 1 , v 2 ,..., v j − 1 , which is impossible because those vectors are linearly independent. Therefore, { v 1 , v 2 ,..., v k } must have been linearly independent after all.
When k \= 2, this says that if v 1 , v 2 are eigenvectors with eigenvalues λ 1 A \= λ 2 , then v 2 is not a multiple of v 1 . In fact, any nonzero multiple cv 1 of v 1 is also an eigenvector with eigenvalue λ 1 :
A
(
cv
1
)=
cAv
1
\=
c
(
λ
1
v
1
)=
λ
1
(
cv
1
)
.
As a consequence of the above [fact]("Fact 6.1.19 Eigenvectors with distinct eigenvalues are linearly independent"), we have the following.
An n × n matrix A has at most n eigenvalues.
# Subsection6\.1.2Eigenspaces
Suppose that A is a square matrix. We already know how to check if a given vector is an eigenvector of A and in that case to find the eigenvalue. Our next goal is to check if a given real number is an eigenvalue of A and in that case to find all of the corresponding eigenvectors. Again this will be straightforward, but more involved. The only missing piece, then, will be to find the eigenvalues of A ; this is the main content of [Section 6.2](https://personal.math.ubc.ca/~tbjw/ila/characteristic-polynomial.html "Section 6.2 The Characteristic Polynomial").
Let A be an n × n matrix, and let λ be a scalar. The eigenvectors with eigenvalue λ , if any, are the nonzero solutions of the equation Av \= λ v . We can rewrite this equation as follows:
Av
\=
λ
v
⇐⇒
Av
−
λ
v
\=
0
⇐⇒
Av
−
λ
I
n
v
\=
0
⇐⇒
(
A
−
λ
I
n
)
v
\=
0\.
Therefore, the eigenvectors of A with eigenvalue λ , if any, are the nontrivial solutions of the matrix equation ( A − λ I n ) v \= 0, i.e., the nonzero vectors in Nul ( A − λ I n ) . If this equation has no nontrivial solutions, then λ is not an eigenvalue of A .
The above observation is important because it says that *finding the eigenvectors for a given eigenvalue means solving a homogeneous system of equations.* For instance, if
A
\=
C
713
−
32
−
3
−
3
−
2
−
1
D
,
then an eigenvector with eigenvalue λ is a nontrivial solution of the matrix equation
C
713
−
32
−
3
−
3
−
2
−
1
DC
x
y
z
D
\=
λ
C
x
y
z
D
.
This translates to the system of equations
E
7
x
\+
y
\+
3
z
\=
λ
x
−
3
x
\+
2
y
−
3
z
\=
λ
y
−
3
x
−
2
y
−
z
\=
λ
z
−−−→
E
(
7
−
λ
)
x
\+
y
\+
3
z
\=
0
−
3
x
\+(
2
−
λ
)
y
−
3
z
\=
0
−
3
x
−
2
y
\+(
−
1
−
λ
)
z
\=
0\.
This is the same as the homogeneous matrix equation
C
7
−
λ
13
−
32
−
λ
−
3
−
3
−
2
−
1
−
λ
DC
x
y
z
D
\=
0,
i.e., ( A − λ I 3 ) v \= 0\.
##### Definition
Let A be an n × n matrix, and let λ be an eigenvalue of A . The *λ \-eigenspace* of A is the solution set of ( A − λ I n ) v \= 0, i.e., the subspace Nul ( A − λ I n ) .
The λ \-eigenspace is a subspace because it is the null space of a matrix, namely, the matrix A − λ I n . This subspace consists of the zero vector and all eigenvectors of A with eigenvalue λ .
##### Note
Since a nonzero subspace is infinite, *every eigenvalue has infinitely many eigenvectors*. (For example, multiplying an eigenvector by a nonzero scalar gives another eigenvector.) On the other hand, there can be at most n linearly independent eigenvectors of an n × n matrix, since R n has dimension n .
[Example(Computing eigenspaces)]("Example 6.1.23 Computing eigenspaces")
[Example(Computing eigenspaces)]("Example 6.1.25 Computing eigenspaces")
[Example(Reflection)]("Example 6.1.27 Reflection")
##### Recipes: Eigenspaces
Let A be an n × n matrix and let λ be a number.
1. λ
is an eigenvalue of
A
if and only if
(
A
−
λ
I
n
)
v
\=
0
has a nontrivial solution, if and only if
Nul
(
A
−
λ
I
n
)
A
\=
{
0
}
.
2. In this case, finding a basis for the
λ
\-eigenspace
of
A
means finding a basis for
Nul
(
A
−
λ
I
n
)
,
which can be done by finding the parametric vector form of the solutions of the homogeneous system of equations
(
A
−
λ
I
n
)
v
\=
0\.
