Linear Algebra. Session 9roquesol/Math_304_Fall_2017_Session_9.pdfSession 9. Abstract Linear Algebra Eigenvalues ansd Eigenvectors. Systems of Linear Di erential Equations Eigenvalues

Abstract Linear Algebra

Linear Algebra. Session 9

Dr. Marco A Roque Sol

08/01/2017

Dr. Marco A Roque Sol Linear Algebra. Session 9

Abstract Linear AlgebraEigenvalues ansd Eigenvectors.Systems of Linear Differential Equations

Eigenvalues and Eigenvectors.


The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x into a new vector x.Given an n × n matrix A we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx

but this is equivalent to say that

Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y


Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y


Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y


Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed

as a linear transformation that maps (or transforms)a given vector x into a new vector x.Given an n × n matrix A we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation

that maps (or transforms)a given vector x into a new vector x.Given an n × n matrix A we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps

(or transforms)a given vector x into a new vector x.Given an n × n matrix A we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)

a given vector x into a new vector x.Given an n × n matrix A we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector

x into a new vector x.Given an n × n matrix A we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x

into a new vector x.Given an n × n matrix A we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x into

a new vector x.Given an n × n matrix A we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x into a new vector x.Given

an n × n matrix A we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x into a new vector x.Given an n × n matrix

A we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x into a new vector x.Given an n × n matrix A

we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x into a new vector x.Given an n × n matrix A we consider

the problem of finding avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x into a new vector x.Given an n × n matrix A we consider the problem

of finding avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x into a new vector x.Given an n × n matrix A we consider the problem of finding

avector x that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x into a new vector x.Given an n × n matrix A we consider the problem of finding avector x

that is transformed into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x into a new vector x.Given an n × n matrix A we consider the problem of finding avector x that is transformed

into a multiple of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x into a new vector x.Given an n × n matrix A we consider the problem of finding avector x that is transformed into a multiple

of itself

Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y


Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y


Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y


Ax = λx

but

this is equivalent to say that

Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y


Ax = λx

but this is equivalent

to say that

Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y


Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y


Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y


Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0





The equation

Ax = y


Ax = λx


Ax = λIx

Ax− λIx = 0

(A− λI) x = 0




The latter equation has nonzero solutions if and only if λ is chosenso that

|A− λI| = 0

This is a polynomial equation of degree n in λ and is called thecharacteristic equation of the matrix A

Values of λ may be either real or complex-valued and are calledeigenvalues of A . The nonzero vectors that are obtained by usingsuch a value of λ are called the eigenvectors corresponding tothat eigenvalue.




The latter equation

has nonzero solutions if and only if λ is chosenso that

|A− λI| = 0






The latter equation has nonzero solutions

if and only if λ is chosenso that

|A− λI| = 0






The latter equation has nonzero solutions if and only if

λ is chosenso that

|A− λI| = 0






The latter equation has nonzero solutions if and only if λ

is chosenso that

|A− λI| = 0






The latter equation has nonzero solutions if and only if λ is chosenso

that

|A− λI| = 0







|A− λI| = 0







|A− λI| = 0







|A− λI| = 0

This is

a polynomial equation of degree n in λ and is called thecharacteristic equation of the matrix A






|A− λI| = 0

This is a polynomial equation

of degree n in λ and is called thecharacteristic equation of the matrix A






|A− λI| = 0

This is a polynomial equation of degree n

in λ and is called thecharacteristic equation of the matrix A






|A− λI| = 0

This is a polynomial equation of degree n in λ

and is called thecharacteristic equation of the matrix A






|A− λI| = 0

This is a polynomial equation of degree n in λ and is called

thecharacteristic equation of the matrix A






|A− λI| = 0

This is a polynomial equation of degree n in λ and is called thecharacteristic equation

of the matrix A






|A− λI| = 0







|A− λI| = 0


Values of λ

may be either real or complex-valued and are calledeigenvalues of A . The nonzero vectors that are obtained by usingsuch a value of λ are called the eigenvectors corresponding tothat eigenvalue.





|A− λI| = 0


Values of λ may be either real or complex-valued and

are calledeigenvalues of A . The nonzero vectors that are obtained by usingsuch a value of λ are called the eigenvectors corresponding tothat eigenvalue.





|A− λI| = 0


Values of λ may be either real or complex-valued and are called

eigenvalues of A . The nonzero vectors that are obtained by usingsuch a value of λ are called the eigenvectors corresponding tothat eigenvalue.





|A− λI| = 0


Values of λ may be either real or complex-valued and are calledeigenvalues

of A . The nonzero vectors that are obtained by usingsuch a value of λ are called the eigenvectors corresponding tothat eigenvalue.





|A− λI| = 0


Values of λ may be either real or complex-valued and are calledeigenvalues of A .

The nonzero vectors that are obtained by usingsuch a value of λ are called the eigenvectors corresponding tothat eigenvalue.





|A− λI| = 0


Values of λ may be either real or complex-valued and are calledeigenvalues of A . The nonzero vectors

that are obtained by usingsuch a value of λ are called the eigenvectors corresponding tothat eigenvalue.





|A− λI| = 0


Values of λ may be either real or complex-valued and are calledeigenvalues of A . The nonzero vectors that are obtained

by usingsuch a value of λ are called the eigenvectors corresponding tothat eigenvalue.





|A− λI| = 0


Values of λ may be either real or complex-valued and are calledeigenvalues of A . The nonzero vectors that are obtained by using

such a value of λ are called the eigenvectors corresponding tothat eigenvalue.





|A− λI| = 0


Values of λ may be either real or complex-valued and are calledeigenvalues of A . The nonzero vectors that are obtained by usingsuch a value

of λ are called the eigenvectors corresponding tothat eigenvalue.





|A− λI| = 0


Values of λ may be either real or complex-valued and are calledeigenvalues of A . The nonzero vectors that are obtained by usingsuch a value of λ

are called the eigenvectors corresponding tothat eigenvalue.





|A− λI| = 0


Values of λ may be either real or complex-valued and are calledeigenvalues of A . The nonzero vectors that are obtained by usingsuch a value of λ are called

the eigenvectors corresponding tothat eigenvalue.





|A− λI| = 0


Values of λ may be either real or complex-valued and are calledeigenvalues of A . The nonzero vectors that are obtained by usingsuch a value of λ are called the eigenvectors

corresponding tothat eigenvalue.





|A− λI| = 0


Values of λ may be either real or complex-valued and are calledeigenvalues of A . The nonzero vectors that are obtained by usingsuch a value of λ are called the eigenvectors corresponding

tothat eigenvalue.





|A− λI| = 0






Facts

a) It is possible to show that if λ1 and λ2 are two eigenvalues of Aandif λ1 6= λ2, then their corresponding eigenvectors x (1) and x (2)

are linearly independent.

This result extends to any set λ1, ..., λk of distinct eigenvalues:their eigenvectors x (1), ..., x (k) are linearly independent. Thus, ifeach eigenvalue of an n×n matrix is simple,then the n eigenvectorsof A , one for each eigenvalue, are linearly independent.




Facts







Facts

a) It is possible to show that if λ1 and λ2

are two eigenvalues of Aandif λ1 6= λ2, then their corresponding eigenvectors x (1) and x (2)






Facts

a) It is possible to show that if λ1 and λ2 are two eigenvalues of Aand

if λ1 6= λ2, then their corresponding eigenvectors x (1) and x (2)






Facts

a) It is possible to show that if λ1 and λ2 are two eigenvalues of Aandif λ1 6= λ2, then

their corresponding eigenvectors x (1) and x (2)






Facts







Facts



This result extends to any set λ1, ..., λk of distinct eigenvalues:

their eigenvectors x (1), ..., x (k) are linearly independent. Thus, ifeach eigenvalue of an n×n matrix is simple,then the n eigenvectorsof A , one for each eigenvalue, are linearly independent.




Facts



This result extends to any set λ1, ..., λk of distinct eigenvalues:their eigenvectors x (1), ..., x (k) are linearly independent.

Thus, ifeach eigenvalue of an n×n matrix is simple,then the n eigenvectorsof A , one for each eigenvalue, are linearly independent.




Facts



This result extends to any set λ1, ..., λk of distinct eigenvalues:their eigenvectors x (1), ..., x (k) are linearly independent. Thus, ifeach eigenvalue of an n×n matrix is simple,

then the n eigenvectorsof A , one for each eigenvalue, are linearly independent.




Facts



This result extends to any set λ1, ..., λk of distinct eigenvalues:their eigenvectors x (1), ..., x (k) are linearly independent. Thus, ifeach eigenvalue of an n×n matrix is simple,then the n eigenvectorsof A ,

one for each eigenvalue, are linearly independent.




Facts







b) On the other hand, if A has one or more repeated eigenvalues,then there may be fewer than n linearly independent eigenvectorsassociated with A, since for a repeated eigenvalue with multiplicitym, we may have q < m linearly independent vectors.

c) In the case of an eigenvalue, λi with multiplicity m, if we canfind m eigenvectors x (i1), x (i2), ..., x (im), linearly independentassociated to λi , we say that the matrix is Non-defective.

d) Otherwise, if we are able to find just x (i1), x (i2), ..., x (iq);q < m linearly independent associated to λi , we say that thematrix is Defective.




b) On the other hand,

if A has one or more repeated eigenvalues,then there may be fewer than n linearly independent eigenvectorsassociated with A, since for a repeated eigenvalue with multiplicitym, we may have q < m linearly independent vectors.






b) On the other hand, if A has one or more repeated eigenvalues,

then there may be fewer than n linearly independent eigenvectorsassociated with A, since for a repeated eigenvalue with multiplicitym, we may have q < m linearly independent vectors.






b) On the other hand, if A has one or more repeated eigenvalues,then there may be fewer than n linearly independent eigenvectorsassociated with A,

since for a repeated eigenvalue with multiplicitym, we may have q < m linearly independent vectors.













c) In the case of an eigenvalue, λi with multiplicity m, if

we canfind m eigenvectors x (i1), x (i2), ..., x (im), linearly independentassociated to λi , we say that the matrix is Non-defective.






c) In the case of an eigenvalue, λi with multiplicity m, if we canfind m eigenvectors x (i1), x (i2), ..., x (im),

linearly independentassociated to λi , we say that the matrix is Non-defective.













d) Otherwise, if we are able to find

just x (i1), x (i2), ..., x (iq);q < m linearly independent associated to λi , we say that thematrix is Defective.






d) Otherwise, if we are able to find just x (i1), x (i2), ..., x (iq);q < m

linearly independent associated to λi , we say that thematrix is Defective.










