Chapter 7 7.8 Repeated Eigenvalues - University of...

§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue

Sample ProblemsHomework

Chapter 7

§7.8 Repeated Eigenvalues

Satya Mandal, KU

Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues



Repeated Eigenvalues

◮ We continue to consider homogeneous linear systems withconstant coefficients:

x′ = Ax A is an n × n matrix with constant entries

(1)

◮ Now, we consider the case, when some of the eigenvaluesare repeated. We will only consider double eigenvalues




Case I: When there are m independent eigenvectorCase II: When there are fewer than m independent eigenvectors

Two Cases of a double eigenvalue

Consider the system (1).

◮ Suppose r is an eigenvalue of the coefficient matrix A ofmultiplicity m ≥ 2.Then one of the following situationsarise:

◮ There are m linearly independent eigenvectors of A,corresponding to the eigenvalue r :

ξ(1), . . . , ξ(m) : i .e. (A − rI )ξ(i) = 0.

◮ There are fewer than m linearly independent eigenvectorsof A, corresponding to the eigenvalue r .





When there are m independent eigenvector

◮ Suppose there are m independent eigenvectorcorresponding to the eigenvalue r : ξ(1), . . . , ξ(m)

◮ Then,

x(1) = ξ(1)ert , . . . , x(m) = ξ(m)

ert

are solutions of (1)

◮ They are linearly independent for all t.◮ They extend to a fundamental set of solutions, with other

n − m solutions corresponding to other eigenvalues of A.





When there are fewer that m independent

eigenvector

We assume m = 2 and there is only one linearly independenteigenvector ξ for r (i. .e. (A − r I)ξ = 0).

◮ Then x(1) = ξert is a solution of (1).

◮ Further, the linear algebraic system

(A − r I)η = ξ has a solution (2)

andx(2) = ξtert + ηert

is a solution of (1). (3)





◮ x(1), x(2) extend to a fundamental set of solutions, with

other n − m = n − 2 solutions corresponding to othereigenvalues of A.

◮ It is interesting to note, by multiplying (2) by (A − r I),we have (A − r I)2η = 0.

◮ Subsequently, we ONLY consider problems witheigenvalues with multiplicity two, with only one linearlyindependent eigenvector.




Sample I Ex 1Sample II Ex 5

Sample I Ex 1

Find the general solution of the following system of equations:

x′ =

(

3 −41 −1

)

x (4)





Computing Eigenvalues

◮ Eigenvalues of the coef. matrix A, are: given by

∣

∣

∣

∣

3 − r −41 −1 − r

∣

∣

∣

∣

= 0 =⇒ (r − 1)2 = 0 =⇒ r = 1





Eigenvectors

◮ Eigenvectors for r = 1 is given by (A− rI )x = 0, which is

(

3 − 1 −41 −1 − 1

)(

x1

x2

)

=

(

00

)

(

2 −41 −2

)(

x1

x2

)

=

(

00

)

=⇒

{

x1 − 2x2 = 00 = 0

◮ Taking x2 = 1, an eigenvector of r = 1 is

ξ =

(

21

)





◮ Correspondingly, a solution of (4) is:

x(1) = ξert =

(

21

)

et

◮ There is no second linearly independent eigenvector.

◮ So, use (3) to compute x(2). We proceed to solve the

equation (A − rI )η = ξ





Compute η

◮ Write down the equation (A − rI )η = ξ as follows:

(

3 − 1 −41 −1 − 1

)(

η1

η2

)

=

(

21

)

(

2 −41 −2

)(

η1

η2

)

=

(

21

)

=⇒

{

η1 − 2η2 = 10 = 0

◮ Taking η1 = 1 a choice of η is

η =

(

10

)





Answer

◮ By (3) another solution of (4) is

x(2) = ξtert + ηert=

(

21

)

tet +

(

10

)

et

◮ So, the general solution is x = c1x(1) + c2x

(2), or

x = c1

(

21

)

et + c2

[(

21

)

tet +

(

10

)

et

]

(5)





◮ Remark. While solving for η we could have taken η1 = 3(or η2 = 1). In that case we would have

η =

(

31

)

In that case,◮ x

(2) would be different.◮ But the general solution (5), would be the same, after

simplification.





Sample II Ex 5

Find the general solution of the following system of equations:

x′ =

1 1 12 1 −10 −1 1

x (6)





Computing Eigenvalues

◮ Eigenvalues of the coef. matrix A, are: given by

∣

∣

∣

∣

∣

∣

1 − r 1 12 1 − r −10 −1 1 − r

∣

∣

∣

∣

∣

∣

= 0

(1− r)

∣

∣

∣

∣

1 − r −1−1 1 − r

∣

∣

∣

∣

−

∣

∣

∣

∣

2 −10 1 − r

∣

∣

∣

∣

+

∣

∣

∣

∣

2 1 − r

0 −1

∣

∣

∣

∣

= 0

◮ −(r − 2)2(r + 1) = 0

◮ So, r = −1, and r = 2 with multiplicity 2.





Eigenvectors

◮ Eigenvectors for r = −1 is given by (A − rI )x = 0:

2 1 12 2 −10 −1 2

x1

x2

x3

=

000

Use TI84 (rref)

1 0 1.50 1 −20 0 0

x1

x2

x3

=

000





◮ Taking x3 = 2 and eigenvector of r = −1 is:

ξ =

−342

◮ So, a solution to (6), corresponding to r = 1 isx(1) = ξert :

x(1) =

−342

e−t





Eigenvectors

◮ Eigenvectors for r = 2 is given by (A − rI )x = 0:

−1 1 12 −1 −10 −1 −1

x1

x2

x3

=

000

Use TI84 (rref)

1 0 00 1 10 0 0

x1

x2

x3

=

000





◮ Taking x1 = 2 and eigenvector of r = 2 is:

ξ =

01−1

◮ Correspondingly, a solution to (6), corresponding to r = 2is x

(2) = ξert :

x(2) =

01−1

e2t

◮ There is no second linearly independent eigenvector.

◮ So, use (3) to compute another solution x(3). We proceed

to solve the equation (A − rI )η = ξ





Compute η

◮ Write down the equation (A − rI )η = ξ as follows:

−1 1 12 −1 −10 −1 −1

η1

η2

η3

=

01−1

Use TI84 (rref)

1 0 00 1 10 0 0

η1

η2

η3

=

110

◮ Taking η3 = 1 a choice of η is

η =

101





Answer

◮ By (3) another solution of (6) is

x(3) = ξtert + ηert=

01−1

te2t +

101

e2t

◮ So, the general solution is x = c1x(1) + c2x

(2) + c3x(2), or

x = c1

−342

e−t + c2

01−1

e2t

+c3

01−1

te2t +

101

e2t




§7.8 Assignments and Homework

◮ Read Example 1, 2 (They are helpful).

◮ Homework: §7.8 See Homework Page


Chapter 7 7.8 Repeated Eigenvalues - University of...

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