3. The dimension of the
λ
\-eigenspace
of
A
is equal to the number of free variables in the system of equations
(
A
−
λ
I
n
)
v
\=
0,
which is the number of columns of
A
−
λ
I
n
without pivots.
4. The eigenvectors with eigenvalue
λ
are the nonzero vectors in
Nul
(
A
−
λ
I
n
)
,
or equivalently, the nontrivial solutions of
(
A
−
λ
I
n
)
v
\=
0\.
We conclude with an observation about the 0 \-eigenspace of a matrix.
##### Fact
Let A be an n × n matrix.
1. The number
0
is an eigenvalue of
A
if and only if
A
is not invertible.
2. In this case, the
0
\-eigenspace
of
A
is
Nul
(
A
)
.
##### Proof
We know that 0 is an eigenvalue of A if and only if Nul ( A − 0 I n )= Nul ( A ) is nonzero, which is equivalent to the noninvertibility of A by the [invertible matrix theorem in Section 4.6]("Invertible Matrix Theorem 4.6.1 "). In this case, the 0 \-eigenspace is by definition Nul ( A − 0 I n )= Nul ( A ) .
Concretely, an eigenvector with eigenvalue 0 is a nonzero vector v such that Av \= 0 v , i.e., such that Av \= 0\. These are exactly the nonzero vectors in the null space of A .
# Subsection6\.1.3The Invertible Matrix Theorem: Addenda
We now have two new ways of saying that a matrix is invertible, so we add them to the [invertible matrix theorem]("Invertible Matrix Theorem 4.6.1 ").
##### Invertible Matrix Theorem
Let A be an n × n matrix, and let T : R n → R n be the matrix transformation T ( x )= Ax . The following statements are equivalent:
1. A
is invertible.
2. A
has
n
pivots.
3. Nul
(
A
)=
{
0
}
.
4. The columns of
A
are linearly independent.
5. The columns of
A
span
R
n
.
6. Ax
\=
b
has a unique solution for each
b
in
R
n
.
7. T
is invertible.
8. T
is one-to-one.
9. T
is onto.
10. det
(
A
)
A
\=
0\.
11. 0
is not an eigenvalue of
A
.
[Comments, corrections or suggestions?(Free GitHub account required)](https://github.com/tbjw/ILAUBC/issues/new?title=Feedback%20on%20eigenvectors&body=In%20reference%20to%20version%205c2a53d525a9556fcdce1be3670584cf9b7b984b) |
| Readable Markdown | ## Section6\.1Eigenvalues and Eigenvectors[¶ permalink](https://personal.math.ubc.ca/~tbjw/ila/eigenvectors.html)
##### Objectives
1. Learn the definition of eigenvector and eigenvalue.
2. Learn to find eigenvectors and eigenvalues geometrically.
3. Learn to decide if a number is an eigenvalue of a matrix, and if so, how to find an associated eigenvector.
4. *Recipe:* find a basis for the
λ
\-eigenspace.
5. *Pictures:* whether or not a vector is an eigenvector, eigenvectors of standard matrix transformations.
6. *Theorem:* the expanded invertible matrix theorem.
7. *Vocabulary:* *eigenspace*.
8. *Essential Vocabulary:* *eigenvector*, *eigenvalue*.
In this section, we define eigenvalues and eigenvectors. These form the most important facet of the structure theory of square matrices. As such, eigenvalues and eigenvectors tend to play a key role in the real-life applications of linear algebra.
## Subsection6\.1.1Eigenvalues and Eigenvectors
Here is the most important definition in this text.
##### Definition
Let A be an n × n matrix.
1. An *eigenvector* of
A
is a *nonzero* vector
v
in
R
n
such that
Av
\=
λ
v
,
for some scalar
λ
.
2. An *eigenvalue* of
A
is a scalar
λ
such that the equation
Av
\=
λ
v
has a *nontrivial* solution.
If Av \= λ v for v A \= 0, we say that λ is the *eigenvalue for* v , and that v is an *eigenvector for* λ .
The German prefix “eigen” roughly translates to “self” or “own”. An eigenvector of A is a vector that is taken to a multiple of itself by the matrix transformation T ( x )= Ax , which perhaps explains the terminology. On the other hand, “eigen” is often translated as “characteristic”; we may think of an eigenvector as describing an intrinsic, or characteristic, property of A .
Eigenvectors are *by definition nonzero*. Eigenvalues may be equal to zero.
We do not consider the zero vector to be an eigenvector: since A 0 \= 0 \= λ 0 for *every* scalar λ , the associated eigenvalue would be undefined.
