5.5 • Solutions 313
3. 51
.
81
A⎡⎤
=⎢⎥
−
⎣⎦
The characteristic polynomial is
2
613,−+
so the eigenvalues of A are
12
8(22) 0,xix−+−− =
so
12
1
4
i
xx
−−
=
with
2
x free. A nice basis vector for the eigenspace is thus
1
1.
4
i−−
⎡⎤
=⎢⎥
⎣⎦
v
⎡
⎢
⎣
⎦
4. 12
.
13
A−
⎡⎤
=⎢⎥
⎣⎦
The characteristic polynomial is
2
45,−+
so the eigenvalues of A are
44
2.
2i
±−
==±
⎡
⎢
⎣
5. 31
.
25
A⎡⎤
=⎢⎥
−
⎣⎦
The characteristic polynomial is
2
817,−+
so the eigenvalues of A are
84
4.
2i
±−
==±
⎡
⎢
⎣
314 CHAPTER 5 • Eigenvalues and Eigen
v
r
6. 75
.
13
A−
⎡⎤
=⎢⎥
⎣⎦
The characteristic p
o
10 4 5.
2i
±−
==±
h
s
7.
31
.
13
⎡⎤
⎢⎥
⎢⎥
⎢⎥
⎢⎥
⎣⎦
−
=A
From Example 6, t
h
Axx
is
22
(3) 1
2
r=
||
=+=
xy-plane and use trigonometry:
Note:
Your students will want to kno
w
matrix of the form ab
ba
−
⎡⎤
⎢⎥
⎣⎦
and simpl
y
a
a
8.
333
.
33 3
A
⎡⎤
⎢⎥
⎢⎥
⎢⎥
⎢⎥
⎣⎦
=−
The eigenvalu
e
o
v
ectors
r
o
lynomial is
2
10 26,−+
so the eigenvalues of A a
r
h
e eigenvalues are
3.±i
The scale factor for the tra
n
2
. For the angle of rotation, plot the point
()(3ab,= ,
w
whether you permit them on an exam to omit calc
u
y
write the eigenvalues
.±abi
A similar question ma
y
e
s are
3(33).i±
The scale factor for the transformati
o
r
e
n
sformation
1)
in the
u
lations for a
y
arise about
o
n
Axx
5.5 • Solutions 315
9. 02
.
20
A⎡⎤
=⎢⎥
−
⎣⎦
The eigenvalues are
2.i±
The scale factor for the transformation
Axx
is
10. 0.5
.
.5 0
A⎡⎤
=⎢⎥
−
⎣⎦
The eigenvalues are
.5 .i±
The scale factor for the transformation
Axx
is
11.
31
.
13
A
⎡⎤
−
=⎢⎥
−−
⎢⎥
⎣⎦
The eigenvalues are
3.i−±
The scale factor for the transformation
Axx
is
12.
33
.
33
A
⎡⎤
−
=⎢⎥
⎢⎥
⎣⎦
The eigenvalues are
3(3).i±
The scale factor for the transformation
Axx
is
13. From Exercise 1,
2,=±i
and the eigenvector 1
1
i−−
⎡
⎤
=
⎢
⎥
⎣
⎦
v corresponds to
2.=−i
Since Re
1
1
−
⎡⎤
=⎢⎥
⎣⎦
v and Im 1,
0
−
⎡
⎤
=
⎢
⎥
⎣
⎦
v take 11
.
10
−−
⎡
⎤
=
⎢
⎥
⎣
⎦
P Then compute
Notes:
The Study Guide points out that the matrix C is described in Theorem 9 and the first column of C
is the real part of the eigenvector corresponding to
,−abi
not
,+abi
as one might expect. Since students
1
−
316 CHAPTER 5 • Eigenvalues and Eigenvectors
14. 33
.
11
A−
⎡⎤
=⎢⎥
⎣⎦
The eigenvalues of A are
2(2),i=±
and the eigenvector 2i
i
⎡⎤
+
=⎢⎥
⎣⎦
v corresponds
to
2(2).i=−
By Theorem 9, 21
[Re Im ] 01
P
⎡
⎤
==
⎢
⎥
⎣
⎦
vv and
⎡
⎢
⎢
⎣
[Re Im ]P==ww
12
10
⎡
⎤
⎢
⎥
⎣
⎡
⎢
⎢
⎣
and
15. 05
.
22
A⎡⎤
=⎢⎥
−
⎣⎦
The eigenvalues of A are
13,i=±
and the eigenvector 3
2
i
i
+
⎡⎤
=⎢⎥
⎣⎦
v corresponds to
13.i=+
By Theorem 9,
[Re Im ]P==vv
31
02
⎡
⎤
⎢
⎥
⎣
⎦and
16. 42
.
16
A−
⎡⎤
=⎢⎥
⎣⎦
The eigenvalues of A are
5,i=±
and the eigenvector 1i
i
+
⎡
⎤
=
⎢
⎥
−
⎣
⎦
v corresponds to
5.i=+
By Theorem 9,
[]
11
Re Im 01
P
⎡
⎤
==
⎢
⎥
−
⎣
⎦
vv and
5.5 • Solutions 317
17. 11 4 .