Example 9.1

Find the eigenvalues and eigenvectors of the matrix

A =

2 −3 −10 −1 0−1 1 2

Solution

The eigenvalues λ and eigenvectors x satisfy the equation(A− λI) x = 0, or




Example 9.1


A =

2 −3 −10 −1 0−1 1 2

Solution





Example 9.1

Find

the eigenvalues and eigenvectors of the matrix

A =

2 −3 −10 −1 0−1 1 2

Solution





Example 9.1

Find the eigenvalues and

eigenvectors of the matrix

A =

2 −3 −10 −1 0−1 1 2

Solution





Example 9.1

Find the eigenvalues and eigenvectors of

the matrix

A =

2 −3 −10 −1 0−1 1 2

Solution





Example 9.1


A =

2 −3 −10 −1 0−1 1 2

Solution





Example 9.1


A =

2 −3 −10 −1 0−1 1 2

Solution





Example 9.1


A =

2 −3 −10 −1 0−1 1 2

Solution





Example 9.1


A =

2 −3 −10 −1 0−1 1 2

Solution

The eigenvalues

λ and eigenvectors x satisfy the equation(A− λI) x = 0, or




Example 9.1


A =

2 −3 −10 −1 0−1 1 2

Solution

The eigenvalues λ and

eigenvectors x satisfy the equation(A− λI) x = 0, or




Example 9.1


A =

2 −3 −10 −1 0−1 1 2

Solution

The eigenvalues λ and eigenvectors x

satisfy the equation(A− λI) x = 0, or




Example 9.1


A =

2 −3 −10 −1 0−1 1 2

Solution

The eigenvalues λ and eigenvectors x satisfy the equation

(A− λI) x = 0, or




Example 9.1


A =

2 −3 −10 −1 0−1 1 2

Solution





(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

=

000

The eigenvalues are the roots of the equation

|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =




(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =




(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =




(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

=

000

The eigenvalues

are the roots of the equation

|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =




(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

=

000

The eigenvalues are the roots

of the equation

|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =




(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =




(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =




(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =




(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =




(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =




−

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

]= 0

The roots are λ1 = −1, λ2 = 1, and λ3 = 3 .

1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




−

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ =

(1 + λ)[(2− λ)2 − 1

]= 0


1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




−

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

]= 0


1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




−

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

]= 0

The roots are

λ1 = −1, λ2 = 1, and λ3 = 3 .

1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




−

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

]= 0


1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




−

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

]= 0


1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




−

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

]= 0


1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




−

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

]= 0


1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




−

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

]= 0


1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




3 −3 −10 0 0−1 1 3

x1x2x3

=

000

We can reduce this to the equivalent system

3 −3 −10 0 01 1 3

=

1 1 30 0 03 −3 −1

=

1 1 30 0 80 0 0

x1x2x3

=

000




3 −3 −10 0 0−1 1 3

x1x2x3

=

000


3 −3 −10 0 01 1 3

=

1 1 30 0 03 −3 −1

=

1 1 30 0 80 0 0

x1x2x3

=

000




3 −3 −10 0 0−1 1 3

x1x2x3

=

000

We can

reduce this to the equivalent system

3 −3 −10 0 01 1 3

=

1 1 30 0 03 −3 −1

=

1 1 30 0 80 0 0

x1x2x3

=

000




3 −3 −10 0 0−1 1 3

x1x2x3

=

000

We can reduce this

to the equivalent system

3 −3 −10 0 01 1 3

=

1 1 30 0 03 −3 −1

=

1 1 30 0 80 0 0

x1x2x3

=

000




3 −3 −10 0 0−1 1 3

x1x2x3

=

000


3 −3 −10 0 01 1 3

=

1 1 30 0 03 −3 −1

=

1 1 30 0 80 0 0

x1x2x3

=

000




3 −3 −10 0 0−1 1 3

x1x2x3

=

000


3 −3 −10 0 01 1 3

=

1 1 30 0 03 −3 −1

=

1 1 30 0 80 0 0

x1x2x3

=

000




3 −3 −10 0 0−1 1 3

x1x2x3

=

000


3 −3 −10 0 01 1 3

=

1 1 30 0 03 −3 −1

=

1 1 30 0 80 0 0

x1x2x3

=

000




3 −3 −10 0 0−1 1 3

x1x2x3

=

000


3 −3 −10 0 01 1 3

=

1 1 30 0 03 −3 −1

=

1 1 30 0 80 0 0

x1x2x3

=

000




The above system is reduced immediately to the equations

x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x2 = α. Therefore we have x1 = −x2 = −α, andx3 = 0 . Thus, we get

x =

−αα0

= α

1−10

; α = real number




The above system

is reduced immediately to the equations

x1 + x2 + 3x3 = 0 8x3 = 0


x =

−αα0

= α

1−10

; α = real number




The above system is reduced immediately to

the equations

x1 + x2 + 3x3 = 0 8x3 = 0


x =

−αα0

= α

1−10

; α = real number





x1 + x2 + 3x3 = 0 8x3 = 0


x =

−αα0

= α

1−10

; α = real number





x1 + x2 + 3x3 = 0 8x3 = 0


x =

−αα0

= α

1−10

; α = real number





x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence,

one of them is freevariable, let’s say x2 = α. Therefore we have x1 = −x2 = −α, andx3 = 0 . Thus, we get

x =

−αα0

= α

1−10

; α = real number





x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence, one of them is freevariable,

let’s say x2 = α. Therefore we have x1 = −x2 = −α, andx3 = 0 . Thus, we get

x =

−αα0

= α

1−10

; α = real number





x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x2 = α.

Therefore we have x1 = −x2 = −α, andx3 = 0 . Thus, we get

x =

−αα0

= α

1−10

; α = real number





x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x2 = α. Therefore we have

x1 = −x2 = −α, andx3 = 0 . Thus, we get

x =

−αα0

= α

1−10

; α = real number





x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x2 = α. Therefore we have x1 = −x2 = −α, and

x3 = 0 . Thus, we get

x =

−αα0

= α

1−10

; α = real number





x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x2 = α. Therefore we have x1 = −x2 = −α, andx3 = 0 .

Thus, we get

x =

−αα0

= α

1−10

; α = real number





x1 + x2 + 3x3 = 0 8x3 = 0


x =

−αα0

= α

1−10

; α = real number





x1 + x2 + 3x3 = 0 8x3 = 0


x =

−αα0

= α

1−10

; α = real number




A particular eigenvector is

x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

=




A particular

eigenvector is

x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

=





x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

=





x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

=





x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

=





x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

=





x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

=

hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

=





x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

=




1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x3 = α. Therefore we have x1 = x3 = α, andx2 = 0 . Thus, we get




1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0





1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0





1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0





1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000

The above system


x1 − 3x2 − x3 = 0 − 2x2 = 0





1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000

The above system is reduced immediately

to the equations

x1 − 3x2 − x3 = 0 − 2x2 = 0





1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0





1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0





1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and

two unknowns. Hence, one of them is freevariable, let’s say x3 = α. Therefore we have x1 = x3 = α, andx2 = 0 . Thus, we get




1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns.

Hence, one of them is freevariable, let’s say x3 = α. Therefore we have x1 = x3 = α, andx2 = 0 . Thus, we get




1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0


one of them is freevariable, let’s say x3 = α. Therefore we have x1 = x3 = α, andx2 = 0 . Thus, we get




1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them

is freevariable, let’s say x3 = α. Therefore we have x1 = x3 = α, andx2 = 0 . Thus, we get




1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them is freevariable,

let’s say x3 = α. Therefore we have x1 = x3 = α, andx2 = 0 . Thus, we get




1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x3 = α.

Therefore we have x1 = x3 = α, andx2 = 0 . Thus, we get




1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x3 = α. Therefore we have

x1 = x3 = α, andx2 = 0 . Thus, we get




1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x3 = α. Therefore we have x1 = x3 = α, and





1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x3 = α. Therefore we have x1 = x3 = α, andx2 = 0 .

Thus, we get




1 −3 −10 −2 00 −2 0

=

1 −3 −10 −2 00 0 0

x1x2x3

=

000


x1 − 3x2 − x3 = 0 − 2x2 = 0




Eigenvalues ansd Eigenvectors.

x =

α0α

= α

101

; α = real number

A particular eigenvector is given by

x(1) =

101

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




x =

α0α

= α

101

; α = real number


x(1) =

101

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




x =

α0α

= α

101

; α = real number

A particular

eigenvector is given by

x(1) =

101

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




x =

α0α

= α

101

; α = real number

A particular eigenvector

is given by

x(1) =

101

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




x =

α0α

= α

101

; α = real number


x(1) =

101

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




x =

α0α

= α

101

; α = real number


x(1) =

101

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




x =

α0α

= α

101

; α = real number


x(1) =

101

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




x =

α0α

= α

101

; α = real number


x(1) =

101

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=




−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore we have x1 = −x3 = −α, andx2 = 0 . Thus, we get




−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0





−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0





−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0





−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0





−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0





−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000

The above system


−x1 − 3x2 − x3 = 0 − 4x2 = 0





−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


to the equations

−x1 − 3x2 − x3 = 0 − 4x2 = 0





−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0





−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0





−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and

two unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore we have x1 = −x3 = −α, andx2 = 0 . Thus, we get




−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns.

Hence, one of them is a freevariable, let’s say x3 = α. Therefore we have x1 = −x3 = −α, andx2 = 0 . Thus, we get




−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0


one of them is a freevariable, let’s say x3 = α. Therefore we have x1 = −x3 = −α, andx2 = 0 . Thus, we get




−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them

is a freevariable, let’s say x3 = α. Therefore we have x1 = −x3 = −α, andx2 = 0 . Thus, we get




−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them is a freevariable,

let’s say x3 = α. Therefore we have x1 = −x3 = −α, andx2 = 0 . Thus, we get




−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them is a freevariable, let’s say x3 = α.

Therefore we have x1 = −x3 = −α, andx2 = 0 . Thus, we get




−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore we have

x1 = −x3 = −α, andx2 = 0 . Thus, we get




−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore we have x1 = −x3 = −α, and





−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore we have x1 = −x3 = −α, andx2 = 0 .