If someone hands you a matrix A and a vector v , it is easy to check if v is an eigenvector of A : simply multiply v by A and see if Av is a scalar multiple of v . On the other hand, given just the matrix A , it is not obvious at all how to find the eigenvectors. We will learn how to do this in [Section 6.2](https://personal.math.ubc.ca/~tbjw/ila/characteristic-polynomial.html "Section 6.2 The Characteristic Polynomial").
[Example(Verifying eigenvectors)]("Example 6.1.4 Verifying eigenvectors")
[Example(Verifying eigenvectors)]("Example 6.1.5 Verifying eigenvectors")
[Example(An eigenvector with eigenvalue 0 )]("Example 6.1.6 An eigenvector with eigenvalue 0")
To say that Av \= λ v means that Av and λ v are *collinear with the origin*. So, an eigenvector of A is a nonzero vector v such that Av and v lie on the same line through the origin. In this case, Av is a scalar multiple of v ; the eigenvalue is the scaling factor.
v Av w Aw 0 v isaneigenvector w isnotaneigenvector
For matrices that arise as the standard matrix of a linear transformation, it is often best to draw a picture, then find the eigenvectors and eigenvalues geometrically by studying which vectors are not moved off of their line. For a transformation that is defined geometrically, it is not necessary even to compute its matrix to find the eigenvectors and eigenvalues.
##### Example(Reflection)
Here is an example of this. Let T : R 2 → R 2 be the linear transformation that reflects over the line L defined by y \= − x , and let A be the matrix for T . We will find the eigenvalues and eigenvectors of A without doing any computations.
This transformation is defined geometrically, so we draw a picture.
L u Au 0
The vector u is not an eigenvector, because Au is not collinear with u and the origin.
L z Az 0
The vector z is not an eigenvector either.
L v Av 0
The vector v is an eigenvector because Av is collinear with v and the origin. The vector Av has the same length as v , but the opposite direction, so the associated eigenvalue is − 1\.
L w Aw 0
The vector w is an eigenvector because Aw is collinear with w and the origin: indeed, Aw is equal to w \! This means that w is an eigenvector with eigenvalue 1\.
It appears that all eigenvectors lie either on L , or on the line perpendicular to L . The vectors on L have eigenvalue 1, and the vectors perpendicular to L have eigenvalue − 1\.
Figure8An eigenvector of
A
is a vector
x
such that
Ax
is collinear with
x
and the origin. Click and drag the head of
x
to convince yourself that all such vectors lie either on
L
,
or on the line perpendicular to
L
.
We will now give five more examples of this nature
[Example(Projection)]("Example 6.1.9 Projection")
[Example(Identity)]("Example 6.1.11 Identity")
[Example(Dilation)]("Example 6.1.13 Dilation")
[Example(Shear)]("Example 6.1.15 Shear")
[Example(Rotation)]("Example 6.1.17 Rotation")
Here we mention one basic fact about eigenvectors.
##### Fact(Eigenvectors with distinct eigenvalues are linearly independent)
Let v 1 , v 2 ,..., v k be eigenvectors of a matrix A , and suppose that the corresponding eigenvalues λ 1 , λ 2 ,..., λ k are *distinct* (all different from each other). Then { v 1 , v 2 ,..., v k } is linearly independent.
When k \= 2, this says that if v 1 , v 2 are eigenvectors with eigenvalues λ 1 A \= λ 2 , then v 2 is not a multiple of v 1 . In fact, any nonzero multiple cv 1 of v 1 is also an eigenvector with eigenvalue λ 1 :
A ( cv 1 )= cAv 1 \= c ( λ 1 v 1 )= λ 1 ( cv 1 ) .
As a consequence of the above [fact]("Fact 6.1.19 Eigenvectors with distinct eigenvalues are linearly independent"), we have the following.
An n × n matrix A has at most n eigenvalues.
## Subsection6\.1.2Eigenspaces
Suppose that A is a square matrix. We already know how to check if a given vector is an eigenvector of A and in that case to find the eigenvalue. Our next goal is to check if a given real number is an eigenvalue of A and in that case to find all of the corresponding eigenvectors. Again this will be straightforward, but more involved. The only missing piece, then, will be to find the eigenvalues of A ; this is the main content of [Section 6.2](https://personal.math.ubc.ca/~tbjw/ila/characteristic-polynomial.html "Section 6.2 The Characteristic Polynomial").
Let A be an n × n matrix, and let λ be a scalar. The eigenvectors with eigenvalue λ , if any, are the nonzero solutions of the equation Av \= λ v . We can rewrite this equation as follows:
Av \= λ v ⇐⇒ Av − λ v \= 0 ⇐⇒ Av − λ I n v \= 0 ⇐⇒ ( A − λ I n ) v \= 0\.