20 5
A−−
⎡⎤
=⎢⎥
⎣⎦
The eigenvalues of A are
34,i=− ±
and the eigenvector
12
5
i
i
+
⎡⎤
=⎢⎥
−
⎣⎦
v corresponds to
34.i=− +
By Theorem 9,
[]
12
Re Im 05
P⎡⎤
==
⎢⎥
−
⎣⎦
vv and
18. 35
.
25
A−
⎡⎤
=⎢⎥
⎣⎦
The eigenvalues of A are
43,i=±
and the eigenvector 3
2
i
i
+
⎡⎤
=⎢⎥
−
⎣⎦
v corresponds to
43.i=+
By Theorem 9,
[]
31
Re Im 02
P
⎡
⎤
==
⎢
⎥
−
⎣
⎦
vv and
1
213531 43
1
032502 34
6
CPAP
−
−
⎡⎤⎡⎤⎡⎤⎡⎤
== =
⎢⎥⎢⎥⎢⎥⎢⎥
−−−
⎣⎦⎣⎦⎣⎦⎣⎦
. There are many choices for the muliple of
⎡
⎢
⎣
⎡
⎢
⎣
19. 152 7 .
56 4
.−.
⎡⎤
=⎢⎥
..
⎣⎦
A The characteristic polynomial is
2
192 1,−. +
so the eigenvalues of A are
96 28 .=. ±. i
To find an eigenvector corresponding to
96 28 ,.−.i
we compute
56 28 7
(96 28) 56 56 28
i
AiI i
.+. −.
⎡⎤
−. −. =
⎢⎥
.−.+.
⎣⎦
318 CHAPTER 5 • Eigenvalues and Eigenvectors
⎡
⎢
⎣
⎡
⎢
⎣
⎡
⎢
⎣
20. 38
.
45
A−−
⎡⎤
=⎢⎥
⎣⎦
The eigenvalues of A are
14,i=±
and the eigenvector 1i
i
+
⎡⎤
=⎢⎥
−
⎣⎦
v corresponds to
14.i=+
By Theorem 9,By Theorem 9,
[]
11
Re Im 01
P
⎡
⎤
==
⎢
⎥
−
⎣
⎦
vv and
⎡
⎢
⎣
⎡
⎢
⎣
21. The first equation in (2) is
12
(3 6) 6 0.−. + . − . =ix x We solve this for
2
x to find that
211
(( 3 6)6) ((1 2)2) .=−.+. /. =−+ /xixix
Letting
1
2,=x we find that 2
12i
⎡
⎤
=
⎢
⎥
−+
⎣
⎦
y is an eigenvector
22. Since
() () ()()===,xxxxxAA
µµ µ µµ
is an eigenvector of A.
23. (a) properties of conjugates and the fact that
TT
=xx
(b)
=xxAA
and A is real
24.
()
TT T
A=λ=λ⋅xxx x xx
because x is an eigenvector. It is easy to see that
T
xx
is real (and positive)
because zz is nonnegative for every complex number z. Since
T
Axx
is real, by Exercise 23, so is
.
Next, write
,=+xu vi
where u and v are real vectors. Then
5.5 • Solutions 319
,=uuA
which shows that the real part of x is an eigenvector of A.
25. Write
Re (Im ),=+xx xi
so that
(Re ) (Im ).=+xx xAA iA
Since A is real, so are
(Re )Ax
and
26. a. If
,=−abi
then
Re Im
()(Re Im )
( Re Im ) ( Im Re )
Av Av
Aabii
ab iab
==− +
=++ −
vv v v
vv vv
b. Let
[]
Re Im .=vvP
By (a),
(Re ) (Im )
ab
APAP
ba
−
⎡⎤ ⎡ ⎤
=, =
⎢⎥ ⎢ ⎥
⎣⎦ ⎣ ⎦
vv
So
27.