Thus, we get




−1 −3 −10 −4 0−1 1 1

=

−1 −3 −10 −4 00 4 0

=

−1 −3 −10 −4 00 0 0

x1x2x3

=

000


−x1 − 3x2 − x3 = 0 − 4x2 = 0





x =

−α0α

= α

−101

; α = real number


x(1) =

−101




x =

−α0α

= α

−101

; α = real number


x(1) =

−101




x =

−α0α

=

α

−101

; α = real number


x(1) =

−101




x =

−α0α

= α

−101

; α = real number


x(1) =

−101




x =

−α0α

= α

−101

; α = real number

A particular

eigenvector is given by

x(1) =

−101




x =

−α0α

= α

−101

; α = real number

A particular eigenvector

is given by

x(1) =

−101




x =

−α0α

= α

−101

; α = real number


x(1) =

−101




x =

−α0α

= α

−101

; α = real number


x(1) =

−101




Thus, the three linearly independent eigenvectors, are

x(1) =

110

x(2) =

101

x(3) =

− 101





x(1) =

110

x(2) =

101

x(3) =

− 101




Thus,

the three linearly independent eigenvectors, are

x(1) =

110

x(2) =

101

x(3) =

− 101




Thus, the three linearly independent

eigenvectors, are

x(1) =

110

x(2) =

101

x(3) =

− 101





x(1) =

110

x(2) =

101

x(3) =

− 101





x(1) =

110

x(2) =

101

x(3) =

− 101





x(1) =

110

x(2) =

101

x(3) =

− 101





x(1) =

110

x(2) =

101

x(3) =

− 101




Example 9.2


A =

1 0 02 1 −23 2 1

Solution

The eigenvalues λ andeigenvectors x satisfy the equation(A− λI) x = 0, or




Example 9.2


A =

1 0 02 1 −23 2 1

Solution





Example 9.2



A =

1 0 02 1 −23 2 1

Solution





Example 9.2


A =

1 0 02 1 −23 2 1

Solution





Example 9.2


A =

1 0 02 1 −23 2 1

Solution





Example 9.2


A =

1 0 02 1 −23 2 1

Solution





Example 9.2


A =

1 0 02 1 −23 2 1

Solution

The eigenvalues

λ andeigenvectors x satisfy the equation(A− λI) x = 0, or




Example 9.2


A =

1 0 02 1 −23 2 1

Solution






Example 9.2


A =

1 0 02 1 −23 2 1

Solution

The eigenvalues λ andeigenvectors x





Example 9.2


A =

1 0 02 1 −23 2 1

Solution

The eigenvalues λ andeigenvectors x satisfy the equation

(A− λI) x = 0, or




Example 9.2


A =

1 0 02 1 −23 2 1

Solution





(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0




(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0




(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0




(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0




(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000

The eigenvalues


|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0




(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000


of the equation

|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0




(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0




(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0




(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ =

(1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0




(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0




(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0




(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) =

(1− λ)(λ2 − 2λ+ 5) = 0




(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0




The roots are λ1 = 1, λ2 = 1 + 2 i , and λ3 = 1− 2 i .

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000




The roots

are λ1 = 1, λ2 = 1 + 2 i , and λ3 = 1− 2 i .

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000

We can


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000

We can reduce this


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

=

0 0 02 0 −23 2 0

x1x2x3

=

000


2 0 −20 0 03 2 0

=

1 0 −10 0 03 2 0

=

1 0 −10 2 40 0 0

x1x2x3

=

000




Solving this system yields the eigenvector

x(1) =

1− 3/2

1

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

=

000




Solving this system

yields the eigenvector

x(1) =

1− 3/2

1

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

=

000





x(1) =

1− 3/2

1

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

=

000





x(1) =

1− 3/2

1

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

=

000





x(1) =

1− 3/2

1

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

=

000





x(1) =

1− 3/2

1

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

=

000





x(1) =

1− 3/2

1

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

=

000





x(1) =

1− 3/2

1

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

=

000





x(1) =

1− 3/2

1

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

=

000





x(1) =

1− 3/2

1

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

=

000





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1 is a free veriable, let’s say x2 = α, x3 = −i α . Hence

x =

0α−i α

= α

01

− i

= α

010

− i

001




The above system


x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0


x =

0α−i α

= α

01

− i

= α

010

− i

001





to the equations

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0


x =

0α−i α

= α

01

− i

= α

010

− i

001





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0


x =

0α−i α

= α

01

− i

= α

010

− i

001





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0


x =

0α−i α

= α

01

− i

= α

010

− i

001





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

Thus, we have

one equation and two unknowns. Hence, one ofthem 1 is a free veriable, let’s say x2 = α, x3 = −i α . Hence

x =

0α−i α

= α

01

− i

= α

010

− i

001





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

Thus, we have one equation and two unknowns. Hence,

one ofthem 1 is a free veriable, let’s say x2 = α, x3 = −i α . Hence

x =

0α−i α

= α

01

− i

= α

010

− i

001





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1 is a free veriable,

let’s say x2 = α, x3 = −i α . Hence

x =

0α−i α

= α

01

− i

= α

010

− i

001





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1 is a free veriable, let’s say x2 = α,

x3 = −i α . Hence

x =

0α−i α

= α

01

− i

= α

010

− i

001





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1 is a free veriable, let’s say x2 = α, x3 = −i α .

Hence

x =

0α−i α

= α

01

− i

= α

010

− i

001





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0


x =

0α−i α

= α

01

− i

= α

010

− i

001





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0


x =

0α−i α

=

α

01

− i

= α

010

− i

001





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0


x =

0α−i α

= α

01

− i

=

α

010

− i

001





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0


x =

0α−i α

= α

01

− i

= α

010

−

i

001





x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0


x =

0α−i α

= α

01

− i

= α

010

− i

001




In this way, there are two real linearly independent eigenvectorsassociated to λ2 = 1 + 2 i , namely,

x(2) =

010

x(3) =

001

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

2 i 0 02 2 i −23 2 2 i

x1x2x3

=

000




In this way,

there are two real linearly independent eigenvectorsassociated to λ2 = 1 + 2 i , namely,

x(2) =

010

x(3) =

001

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

2 i 0 02 2 i −23 2 2 i

x1x2x3

=

000




In this way, there are two real

linearly independent eigenvectorsassociated to λ2 = 1 + 2 i , namely,

x(2) =

010

x(3) =

001

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

2 i 0 02 2 i −23 2 2 i

x1x2x3

=

000




In this way, there are two real linearly independent

eigenvectorsassociated to λ2 = 1 + 2 i , namely,

x(2) =

010

x(3) =

001

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

2 i 0 02 2 i −23 2 2 i

x1x2x3

=

000




In this way, there are two real linearly independent eigenvectorsassociated

to λ2 = 1 + 2 i , namely,

x(2) =

010

x(3) =

001

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

2 i 0 02 2 i −23 2 2 i

x1x2x3

=

000





x(2) =

010

x(3) =

001

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

2 i 0 02 2 i −23 2 2 i

x1x2x3

=

000





x(2) =

010

x(3) =

001

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

2 i 0 02 2 i −23 2 2 i

x1x2x3

=

000





x(2) =

010

x(3) =

001

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

2 i 0 02 2 i −23 2 2 i

x1x2x3

=

000





x(2) =

010

x(3) =

001

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

2 i 0 02 2 i −23 2 2 i

x1x2x3

=

000





x(2) =

010

x(3) =

001

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

2 i 0 02 2 i −23 2 2 i

x1x2x3

=

000





x(2) =

010

x(3) =

001

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

2 i 0 02 2 i −23 2 2 i

x1x2x3

=

000





x(2) =

010

x(3) =

001

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

=

2 i 0 02 2 i −23 2 2 i

x1x2x3

=

000





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1, is a free variable, let’s say x2 = α, x3 = i α . Thus wehave

x =

0αi α

= α

01i

= α

010

+ i

001




The above system


x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0


x =

0αi α

= α

01i

= α

010

+ i

001





to the equations

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0


x =

0αi α

= α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0


x =

0αi α

= α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0


x =

0αi α

= α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus,

we have one equation and two unknowns. Hence, one ofthem 1, is a free variable, let’s say x2 = α, x3 = i α . Thus wehave

x =

0αi α

= α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and

two unknowns. Hence, one ofthem 1, is a free variable, let’s say x2 = α, x3 = i α . Thus wehave

x =

0αi α

= α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and two unknowns.

Hence, one ofthem 1, is a free variable, let’s say x2 = α, x3 = i α . Thus wehave

x =

0αi α

= α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and two unknowns. Hence,

one ofthem 1, is a free variable, let’s say x2 = α, x3 = i α . Thus wehave

x =

0αi α

= α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1,

is a free variable, let’s say x2 = α, x3 = i α . Thus wehave

x =

0αi α

= α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1, is a free variable,

let’s say x2 = α, x3 = i α . Thus wehave

x =

0αi α

= α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1, is a free variable, let’s say x2 = α,

x3 = i α . Thus wehave

x =

0αi α

= α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1, is a free variable, let’s say x2 = α, x3 = i α .

Thus wehave

x =

0αi α

= α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0


x =

0αi α

= α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0


x =

0αi α

=

α

01i

= α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0


x =

0αi α

= α

01i

=

α

010

+ i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0


x =

0αi α

= α

01i

= α

010

+

i

001





x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0


x =

0αi α

= α

01i

= α

010

+ i

001




In this way, there is two real linearly independent eigenvectorsassociated to λ2 = 1 + 2 i , namely,

x(2) =

010

x(3) =

001

Hence, we have three linearly independent eigenvectors, namely

x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001

that is, the matrix A is Non-defective




In this way,

there is two real linearly independent eigenvectorsassociated to λ2 = 1 + 2 i , namely,

x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001





In this way, there is

two real linearly independent eigenvectorsassociated to λ2 = 1 + 2 i , namely,

x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001





In this way, there is two real

linearly independent eigenvectorsassociated to λ2 = 1 + 2 i , namely,

x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001





In this way, there is two real linearly independent

eigenvectorsassociated to λ2 = 1 + 2 i , namely,

x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001





In this way, there is two real linearly independent eigenvectorsassociated

to λ2 = 1 + 2 i , namely,

x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001






x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001






x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001






x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001






x(2) =

010

x(3) =

001

Hence,

we have three linearly independent eigenvectors, namely

x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001






x(2) =

010

x(3) =

001

Hence, we have three

linearly independent eigenvectors, namely

x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001






x(2) =

010

x(3) =

001

Hence, we have three linearly independent

eigenvectors, namely

x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001






x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001






x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001






x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001






x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001






x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001

that is,

the matrix A is Non-defective





x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001

that is, the matrix A

is Non-defective





x(2) =

010

x(3) =

001


x(1) =

1− 3/2

1

x(2) =

010

x(3) =

001





Example 9.3


A =

0 1 11 0 11 1 0

Solution





Example 9.3


A =

0 1 11 0 11 1 0

Solution





Example 9.3

Find


A =

0 1 11 0 11 1 0

Solution





Example 9.3



A =

0 1 11 0 11 1 0

Solution





Example 9.3

Find the eigenvalues and eigenvectors

of the matrix

A =

0 1 11 0 11 1 0

Solution





Example 9.3


A =

0 1 11 0 11 1 0

Solution





Example 9.3


A =

0 1 11 0 11 1 0

Solution





Example 9.3


A =

0 1 11 0 11 1 0

Solution





Example 9.3


A =

0 1 11 0 11 1 0

Solution






Example 9.3


A =

0 1 11 0 11 1 0

Solution






Example 9.3


A =

0 1 11 0 11 1 0

Solution

The eigenvalues λ and eigenvectors

x satisfy the equation(A− λI) x = 0, or




Example 9.3


A =

0 1 11 0 11 1 0

Solution






Example 9.3


A =

0 1 11 0 11 1 0

Solution


(A− λI) x = 0, or




Example 9.3


A =

0 1 11 0 11 1 0

Solution





(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0




(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0




(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

=

000

The eigenvalues


|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0




(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

=

000


of the equation

|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0




(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0




(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0




(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0




(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0




(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ =

− λ3 + 3λ2 + 2 = 0




(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0




The roots are λ1 = 2, λ2 = −1, and λ3 = −1 .

