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§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
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
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
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
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 .
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Case I: When there are m independent eigenvectorCase II: When there are fewer than m independent eigenvectors
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.
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Case I: When there are m independent eigenvectorCase II: When there are fewer than m independent eigenvectors
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)
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Case I: When there are m independent eigenvectorCase II: When there are fewer than m independent eigenvectors
◮ 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.
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
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)
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
Computing Eigenvalues
◮ Eigenvalues of the coef. matrix A, are: given by
∣
∣
∣
∣
3 − r −41 −1 − r
∣
∣
∣
∣
= 0 =⇒ (r − 1)2 = 0 =⇒ r = 1
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
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
)
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
◮ 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
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
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
)
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
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)
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 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.
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
Sample II Ex 5
Find the general solution of the following system of equations:
x′ =
1 1 12 1 −10 −1 1
x (6)
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
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.
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
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
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
◮ 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
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
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
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
◮ 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
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
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
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues
§7.8 HL System and Repeated EigenvaluesTwo Cases of a double eigenvalue
Sample ProblemsHomework
Sample I Ex 1Sample II Ex 5
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
Satya Mandal, KU Chapter 7 §7.8 Repeated Eigenvalues