Therefore, the eigenvectors of A with eigenvalue λ , if any, are the nontrivial solutions of the matrix equation ( A − λ I n ) v \= 0, i.e., the nonzero vectors in Nul ( A − λ I n ) . If this equation has no nontrivial solutions, then λ is not an eigenvalue of A .
The above observation is important because it says that *finding the eigenvectors for a given eigenvalue means solving a homogeneous system of equations.* For instance, if
A \= C 713 − 32 − 3 − 3 − 2 − 1 D ,
then an eigenvector with eigenvalue λ is a nontrivial solution of the matrix equation
C 713 − 32 − 3 − 3 − 2 − 1 DC x y z D \= λ C x y z D .
This translates to the system of equations
E 7 x \+ y \+ 3 z \= λ x − 3 x \+ 2 y − 3 z \= λ y − 3 x − 2 y − z \= λ z −−−→ E ( 7 − λ ) x \+ y \+ 3 z \= 0 − 3 x \+( 2 − λ ) y − 3 z \= 0 − 3 x − 2 y \+( − 1 − λ ) z \= 0\.
This is the same as the homogeneous matrix equation
C 7 − λ 13 − 32 − λ − 3 − 3 − 2 − 1 − λ DC x y z D \= 0,
i.e., ( A − λ I 3 ) v \= 0\.
##### Definition
Let A be an n × n matrix, and let λ be an eigenvalue of A . The *λ \-eigenspace* of A is the solution set of ( A − λ I n ) v \= 0, i.e., the subspace Nul ( A − λ I n ) .
The λ \-eigenspace is a subspace because it is the null space of a matrix, namely, the matrix A − λ I n . This subspace consists of the zero vector and all eigenvectors of A with eigenvalue λ .
[Example(Computing eigenspaces)]("Example 6.1.23 Computing eigenspaces")
[Example(Computing eigenspaces)]("Example 6.1.25 Computing eigenspaces")
[Example(Reflection)]("Example 6.1.27 Reflection")
##### Recipes: Eigenspaces
Let A be an n × n matrix and let λ be a number.
1. λ
is an eigenvalue of
A
if and only if
(
A
−
λ
I
n
)
v
\=
0
has a nontrivial solution, if and only if
Nul
(
A
−
λ
I
n
)
A
\=
{
0
}
.
2. In this case, finding a basis for the
λ
\-eigenspace
of
A
means finding a basis for
Nul
(
A
−
λ
I
n
)
,
which can be done by finding the parametric vector form of the solutions of the homogeneous system of equations
(
A
−
λ
I
n
)
v
\=
0\.
3. The dimension of the
λ
\-eigenspace
of
A
is equal to the number of free variables in the system of equations
(
A
−
λ
I
n
)
v
\=
0,
which is the number of columns of
A
−
λ
I
n
without pivots.
4. The eigenvectors with eigenvalue
λ
are the nonzero vectors in
Nul
(
A
−
λ
I
n
)
,
or equivalently, the nontrivial solutions of
(
A
−
λ
I
n
)
v
\=
0\.
We conclude with an observation about the 0 \-eigenspace of a matrix.
##### Fact
Let A be an n × n matrix.
1. The number
0
is an eigenvalue of
A
if and only if
A
is not invertible.
2. In this case, the
0
\-eigenspace
of
A
is
Nul
(
A
)
.
##### Proof
We know that 0 is an eigenvalue of A if and only if Nul ( A − 0 I n )= Nul ( A ) is nonzero, which is equivalent to the noninvertibility of A by the [invertible matrix theorem in Section 4.6]("Invertible Matrix Theorem 4.6.1 "). In this case, the 0 \-eigenspace is by definition Nul ( A − 0 I n )= Nul ( A ) .
Concretely, an eigenvector with eigenvalue 0 is a nonzero vector v such that Av \= 0 v , i.e., such that Av \= 0\. These are exactly the nonzero vectors in the null space of A .
## Subsection6\.1.3The Invertible Matrix Theorem: Addenda
We now have two new ways of saying that a matrix is invertible, so we add them to the [invertible matrix theorem]("Invertible Matrix Theorem 4.6.1 ").
##### Invertible Matrix Theorem
Let A be an n × n matrix, and let T : R n → R n be the matrix transformation T ( x )= Ax . The following statements are equivalent:
1. A
is invertible.
2. A
has
n
pivots.
3. Nul
(
A
)=
{
0
}
.
4. The columns of
A
are linearly independent.
5. The columns of
A
span
R
n
.
6. Ax
\=
b
has a unique solution for each
b
in
R
n
.
7. T
is invertible.
8. T
is one-to-one.
9. T
is onto.
10. det
(
A
)
A
\=
0\.
11. 0
is not an eigenvalue of
A
. |
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