1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

=

−2 1 11 −2 11 1 −2

x1x2x3

=

000

We can reduce this to the equivalent system2 −1 −1

0 1 −10 0 0

x1x2x3

=

000




The roots

are λ1 = 2, λ2 = −1, and λ3 = −1 .

1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

=

−2 1 11 −2 11 1 −2

x1x2x3

=

000


0 1 −10 0 0

x1x2x3

=

000





1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

=

−2 1 11 −2 11 1 −2

x1x2x3

=

000


0 1 −10 0 0

x1x2x3

=

000





1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

=

−2 1 11 −2 11 1 −2

x1x2x3

=

000


0 1 −10 0 0

x1x2x3

=

000





1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

=

−2 1 11 −2 11 1 −2

x1x2x3

=

000


0 1 −10 0 0

x1x2x3

=

000





1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

=

−2 1 11 −2 11 1 −2

x1x2x3

=

000


0 1 −10 0 0

x1x2x3

=

000





1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

=

−2 1 11 −2 11 1 −2

x1x2x3

=

000

We can reduce this to the equivalent system2 −1 −10 1 −10 0 0

x1x2x3

=

000





1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

=

−2 1 11 −2 11 1 −2

x1x2x3

=

000

We can reduce this

to the equivalent system2 −1 −10 1 −10 0 0

x1x2x3

=

000





1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

=

−2 1 11 −2 11 1 −2

x1x2x3

=

000


2 −1 −10 1 −10 0 0

x1x2x3

=

000





1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

=

−2 1 11 −2 11 1 −2

x1x2x3

=

000


0 1 −10 0 0

x1x2x3

=

000





2x1 − x2 − x3 = 0 x2 − x3 = 0

Two equations and three unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore x2 = x3 = α andx1 = (x2 + x3)/2 = α. Thus we have

x =

ααα

= α

111

; α = real number




The above system


2x1 − x2 − x3 = 0 x2 − x3 = 0


x =

ααα

= α

111

; α = real number




The above system is reduced

immediately to the equations

2x1 − x2 − x3 = 0 x2 − x3 = 0


x =

ααα

= α

111

; α = real number





the equations

2x1 − x2 − x3 = 0 x2 − x3 = 0


x =

ααα

= α

111

; α = real number





2x1 − x2 − x3 = 0 x2 − x3 = 0


x =

ααα

= α

111

; α = real number





2x1 − x2 − x3 = 0 x2 − x3 = 0


x =

ααα

= α

111

; α = real number





2x1 − x2 − x3 = 0 x2 − x3 = 0

Two equations and three unknowns. Hence,

one of them is a freevariable, let’s say x3 = α. Therefore x2 = x3 = α andx1 = (x2 + x3)/2 = α. Thus we have

x =

ααα

= α

111

; α = real number





2x1 − x2 − x3 = 0 x2 − x3 = 0

Two equations and three unknowns. Hence, one of them is a freevariable,

let’s say x3 = α. Therefore x2 = x3 = α andx1 = (x2 + x3)/2 = α. Thus we have

x =

ααα

= α

111

; α = real number





2x1 − x2 − x3 = 0 x2 − x3 = 0

Two equations and three unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore

x2 = x3 = α andx1 = (x2 + x3)/2 = α. Thus we have

x =

ααα

= α

111

; α = real number





2x1 − x2 − x3 = 0 x2 − x3 = 0

Two equations and three unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore x2 = x3 = α andx1 = (x2 + x3)/2 = α.

Thus we have

x =

ααα

= α

111

; α = real number





2x1 − x2 − x3 = 0 x2 − x3 = 0


x =

ααα

= α

111

; α = real number





2x1 − x2 − x3 = 0 x2 − x3 = 0


x =

ααα

= α

111

; α = real number




In particular, we have the eigenvector

x(1) =

111

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

=

1 1 11 1 11 1 1

x1x2x3

=

000




In particular,

we have the eigenvector

x(1) =

111

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

=

1 1 11 1 11 1 1

x1x2x3

=

000




In particular, we have

the eigenvector

x(1) =

111

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

=

1 1 11 1 11 1 1

x1x2x3

=

000





x(1) =

111

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

=

1 1 11 1 11 1 1

x1x2x3

=

000





x(1) =

111

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

=

1 1 11 1 11 1 1

x1x2x3

=

000





x(1) =

111

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

=

1 1 11 1 11 1 1

x1x2x3

=

000





x(1) =

111

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

=

1 1 11 1 11 1 1

x1x2x3

=

000





x(1) =

111

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

=

1 1 11 1 11 1 1

x1x2x3

=

000




The above system is reduced immediately to the single equation

x1 + x2 + x3 = 0

One equation and three unknowns. Hence, two of them are freeveriables, let’s say x1 = α, x2 = β, and x3 = −α− β . Thus wehave

x =

αβ

−α− β

= α

10−1

+ β

01−1

; α, β = real numbers




The above system

is reduced immediately to the single equation

x1 + x2 + x3 = 0


x =

αβ

−α− β

= α

10−1

+ β

01−1






immediately to the single equation

x1 + x2 + x3 = 0


x =

αβ

−α− β

= α

10−1

+ β

01−1






the single equation

x1 + x2 + x3 = 0


x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0


x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0


x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0

One equation and

three unknowns. Hence, two of them are freeveriables, let’s say x1 = α, x2 = β, and x3 = −α− β . Thus wehave

x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0

One equation and three unknowns.

Hence, two of them are freeveriables, let’s say x1 = α, x2 = β, and x3 = −α− β . Thus wehave

x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0

One equation and three unknowns. Hence,

two of them are freeveriables, let’s say x1 = α, x2 = β, and x3 = −α− β . Thus wehave

x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0

One equation and three unknowns. Hence, two of them

are freeveriables, let’s say x1 = α, x2 = β, and x3 = −α− β . Thus wehave

x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0

One equation and three unknowns. Hence, two of them are freeveriables,

let’s say x1 = α, x2 = β, and x3 = −α− β . Thus wehave

x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0

One equation and three unknowns. Hence, two of them are freeveriables, let’s say

x1 = α, x2 = β, and x3 = −α− β . Thus wehave

x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0

One equation and three unknowns. Hence, two of them are freeveriables, let’s say x1 = α,

x2 = β, and x3 = −α− β . Thus wehave

x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0

One equation and three unknowns. Hence, two of them are freeveriables, let’s say x1 = α, x2 = β, and

x3 = −α− β . Thus wehave

x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0

One equation and three unknowns. Hence, two of them are freeveriables, let’s say x1 = α, x2 = β, and x3 = −α− β .

Thus wehave

x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0


x =

αβ

−α− β

= α

10−1

+ β

01−1






x1 + x2 + x3 = 0


x =

αβ

−α− β

= α

10−1

+ β

01−1





In this way two linearly independent eigenvectors associated toλ2 = −1 are ( α = β = 1 )

x(2) =

10

− 1

x(3) =

01

− 1


x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1




In this way

two linearly independent eigenvectors associated toλ2 = −1 are ( α = β = 1 )

x(2) =

10

− 1

x(3) =

01

− 1


x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1




In this way two linearly

independent eigenvectors associated toλ2 = −1 are ( α = β = 1 )

x(2) =

10

− 1

x(3) =

01

− 1


x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1




In this way two linearly independent eigenvectors

associated toλ2 = −1 are ( α = β = 1 )

x(2) =

10

− 1

x(3) =

01

− 1


x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1




In this way two linearly independent eigenvectors associated toλ2 = −1 are

( α = β = 1 )

x(2) =

10

− 1

x(3) =

01

− 1


x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1





x(2) =

10

− 1

x(3) =

01

− 1


x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1





x(2) =

10

− 1

x(3) =

01

− 1


x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1





x(2) =

10

− 1

x(3) =

01

− 1


x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1





x(2) =

10

− 1

x(3) =

01

− 1

Thus,

the three linearly independent eigenvectors, are

x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1





x(2) =

10

− 1

x(3) =

01

− 1

Thus, the three

linearly independent eigenvectors, are

x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1





x(2) =

10

− 1

x(3) =

01

− 1

Thus, the three linearly independent

eigenvectors, are

x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1





x(2) =

10

− 1

x(3) =

01

− 1


x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1





x(2) =

10

− 1

x(3) =

01

− 1


x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1





x(2) =

10

− 1

x(3) =

01

− 1


x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1





x(2) =

10

− 1

x(3) =

01

− 1


x(1) =

111

x(2) =

10

− 1

x(3) =

01

− 1




Example 9.4


A =

4 6 61 3 21 −5 −2

Solution





Example 9.4


A =

4 6 61 3 21 −5 −2

Solution





Example 9.4

Find


A =

4 6 61 3 21 −5 −2

Solution





Example 9.4



A =

4 6 61 3 21 −5 −2

Solution





Example 9.4


A =

4 6 61 3 21 −5 −2

Solution





Example 9.4


A =

4 6 61 3 21 −5 −2

Solution





Example 9.4


A =

4 6 61 3 21 −5 −2

Solution





Example 9.4


A =

4 6 61 3 21 −5 −2

Solution

The eigenvalues

λ and eigenvectors x satisfy the equation(A− λI) x = 0, or




Example 9.4


A =

4 6 61 3 21 −5 −2

Solution






Example 9.4


A =

4 6 61 3 21 −5 −2

Solution






Example 9.4


A =

4 6 61 3 21 −5 −2

Solution






Example 9.4


A =

4 6 61 3 21 −5 −2

Solution


(A− λI) x = 0, or




Example 9.4


A =

4 6 61 3 21 −5 −2

Solution





(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000

The eigenvalues


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


of the equation

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ =

− λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0




(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

000


|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

−

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0Dr. Marco A Roque Sol Linear Algebra. Session 9



The roots are λ1 = 1, λ2 = 2, and λ3 = 2 .

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000




The roots

are λ1 = 1, λ2 = 2, and λ3 = 2 .

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000

We can


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000

We can reduce this


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

=

3 6 61 2 2−1 −5 −3

x1x2x3

=

000


3 6 61 2 20 3 1

=

1 2 21 2 20 3 1

=

1 2 21 −3 −10 0 0

x1x2x3

=

000





x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

=

2 6 61 1 2−1 −5 −4

x1x2x3

=

2 6 61 1 2−1 −5 −4

=

1 1 20 4 20 −4 −2

=

1 1 20 4 20 0 0

x1x2x3

=

000




Solving this system

yields the eigenvector

x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

=

2 6 61 1 2−1 −5 −4

x1x2x3

=

2 6 61 1 2−1 −5 −4

=

1 1 20 4 20 −4 −2

=

1 1 20 4 20 0 0

x1x2x3

=

000





x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

=

2 6 61 1 2−1 −5 −4

x1x2x3

=

2 6 61 1 2−1 −5 −4

=

1 1 20 4 20 −4 −2

=

1 1 20 4 20 0 0

x1x2x3

=

000





x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

=

2 6 61 1 2−1 −5 −4

x1x2x3

=

2 6 61 1 2−1 −5 −4

=

1 1 20 4 20 −4 −2

=

1 1 20 4 20 0 0

x1x2x3

=

000





x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

=

2 6 61 1 2−1 −5 −4

x1x2x3

=

2 6 61 1 2−1 −5 −4

=

1 1 20 4 20 −4 −2

=

1 1 20 4 20 0 0

x1x2x3

=

000





x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

=

2 6 61 1 2−1 −5 −4

x1x2x3

=

2 6 61 1 2−1 −5 −4

=

1 1 20 4 20 −4 −2

=

1 1 20 4 20 0 0

x1x2x3

=

000





x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

=

2 6 61 1 2−1 −5 −4

x1x2x3

=

2 6 61 1 2−1 −5 −4

=

1 1 20 4 20 −4 −2

=

1 1 20 4 20 0 0

x1x2x3

=

000





x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

=

2 6 61 1 2−1 −5 −4

x1x2x3

=

2 6 61 1 2−1 −5 −4

=

1 1 20 4 20 −4 −2

=

1 1 20 4 20 0 0

x1x2x3

=

000





x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

=

2 6 61 1 2−1 −5 −4

x1x2x3

=

2 6 61 1 2−1 −5 −4

=

1 1 20 4 20 −4 −2

=

1 1 20 4 20 0 0

x1x2x3

=

000





x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

=

2 6 61 1 2−1 −5 −4

x1x2x3

=

2 6 61 1 2−1 −5 −4

=

1 1 20 4 20 −4 −2

=

1 1 20 4 20 0 0

x1x2x3

=

000





x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

=

2 6 61 1 2−1 −5 −4

x1x2x3

=

2 6 61 1 2−1 −5 −4

=

1 1 20 4 20 −4 −2

=

1 1 20 4 20 0 0

x1x2x3

=

000





x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

=

2 6 61 1 2−1 −5 −4

x1x2x3

=

2 6 61 1 2−1 −5 −4

=

1 1 20 4 20 −4 −2

=

1 1 20 4 20 0 0

x1x2x3

=

000





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

Two equations and three unknowns. Hence, one of them, is a freevariable, let’s say x3 = α, x2 = 1

2α, and x3 = −2x3 − x2 = −3α .Thus we have

x =

− 3α12α

− 3α

= α

− 312

− 3




The above system


x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0



x =

− 3α12α

− 3α

= α

− 312

− 3





immediately to the equations

x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0



x =

− 3α12α

− 3α

= α

− 312

− 3





the equations

x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0



x =

− 3α12α

− 3α

= α

− 312

− 3





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0



x =

− 3α12α

− 3α

= α

− 312

− 3





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0



x =

− 3α12α

− 3α

= α

− 312

− 3





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

Two equations and

three unknowns. Hence, one of them, is a freevariable, let’s say x3 = α, x2 = 1


x =

− 3α12α

− 3α

= α

− 312

− 3





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

Two equations and three unknowns.

Hence, one of them, is a freevariable, let’s say x3 = α, x2 = 1


x =

− 3α12α

− 3α

= α

− 312

− 3





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

Two equations and three unknowns. Hence,

one of them, is a freevariable, let’s say x3 = α, x2 = 1


x =

− 3α12α

− 3α

= α

− 312

− 3





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

Two equations and three unknowns. Hence, one of them,

is a freevariable, let’s say x3 = α, x2 = 1


x =

− 3α12α

− 3α

= α

− 312

− 3





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

Two equations and three unknowns. Hence, one of them, is a freevariable,

let’s say x3 = α, x2 = 12α, and x3 = −2x3 − x2 = −3α .

Thus we have

x =

− 3α12α

− 3α

= α

− 312

− 3





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

Two equations and three unknowns. Hence, one of them, is a freevariable, let’s say x3 = α,

x2 = 12α, and x3 = −2x3 − x2 = −3α .

Thus we have

x =

− 3α12α

− 3α

= α

− 312

− 3





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0


2α, and

x3 = −2x3 − x2 = −3α .Thus we have

x =

− 3α12α

− 3α

= α

− 312

− 3





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0


2α, and x3 = −2x3 − x2 = −3α .

Thus we have

x =

− 3α12α

− 3α

= α

− 312

− 3





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0



x =

− 3α12α

− 3α

= α

− 312

− 3





x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0



x =

− 3α12α

− 3α

= α

− 312

− 3




In this way, there is one linearly independent eigenvectorassociatedtoλ2,3 = 2 , namely,

x(2) =

31

− 2

Therefore, there are just two linearlyindependent eigenvectors

x(1) =

111

x(2) =

41

− 3

that is, the matrix A is defective




In this way,

there is one linearly independent eigenvectorassociatedtoλ2,3 = 2 , namely,

x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3





In this way, there is one

linearly independent eigenvectorassociatedtoλ2,3 = 2 , namely,

x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3





In this way, there is one linearly independent

eigenvectorassociatedtoλ2,3 = 2 , namely,

x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3





In this way, there is one linearly independent eigenvector

associatedtoλ2,3 = 2 , namely,

x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3





In this way, there is one linearly independent eigenvectorassociatedto

λ2,3 = 2 , namely,

x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3






x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3






x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3






x(2) =

31

− 2

Therefore,

there are just two linearlyindependent eigenvectors

x(1) =

111

x(2) =

41

− 3






x(2) =

31

− 2

Therefore, there are

just two linearlyindependent eigenvectors

x(1) =

111

x(2) =

41

− 3






x(2) =

31

− 2

Therefore, there are just two linearly

independent eigenvectors

x(1) =

111

x(2) =

41

− 3






x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3






x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3






x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3






x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3






x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3

that is,

the matrix A is defective





x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3

that is, the matrix A

is defective





x(2) =

31

− 2


x(1) =

111

x(2) =

41

− 3





OBS

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v

Finally, let’s introduce another concept

The Dot Product in Rn

For y = (y1, y2, ..., yn), x = (x1, x2, ..., xn) ∈ R, define the dotproduct or inner product or scalar product.as




OBS








OBS

Let A

be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v







OBS

Let A be a real-valued

n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v







OBS

Let A be a real-valued n × n matrix.

If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v







OBS

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I )

arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v







OBS

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors

of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v







OBS

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A

with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v







OBS

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then,

x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v







OBS

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I )

are two real eigenvectors for Awith eigenvalues λ = u ± i v







OBS

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors

for Awith eigenvalues λ = u ± i v







OBS

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for A

with eigenvalues λ = u ± i v







OBS








OBS


Finally,

let’s introduce another concept






OBS


Finally, let’s introduce

another concept






OBS








OBS








OBS








OBS




For

y = (y1, y2, ..., yn), x = (x1, x2, ..., xn) ∈ R, define the dotproduct or inner product or scalar product.as




OBS




For y = (y1, y2, ..., yn), x = (x1, x2, ..., xn) ∈ R,

define the dotproduct or inner product or scalar product.as




OBS




For y = (y1, y2, ..., yn), x = (x1, x2, ..., xn) ∈ R, define the dotproduct

or inner product or scalar product.as




OBS




For y = (y1, y2, ..., yn), x = (x1, x2, ..., xn) ∈ R, define the dotproduct or inner product or

scalar product.as




OBS








x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS

a) x and y are said to be orthogonal if < x, y >= 0 .

b) Orthogonal nonzero vectors are linearly independent.




x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS






x · y =

< x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS






x · y = < x, y >=

(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS






x · y = < x, y >=(x1 x2 . . . xn

)

y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS






x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

=

x1y1 + x2y2 + ...+ xnyn

OBS






x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS






x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS






x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS

a) x and y

are said to be orthogonal if < x, y >= 0 .





x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS

a) x and y are said to be

orthogonal if < x, y >= 0 .





x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS

a) x and y are said to be orthogonal

if < x, y >= 0 .





x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS






x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS


b) Orthogonal

nonzero vectors are linearly independent.




x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS


b) Orthogonal nonzero vectors





x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

OBS






Theorem

Let A be an n × n matrix. If A is symetric,( A = AT ) then

1) All eigenvalues are real.

2) A is always Nondefective.

3) The eigenvectors corresponding to different eigenvalues areorthogonal, thus if λ1, λ2, ..., λn are all simple, v1, v2, ..., vn forman orthogonal set.




Theorem








Theorem

Let A

be an n × n matrix. If A is symetric,( A = AT ) then







Theorem

Let A be an n × n matrix.

If A is symetric,( A = AT ) then







Theorem

Let A be an n × n matrix. If A

is symetric,( A = AT ) then







Theorem

Let A be an n × n matrix. If A is symetric,

( A = AT ) then







Theorem








Theorem








Theorem








Theorem








Theorem




3) The eigenvectors

corresponding to different eigenvalues areorthogonal, thus if λ1, λ2, ..., λn are all simple, v1, v2, ..., vn forman orthogonal set.




Theorem




3) The eigenvectors corresponding to

different eigenvalues areorthogonal, thus if λ1, λ2, ..., λn are all simple, v1, v2, ..., vn forman orthogonal set.




Theorem




3) The eigenvectors corresponding to different eigenvalues

areorthogonal, thus if λ1, λ2, ..., λn are all simple, v1, v2, ..., vn forman orthogonal set.




Theorem




3) The eigenvectors corresponding to different eigenvalues areorthogonal, thus

if λ1, λ2, ..., λn are all simple, v1, v2, ..., vn forman orthogonal set.




Theorem




3) The eigenvectors corresponding to different eigenvalues areorthogonal, thus if λ1, λ2, ..., λn

are all simple, v1, v2, ..., vn forman orthogonal set.




Theorem




3) The eigenvectors corresponding to different eigenvalues areorthogonal, thus if λ1, λ2, ..., λn are all simple,

v1, v2, ..., vn forman orthogonal set.




Theorem







Systems of Linear Differential Equations.

The general theory of a system of n first order linear equations

x ′1 = p11x1 + p12x2 + . . .+ p1nxn + g1(t)x ′2 = p21x1 + p22x2 + . . .+ p2nxn + g2(t)...

...x ′n = pn1x1 + pn2x2 + . . .+ pnnxn + gn(t)

or

X′ = P(t)X + g(t)

closely parallels that of a single linear equation of nth order.




The general theory

of a system of n first order linear equations



or

X′ = P(t)X + g(t)





The general theory of a system

of n first order linear equations



or

X′ = P(t)X + g(t)





The general theory of a system of n first order

linear equations



or

X′ = P(t)X + g(t)








or

X′ = P(t)X + g(t)








or

X′ = P(t)X + g(t)








or

X′ = P(t)X + g(t)








or

X′ = P(t)X + g(t)








or

X′ = P(t)X + g(t)








or

X′ = P(t)X + g(t)

closely parallels

that of a single linear equation of nth order.







or

X′ = P(t)X + g(t)

closely parallels that of a single

linear equation of nth order.







or

X′ = P(t)X + g(t)

closely parallels that of a single linear equation

of nth order.







or

X′ = P(t)X + g(t)





we assume that P and g are continuous on some intervalα < t < β; that is, each of the scalar functionsp11, ..., pnn, g1, ..., gn is continuous there.

Theorem

If the vector functions x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2) is also a solution for any constants c1 and c2.

This is the principle of superposition




we assume that

P and g are continuous on some intervalα < t < β; that is, each of the scalar functionsp11, ..., pnn, g1, ..., gn is continuous there.

Theorem






we assume that P and g

are continuous on some intervalα < t < β; that is, each of the scalar functionsp11, ..., pnn, g1, ..., gn is continuous there.

Theorem






we assume that P and g are continuous

on some intervalα < t < β; that is, each of the scalar functionsp11, ..., pnn, g1, ..., gn is continuous there.

Theorem






we assume that P and g are continuous on some interval

α < t < β; that is, each of the scalar functionsp11, ..., pnn, g1, ..., gn is continuous there.

Theorem






we assume that P and g are continuous on some intervalα < t < β;

that is, each of the scalar functionsp11, ..., pnn, g1, ..., gn is continuous there.

Theorem






we assume that P and g are continuous on some intervalα < t < β; that is,

each of the scalar functionsp11, ..., pnn, g1, ..., gn is continuous there.

Theorem






we assume that P and g are continuous on some intervalα < t < β; that is, each of

the scalar functionsp11, ..., pnn, g1, ..., gn is continuous there.

Theorem






we assume that P and g are continuous on some intervalα < t < β; that is, each of the scalar functions

p11, ..., pnn, g1, ..., gn is continuous there.

Theorem






we assume that P and g are continuous on some intervalα < t < β; that is, each of the scalar functionsp11, ..., pnn, g1, ..., gn

is continuous there.

Theorem







Theorem







Theorem







Theorem

If

the vector functions x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2) is also a solution for any constants c1 and c2.






Theorem

If the vector functions

x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2) is also a solution for any constants c1 and c2.






Theorem

If the vector functions x(1) and x(2)

are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2) is also a solution for any constants c1 and c2.






Theorem

If the vector functions x(1) and x(2) are solutions

of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2) is also a solution for any constants c1 and c2.






Theorem

If the vector functions x(1) and x(2) are solutions of thehomogeneus system

( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2) is also a solution for any constants c1 and c2.






Theorem

If the vector functions x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 )

then the linear combinationc1x(1) + c2x(2) is also a solution for any constants c1 and c2.






Theorem

If the vector functions x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combination

c1x(1) + c2x(2) is also a solution for any constants c1 and c2.






Theorem

If the vector functions x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2)

is also a solution for any constants c1 and c2.






Theorem

If the vector functions x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2) is also

a solution for any constants c1 and c2.






Theorem

If the vector functions x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2) is also a solution

for any constants c1 and c2.






Theorem







Theorem


This is

the principle of superposition





Theorem






By repeated application of Theorem, we can conclude that ifx(1), ..., x(k) are solutions of the homogeneous system, then

c1x(1) + ...+ ckx(k)

is also a solution for any constants c1, ..., ck .

Theorem

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)




By repeated application

of Theorem, we can conclude that ifx(1), ..., x(k) are solutions of the homogeneous system, then

c1x(1) + ...+ ckx(k)


Theorem


φ(t) = c1x(1) + ...+ ckx(k)




By repeated application of Theorem, we can conclude

that ifx(1), ..., x(k) are solutions of the homogeneous system, then

c1x(1) + ...+ ckx(k)


Theorem


φ(t) = c1x(1) + ...+ ckx(k)




By repeated application of Theorem, we can conclude that ifx(1), ..., x(k)

are solutions of the homogeneous system, then

c1x(1) + ...+ ckx(k)


Theorem


φ(t) = c1x(1) + ...+ ckx(k)




By repeated application of Theorem, we can conclude that ifx(1), ..., x(k) are solutions

of the homogeneous system, then

c1x(1) + ...+ ckx(k)


Theorem


φ(t) = c1x(1) + ...+ ckx(k)




By repeated application of Theorem, we can conclude that ifx(1), ..., x(k) are solutions of the homogeneous system,

then

c1x(1) + ...+ ckx(k)


Theorem


φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem


φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem


φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)

is also

a solution for any constants c1, ..., ck .

Theorem


φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)

is also a solution

for any constants c1, ..., ck .

Theorem


φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem


φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem


φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If

the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions

x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n)

are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n) are

linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n) are linearly independentsolutions

of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system

for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point

in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the interval

α < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β,

then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution

x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t)

of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem

can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed

as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of

x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n)

inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)





c1x(1) + ...+ ckx(k)


Theorem


φ(t) = c1x(1) + ...+ ckx(k)




If the constants c1, ..., cn are thought of as arbitrary, then theabove equation includes all solutions of the system, and it iscustomary to call it the general solution.

Any set of solutions x(1), ..., x(n) of the homogeneus system that islinearly independent at each point in the interval α < t < β is saidto be a fundamental set of solutions for that interval.

Theorem

If x(1), ..., x(n) are solutions of the homogeneus system on theinterval α < t < β, then in this interval W [x(1), ..., x(n)] given by




If

the constants c1, ..., cn are thought of as arbitrary, then theabove equation includes all solutions of the system, and it iscustomary to call it the general solution.


Theorem





If the constants c1, ..., cn

are thought of as arbitrary, then theabove equation includes all solutions of the system, and it iscustomary to call it the general solution.


Theorem





If the constants c1, ..., cn are thought of

as arbitrary, then theabove equation includes all solutions of the system, and it iscustomary to call it the general solution.


Theorem





If the constants c1, ..., cn are thought of as arbitrary,

then theabove equation includes all solutions of the system, and it iscustomary to call it the general solution.


Theorem





If the constants c1, ..., cn are thought of as arbitrary, then theabove equation

includes all solutions of the system, and it iscustomary to call it the general solution.


Theorem





If the constants c1, ..., cn are thought of as arbitrary, then theabove equation includes all solutions

of the system, and it iscustomary to call it the general solution.


Theorem





If the constants c1, ..., cn are thought of as arbitrary, then theabove equation includes all solutions of the system, and

it iscustomary to call it the general solution.


Theorem





If the constants c1, ..., cn are thought of as arbitrary, then theabove equation includes all solutions of the system, and it iscustomary

to call it the general solution.


Theorem





If the constants c1, ..., cn are thought of as arbitrary, then theabove equation includes all solutions of the system, and it iscustomary to call it

the general solution.


Theorem







Theorem






Any set

of solutions x(1), ..., x(n) of the homogeneus system that islinearly independent at each point in the interval α < t < β is saidto be a fundamental set of solutions for that interval.

Theorem






Any set of solutions

x(1), ..., x(n) of the homogeneus system that islinearly independent at each point in the interval α < t < β is saidto be a fundamental set of solutions for that interval.

Theorem






Any set of solutions x(1), ..., x(n) of the homogeneus system

that islinearly independent at each point in the interval α < t < β is saidto be a fundamental set of solutions for that interval.

Theorem






Any set of solutions x(1), ..., x(n) of the homogeneus system that is

linearly independent at each point in the interval α < t < β is saidto be a fundamental set of solutions for that interval.

Theorem






Any set of solutions x(1), ..., x(n) of the homogeneus system that islinearly independent

at each point in the interval α < t < β is saidto be a fundamental set of solutions for that interval.

Theorem






Any set of solutions x(1), ..., x(n) of the homogeneus system that islinearly independent at each point

in the interval α < t < β is saidto be a fundamental set of solutions for that interval.

Theorem






Any set of solutions x(1), ..., x(n) of the homogeneus system that islinearly independent at each point in the interval

α < t < β is saidto be a fundamental set of solutions for that interval.

Theorem






Any set of solutions x(1), ..., x(n) of the homogeneus system that islinearly independent at each point in the interval α < t < β

is saidto be a fundamental set of solutions for that interval.

Theorem






Any set of solutions x(1), ..., x(n) of the homogeneus system that islinearly independent at each point in the interval α < t < β is saidto be

a fundamental set of solutions for that interval.

Theorem






Any set of solutions x(1), ..., x(n) of the homogeneus system that islinearly independent at each point in the interval α < t < β is saidto be a fundamental set

of solutions for that interval.

Theorem






Any set of solutions x(1), ..., x(n) of the homogeneus system that islinearly independent at each point in the interval α < t < β is saidto be a fundamental set of solutions

for that interval.

Theorem







Theorem







Theorem







Theorem

If

x(1), ..., x(n) are solutions of the homogeneus system on theinterval α < t < β, then in this interval W [x(1), ..., x(n)] given by






Theorem

If x(1), ..., x(n)

are solutions of the homogeneus system on theinterval α < t < β, then in this interval W [x(1), ..., x(n)] given by






Theorem

If x(1), ..., x(n) are solutions

of the homogeneus system on theinterval α < t < β, then in this interval W [x(1), ..., x(n)] given by






Theorem

If x(1), ..., x(n) are solutions of the homogeneus system

on theinterval α < t < β, then in this interval W [x(1), ..., x(n)] given by






Theorem

If x(1), ..., x(n) are solutions of the homogeneus system on theinterval

α < t < β, then in this interval W [x(1), ..., x(n)] given by






Theorem

If x(1), ..., x(n) are solutions of the homogeneus system on theinterval α < t < β,

then in this interval W [x(1), ..., x(n)] given by






Theorem

If x(1), ..., x(n) are solutions of the homogeneus system on theinterval α < t < β, then in this interval

W [x(1), ..., x(n)] given by






Theorem

If x(1), ..., x(n) are solutions of the homogeneus system on theinterval α < t < β, then in this interval W [x(1), ..., x(n)]

given by






Theorem





W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

(n)1

x(1)2 x

(2)2 . . . x

(n)2

...... . . .

...

x(1)n x

(2)n . . . x

(n)n

∣∣∣∣∣∣∣∣∣∣either is identically zero or else never vanishes. To prove thistheorem is necessary to establish that

dW

dt= [p11 + p22 + ...+ pnn]W

Hence

W (t) = ce∫[p11+p22+...+pnn]dt




W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

(n)1

x(1)2 x

(2)2 . . . x

(n)2

...... . . .

...

x(1)n x

(2)n . . . x

(n)n

∣∣∣∣∣∣∣∣∣∣

either is identically zero or else never vanishes. To prove thistheorem is necessary to establish that

dW

dt= [p11 + p22 + ...+ pnn]W

Hence

W (t) = ce∫[p11+p22+...+pnn]dt




W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

(n)1

x(1)2 x

(2)2 . . . x

(n)2

...... . . .

...

x(1)n x

(2)n . . . x

(n)n

∣∣∣∣∣∣∣∣∣∣either

is identically zero or else never vanishes. To prove thistheorem is necessary to establish that

dW

dt= [p11 + p22 + ...+ pnn]W

Hence

W (t) = ce∫[p11+p22+...+pnn]dt




W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

(n)1

x(1)2 x

(2)2 . . . x

(n)2

...... . . .

...

x(1)n x

(2)n . . . x

(n)n

∣∣∣∣∣∣∣∣∣∣either is identically zero or

else never vanishes. To prove thistheorem is necessary to establish that

dW

dt= [p11 + p22 + ...+ pnn]W

Hence

W (t) = ce∫[p11+p22+...+pnn]dt




W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

(n)1

x(1)2 x

(2)2 . . . x

(n)2

...... . . .

...

x(1)n x

(2)n . . . x

(n)n

∣∣∣∣∣∣∣∣∣∣either is identically zero or else never vanishes.

To prove thistheorem is necessary to establish that

dW

dt= [p11 + p22 + ...+ pnn]W

Hence

W (t) = ce∫[p11+p22+...+pnn]dt




W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

(n)1

x(1)2 x

(2)2 . . . x

(n)2

...... . . .

...

x(1)n x

(2)n . . . x

(n)n

∣∣∣∣∣∣∣∣∣∣either is identically zero or else never vanishes. To prove

thistheorem is necessary to establish that

dW

dt= [p11 + p22 + ...+ pnn]W

Hence

W (t) = ce∫[p11+p22+...+pnn]dt




W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

(n)1

x(1)2 x

(2)2 . . . x

(n)2

...... . . .

...

x(1)n x

(2)n . . . x

(n)n

∣∣∣∣∣∣∣∣∣∣either is identically zero or else never vanishes. To prove thistheorem

is necessary to establish that

dW

dt= [p11 + p22 + ...+ pnn]W

Hence

W (t) = ce∫[p11+p22+...+pnn]dt




W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

(n)1

x(1)2 x

(2)2 . . . x

(n)2

...... . . .

...

x(1)n x

(2)n . . . x

(n)n


dW

dt= [p11 + p22 + ...+ pnn]W

Hence

W (t) = ce∫[p11+p22+...+pnn]dt




W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

(n)1

x(1)2 x

(2)2 . . . x

(n)2

...... . . .

...

x(1)n x

(2)n . . . x

(n)n


dW

dt= [p11 + p22 + ...+ pnn]W

Hence

W (t) = ce∫[p11+p22+...+pnn]dt




W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

(n)1

x(1)2 x

(2)2 . . . x

(n)2

...... . . .

...

x(1)n x

(2)n . . . x

(n)n


dW

dt= [p11 + p22 + ...+ pnn]W

Hence

W (t) = ce∫[p11+p22+...+pnn]dt




W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

(n)1

x(1)2 x

(2)2 . . . x

(n)2

...... . . .

...

x(1)n x

(2)n . . . x

(n)n


dW

dt= [p11 + p22 + ...+ pnn]W

Hence

W (t) = ce∫[p11+p22+...+pnn]dt




Theorem

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point in α < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem


e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem

Let

x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point in α < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem

Let x(1), ..., x(n)

be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point in α < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem

Let x(1), ..., x(n) be the solutions

of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point in α < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem

Let x(1), ..., x(n) be the solutions of the homogeneus system

thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point in α < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy

the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point in α < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions

x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point in α < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1),

x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point in α < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),...,

x(n)(t0) = e(n), respectively, where t0 is any point in α < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively,

where t0 is any point in α < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0

is any point in α < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point in

α < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem


e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem


e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem


e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Theorem


e(1) =

10...0

e(2) =

01...0

· · ·

e(n) =

00...1




Theorem


e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1




Then, x(1), ..., x(n) form a fundamental set of solutions of thehomogeneous system.

Finally in the case that the solution is complex-valued, we have thefollowing result.

Theorem

Consider the homogeneous system

X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and its imaginary part v(t) are also solutions of this equation.




Then,

x(1), ..., x(n) form a fundamental set of solutions of thehomogeneous system.


Theorem


X′ = P(t)X





Then, x(1), ..., x(n)

form a fundamental set of solutions of thehomogeneous system.


Theorem


X′ = P(t)X





Then, x(1), ..., x(n) form a fundamental set

of solutions of thehomogeneous system.


Theorem


X′ = P(t)X





Then, x(1), ..., x(n) form a fundamental set of solutions

of thehomogeneous system.


Theorem


X′ = P(t)X







Theorem


X′ = P(t)X






Finally

in the case that the solution is complex-valued, we have thefollowing result.

Theorem


X′ = P(t)X






Finally in the case

that the solution is complex-valued, we have thefollowing result.

Theorem


X′ = P(t)X






Finally in the case that the solution

is complex-valued, we have thefollowing result.

Theorem


X′ = P(t)X






Finally in the case that the solution is complex-valued,

we have thefollowing result.

Theorem


X′ = P(t)X






Finally in the case that the solution is complex-valued, we have

thefollowing result.

Theorem


X′ = P(t)X







Theorem


X′ = P(t)X







Theorem


X′ = P(t)X







Theorem

Consider

the homogeneous system

X′ = P(t)X







Theorem


X′ = P(t)X







Theorem


X′ = P(t)X







Theorem


X′ = P(t)X

where each element

of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and its imaginary part v(t) are also solutions of this equation.






Theorem


X′ = P(t)X

where each element of P

is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and its imaginary part v(t) are also solutions of this equation.






Theorem


X′ = P(t)X

where each element of P is a real-valued

continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and its imaginary part v(t) are also solutions of this equation.






Theorem


X′ = P(t)X

where each element of P is a real-valued continuous function.

Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and its imaginary part v(t) are also solutions of this equation.






Theorem


X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t)

is a complex-valued solution, then its real partu(t) and its imaginary part v(t) are also solutions of this equation.






Theorem


X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution,

then its real partu(t) and its imaginary part v(t) are also solutions of this equation.






Theorem


X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real part

u(t) and its imaginary part v(t) are also solutions of this equation.






Theorem


X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and

its imaginary part v(t) are also solutions of this equation.






Theorem


X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and its imaginary part

v(t) are also solutions of this equation.






Theorem


X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and its imaginary part v(t)

are also solutions of this equation.






Theorem


X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and its imaginary part v(t) are also solutions

of this equation.






Theorem


X′ = P(t)X





We will concentrate most of our attention on systems ofhomogeneous linear differential equations with constant coefficients

x′ = Ax

where A is a constant n × n matrix. Unless stated otherwise, wewill assume further that all the elements of A are real (rather thancomplex) numbers.

The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.




We will concentrate

most of our attention on systems ofhomogeneous linear differential equations with constant coefficients

x′ = Ax






We will concentrate most of our attention

on systems ofhomogeneous linear differential equations with constant coefficients

x′ = Ax






We will concentrate most of our attention on systems of

homogeneous linear differential equations with constant coefficients

x′ = Ax






We will concentrate most of our attention on systems ofhomogeneous linear

differential equations with constant coefficients

x′ = Ax






We will concentrate most of our attention on systems ofhomogeneous linear differential equations

with constant coefficients

x′ = Ax







x′ = Ax







x′ = Ax







x′ = Ax

where A

is a constant n × n matrix. Unless stated otherwise, wewill assume further that all the elements of A are real (rather thancomplex) numbers.






x′ = Ax

where A is a constant

n × n matrix. Unless stated otherwise, wewill assume further that all the elements of A are real (rather thancomplex) numbers.






x′ = Ax

where A is a constant n × n matrix.

Unless stated otherwise, wewill assume further that all the elements of A are real (rather thancomplex) numbers.






x′ = Ax

where A is a constant n × n matrix. Unless stated otherwise,

wewill assume further that all the elements of A are real (rather thancomplex) numbers.






x′ = Ax

where A is a constant n × n matrix. Unless stated otherwise, wewill assume further

that all the elements of A are real (rather thancomplex) numbers.






x′ = Ax

where A is a constant n × n matrix. Unless stated otherwise, wewill assume further that all the elements

of A are real (rather thancomplex) numbers.






x′ = Ax

where A is a constant n × n matrix. Unless stated otherwise, wewill assume further that all the elements of A

are real (rather thancomplex) numbers.






x′ = Ax

where A is a constant n × n matrix. Unless stated otherwise, wewill assume further that all the elements of A are real

(rather thancomplex) numbers.






x′ = Ax

where A is a constant n × n matrix. Unless stated otherwise, wewill assume further that all the elements of A are real (rather thancomplex)

numbers.






x′ = Ax







x′ = Ax


The case

n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2

is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and

lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself

tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization

in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane,

called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane.

Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating

Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax

at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number

of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and

plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting

theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors,

we obtain a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain

a direction field of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field

of tangent vectors tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors

tosolutions of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions

of the system of differential equations.





x′ = Ax


The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system

of differential equations.





x′ = Ax






A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative

understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding

of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior

of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions

can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usually

be gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field.

More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise

information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information results

from including in the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in

the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of

some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or

trajectories.A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.

A plot that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot

that shows a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows

a representative sample of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows a representative sample

of trajectories for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories

for agiven system is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system

is called a phase portrait .




A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot of some solution curves, or trajectories.A plot that shows a representative sample of trajectories for agiven system is called

a phase portrait .












Now, for the system

x′ = Ax

we look for solutions of the form

x = veλt

where the exponent λ and the vector v are to be determined.Substituting x in the system gives

λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt


λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt


λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax

we look

for solutions of the form

x = veλt


λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax

we look for solutions

of the form

x = veλt


λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt


λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt


λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt

where the exponent

λ and the vector v are to be determined.Substituting x in the system gives

λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt

where the exponent λ and

the vector v are to be determined.Substituting x in the system gives

λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt

where the exponent λ and the vector v

are to be determined.Substituting x in the system gives

λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt

where the exponent λ and the vector v are to be

determined.Substituting x in the system gives

λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt

where the exponent λ and the vector v are to be determined.

Substituting x in the system gives

λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt

where the exponent λ and the vector v are to be determined.Substituting x

in the system gives

λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt


λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt


λveλt = Aveλt

(A− λI) v = 0




Now, for the system

x′ = Ax


x = veλt


λveλt = Aveλt

(A− λI) v = 0




Thus, to solve the system of differential equations, we must solvethe above system of algebraic equations. That is, we need to findthe eigenvalues and eigenvectors of the matrix A.

If we assume that A is a real-valued matrix, then we must considerthe following possibilities for the eigenvalues of A:

1. All eigenvalues are real and different from each other.

2. Some eigenvalues occur in complex conjugate pairs.

3. Some eigenvalues, either real or complex, are repeated.




Thus,

to solve the system of differential equations, we must solvethe above system of algebraic equations. That is, we need to findthe eigenvalues and eigenvectors of the matrix A.








Thus, to solve

the system of differential equations, we must solvethe above system of algebraic equations. That is, we need to findthe eigenvalues and eigenvectors of the matrix A.








Thus, to solve the system

of differential equations, we must solvethe above system of algebraic equations. That is, we need to findthe eigenvalues and eigenvectors of the matrix A.








Thus, to solve the system of differential equations,

we must solvethe above system of algebraic equations. That is, we need to findthe eigenvalues and eigenvectors of the matrix A.








Thus, to solve the system of differential equations, we must solve

the above system of algebraic equations. That is, we need to findthe eigenvalues and eigenvectors of the matrix A.








Thus, to solve the system of differential equations, we must solvethe above system

of algebraic equations. That is, we need to findthe eigenvalues and eigenvectors of the matrix A.








Thus, to solve the system of differential equations, we must solvethe above system of algebraic equations.

That is, we need to findthe eigenvalues and eigenvectors of the matrix A.








Thus, to solve the system of differential equations, we must solvethe above system of algebraic equations. That is,

we need to findthe eigenvalues and eigenvectors of the matrix A.








Thus, to solve the system of differential equations, we must solvethe above system of algebraic equations. That is, we need to find

the eigenvalues and eigenvectors of the matrix A.








Thus, to solve the system of differential equations, we must solvethe above system of algebraic equations. That is, we need to findthe eigenvalues and

eigenvectors of the matrix A.








Thus, to solve the system of differential equations, we must solvethe above system of algebraic equations. That is, we need to findthe eigenvalues and eigenvectors

of the matrix A.

















If we assume

that A is a real-valued matrix, then we must considerthe following possibilities for the eigenvalues of A:








If we assume that A

is a real-valued matrix, then we must considerthe following possibilities for the eigenvalues of A:








If we assume that A is a real-valued matrix,

then we must considerthe following possibilities for the eigenvalues of A:








If we assume that A is a real-valued matrix, then we must consider

the following possibilities for the eigenvalues of A:








If we assume that A is a real-valued matrix, then we must considerthe following possibilities

for the eigenvalues of A:








If we assume that A is a real-valued matrix, then we must considerthe following possibilities for the eigenvalues

of A:

















1. All eigenvalues

are real and different from each other.








1. All eigenvalues are real and

different from each other.








1. All eigenvalues are real and different

from each other.

















2. Some eigenvalues

occur in complex conjugate pairs.








2. Some eigenvalues occur in

complex conjugate pairs.

















3. Some eigenvalues,

either real or complex, are repeated.








3. Some eigenvalues, either real or

complex, are repeated.








3. Some eigenvalues, either real or complex,

are repeated.












Example 9.5

Consider the system

x′ = Ax =

(1 14 1

)x

Solution

Let’s find the eigenvalues of the matrix A




Example 9.5

Consider the system

x′ = Ax =

(1 14 1

)x

Solution





Example 9.5

Consider

the system

x′ = Ax =

(1 14 1

)x

Solution





Example 9.5

Consider the system

x′ = Ax =

(1 14 1

)x

Solution





Example 9.5

Consider the system

x′ = Ax =

(1 14 1

)x

Solution





Example 9.5

Consider the system

x′ = Ax =

(1 14 1

)x

Solution





Example 9.5

Consider the system

x′ = Ax =

(1 14 1

)x

Solution





Example 9.5

Consider the system

x′ = Ax =

(1 14 1

)x

Solution

Let’s find

the eigenvalues of the matrix A




Example 9.5

Consider the system

x′ = Ax =

(1 14 1

)x

Solution

Let’s find the eigenvalues

of the matrix A




Example 9.5

Consider the system

x′ = Ax =

(1 14 1

)x

Solution

Let’s find the eigenvalues of the matrix

A




Example 9.5

Consider the system

x′ = Ax =

(1 14 1

)x

Solution





|A− λI| =

∣∣∣∣1− λ 14 1− λ

∣∣∣∣ = 0

(1− λ)2 − 4 = 0 =⇒

(λ2 − 2λ− 3 = (λ− 3)(λ+ 1) = 0 =⇒




|A− λI| =

∣∣∣∣1− λ 14 1− λ

∣∣∣∣ = 0

(1− λ)2 − 4 = 0 =⇒

(λ2 − 2λ− 3 = (λ− 3)(λ+ 1) = 0 =⇒




|A− λI| =

∣∣∣∣1− λ 14 1− λ

∣∣∣∣ = 0

(1− λ)2 − 4 = 0 =⇒

(λ2 − 2λ− 3 = (λ− 3)(λ+ 1) = 0 =⇒




|A− λI| =

∣∣∣∣1− λ 14 1− λ

∣∣∣∣ = 0

(1− λ)2 − 4 = 0 =⇒

(λ2 − 2λ− 3 = (λ− 3)(λ+ 1) = 0 =⇒




|A− λI| =

∣∣∣∣1− λ 14 1− λ

∣∣∣∣ = 0

(1− λ)2 − 4 = 0 =⇒

(λ2 − 2λ− 3 = (λ− 3)(λ+ 1) = 0 =⇒




λ1 = 3, λ2 = −1

If λ1 = 3, then the system reduces to the single equation

−2v1 + v2 = 0, =⇒ v2 = 2v1

and a corresponding eigenvector is

v(1) =

(12

)




λ1 = 3, λ2 = −1


−2v1 + v2 = 0, =⇒ v2 = 2v1


v(1) =

(12

)




λ1 = 3, λ2 = −1

If λ1 = 3,

then the system reduces to the single equation

−2v1 + v2 = 0, =⇒ v2 = 2v1


v(1) =

(12

)




λ1 = 3, λ2 = −1

If λ1 = 3, then the system

reduces to the single equation

−2v1 + v2 = 0, =⇒ v2 = 2v1


v(1) =

(12

)




λ1 = 3, λ2 = −1

If λ1 = 3, then the system reduces to

the single equation

−2v1 + v2 = 0, =⇒ v2 = 2v1


v(1) =

(12

)




λ1 = 3, λ2 = −1


−2v1 + v2 = 0, =⇒ v2 = 2v1


v(1) =

(12

)




λ1 = 3, λ2 = −1


−2v1 + v2 = 0, =⇒ v2 = 2v1


v(1) =

(12

)




λ1 = 3, λ2 = −1


−2v1 + v2 = 0, =⇒ v2 = 2v1

and

a corresponding eigenvector is

v(1) =

(12

)




λ1 = 3, λ2 = −1


−2v1 + v2 = 0, =⇒ v2 = 2v1

and a corresponding

eigenvector is

v(1) =

(12

)




λ1 = 3, λ2 = −1


−2v1 + v2 = 0, =⇒ v2 = 2v1


v(1) =

(12

)




λ1 = 3, λ2 = −1


−2v1 + v2 = 0, =⇒ v2 = 2v1


v(1) =

(12

)




Similarly, corresponding to λ2 = −1, we find that a correspondingeigenvector is

v(2) =

(1

− 2

)The corresponding solutions of the differential equation are

x(1) =

(12

)e3t ; x(2) =

(1

− 2

)e−t




Similarly,

corresponding to λ2 = −1, we find that a correspondingeigenvector is

v(2) =

(1

− 2


x(1) =

(12

)e3t ; x(2) =

(1

− 2

)e−t




Similarly, corresponding to λ2 = −1,

we find that a correspondingeigenvector is

v(2) =

(1

− 2


x(1) =

(12

)e3t ; x(2) =

(1

− 2

)e−t




Similarly, corresponding to λ2 = −1, we find

that a correspondingeigenvector is

v(2) =

(1

− 2


x(1) =

(12

)e3t ; x(2) =

(1

− 2

)e−t




Similarly, corresponding to λ2 = −1, we find that a correspondingeigenvector

is

v(2) =

(1

− 2


x(1) =

(12

)e3t ; x(2) =

(1

− 2

)e−t





v(2) =

(1

− 2


x(1) =

(12

)e3t ; x(2) =

(1

− 2

)e−t





v(2) =

(1

− 2

)

The corresponding solutions of the differential equation are

x(1) =

(12

)e3t ; x(2) =

(1

− 2

)e−t





v(2) =

(1

− 2

)The corresponding solutions

of the differential equation are

x(1) =

(12

)e3t ; x(2) =

(1

− 2

)e−t





v(2) =

(1

− 2

)The corresponding solutions of the differential equation

are

x(1) =

(12

)e3t ; x(2) =

(1

− 2

)e−t





v(2) =

(1

− 2


x(1) =

(12

)e3t ; x(2) =

(1

− 2

)e−t





v(2) =

(1

− 2


x(1) =

(12

)e3t ;

x(2) =

(1

− 2

)e−t





v(2) =

(1

− 2


x(1) =

(12

)e3t ; x(2) =

(1

− 2

)e−t




The Wronskian of these solutions is

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

Hence, the solutions x(1) and x(2) form a fundamental set, and thegeneral solution of the system is

x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t




The Wronskian

of these solutions is

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0


x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t




The Wronskian of these solutions

is

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0


x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0


x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0


x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ =

− 4e2t 6= 0


x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0


x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

Hence,

the solutions x(1) and x(2) form a fundamental set, and thegeneral solution of the system is

x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

Hence, the solutions

x(1) and x(2) form a fundamental set, and thegeneral solution of the system is

x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

Hence, the solutions x(1) and

x(2) form a fundamental set, and thegeneral solution of the system is

x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

Hence, the solutions x(1) and x(2)

form a fundamental set, and thegeneral solution of the system is

x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

Hence, the solutions x(1) and x(2) form a fundamental set,

and thegeneral solution of the system is

x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

Hence, the solutions x(1) and x(2) form a fundamental set, and thegeneral solution

of the system is

x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0


x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0


x =

c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0


x = c1x(1) + c2x(2) =

c1

(12

)e3t + c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0


x = c1x(1) + c2x(2) = c1

(12

)e3t +

c2

(1

− 2

)e−t





W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0


x = c1x(1) + c2x(2) = c1

(12

)e3t + c2

(1

− 2

)e−t


Linear Algebra. Session 9roquesol/Math_304_Fall_2017_Session_9.pdfSession 9. Abstract Linear Algebra Eigenvalues ansd Eigenvectors. Systems of Linear Di erential Equations Eigenvalues

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