26 33 23 20
68113
[] 14 19 16 3
A
⎡⎤
⎢⎥
−−−−
⎢⎥
=⎢⎥
−−−
M
The MATLAB command [V D] = eig(A) returns
V = -0.5709 – 0.4172i -0.5709 + 0.4172i -0.3708 + 0.0732i -0.3708 – 0.0732i
0.4941 – 0.0769i 0.4941 + 0.0769i 0.4440 + 0.2976i 0.4440 – 0.2976i
The eigenvalues of A are the elements listed on the diagonal of D. The corresponding
eigenvectors are listed in the corresponding columns of V. To get a nice eigenvector for
⎢
⎢
⎣
1
i
i
−+
⎡
⎤
⎢
⎥
−
⎢
⎥
320 CHAPTER 5 • Eigenvalues and Eigenvectors
1
2
i
i
−−
⎡⎤
⎢⎥
⎢⎥
Hence by Theorem 9,
1122
1111
0102
Re Im Re Im 100 2
0020
P⎡⎤
⎢⎥
⎣⎦
−−−
⎡
⎤
⎢
⎥
−
⎢
⎥
==
⎢
⎥
−
⎢
⎥
⎣
⎦
vvvv
and
25 00
52 00
−
⎡⎤
⎢⎥
−−
⎢⎥
1
−
7112017
20 40 86 74
⎡⎤
⎢⎥
−−−−
⎢⎥
V= -0.2132 + 0.2132i -0.2132 – 0.2132i 0.1085 – 0.3254i 0.1085 + 0.3254i
0 – 0.8528i 0 + 0.8528i 0.3254 + 0.1085i 0.3254 – 0.1085i
D = 2.0000 + 5.0000i 0 0 0
0 2.0000 – 5.0000i 0 0
5.6 • Solutions 321
⎢
⎢
⎢
⎣
11 2 0
−
⎡
⎤
⎢
⎥
5.6 SOLUTIONS
1. The exercise does not specify the matrix A, but only lists the eigenvalues 3 and 1/3, and the
corresponding eigenvectors
1
1
1
⎡⎤
=⎢⎥
⎣⎦
v and
2
1.
1
−
⎡
⎤
=
⎢
⎥
⎣
⎦
v Also,
0
9.
1
⎡
⎤
=
⎢
⎥
⎣
⎦
x
a. To find the action of A on
0
,x express
0
x in terms of
1
v and
2
.v That is, find
1
c and
2
c such
that
01122
.=+xvvcc
This is certainly possible because the eigenvectors
1
v and
2
v are linearly
2
v are eigenvectors (for the eigenvalues 3 and 1/3):
⎡
⎢
⎣
b. Each time A acts on a linear combination of
1
v and
2
,v the
1
v term is multiplied by the
eigenvalue 3 and the
v term is multiplied by the eigenvalue 1/3:
2. The vectors
123
123
013
357
−
⎡⎤ ⎡⎤ ⎡⎤
⎢⎥ ⎢⎥ ⎢⎥
=,=,=−
⎢⎥ ⎢⎥ ⎢⎥
⎢⎥ ⎢⎥ ⎢⎥
−−
⎣⎦ ⎣⎦ ⎣⎦
vv v
are eigenvectors of a
33×
matrix A, corresponding to
322 CHAPTER 5 • Eigenvalues and Eigenvectors
−
⎣⎦
3.
2
54
det( ) ( 5 )(1 1 ) 08 1 6 63.
211
..
⎡⎤
=,−=.−.−+.=−.+.
⎢⎥
−. .
⎣⎦
AAI
λλλλλ
This characteristic polynomial
factors as
(9)(7),−. −.
λλ
so the eigenvalues are .9 and .7. If
1
v and
2
v denote corresponding
eigenvectors, and if
01122
,=+xvvcc
then
4.
2
54
det( ) ( 5 )(1 1 ) ( 4)( 125) 1 6 6.
125 1 1
..
⎡⎤
=,−=.−.−−..=−.+.
⎢⎥
−. .
⎣⎦
AAI
λλλ λλ
This characteristic
polynomial factors as
(1)( 6),−−.
λλ
so the eigenvalues are 1 and .6. For the eigenvalue 1, solve
5 40 540
()0 .
125 1 0 0 0 0
−. . −
⎡⎤⎡⎤
−=:
⎢⎥⎢⎥
−. .
⎣⎦⎣⎦
x∼AI A basis for the eigenspace is
1
4.
5
⎡⎤
=⎢⎥
⎣⎦
v Let
2
v be an
5.
2
43
det( ) 1 6 5775.
325 1 2
..
⎡⎤
=,−=−.+.
⎢⎥
−. .
⎣⎦
AAI
λλ λ
The quadratic formula provides the roots of the
characteristic equation:
Because one eigenvalue is larger than one, both populations grow in size. Their relative sizes are
determined eventually by the entries in the eigenvector corresponding to 1.05. Solve
(105) :−. =x0AI
5.6 • Solutions 323
Eventually, there will be about 6 spotted owls for every 13 (thousand) flying squirrels.
6. When 43
5,
512
..
⎡⎤
=. , =⎢⎥
−. .
⎣⎦
pA and
2
det( ) 1 6 63 ( 9)( 7).−=−.+.=−. −.AI
λλ λ λ λ
The eigenvalues of A are .9 and .7, both less than 1 in magnitude. The origin is an attractor for the
dynamical system and each trajectory tends toward 0. So both populations of owls and squirrels
eventually perish.
By the quadratic formula,
2
16 16 4(48 3 )
2
p
λ
.± . − . +.
=
7. a. The matrix A in Exercise 1 has eigenvalues 3 and 1/3. Since
31
||
>
and
13 1,
|
/
|
<
the origin is a
saddle point.
b. The direction of greatest attraction is determined by
2
1,
1
−
⎡
⎤
=
⎢
⎥
⎣
⎦
v the eigenvector corresponding to
324 CHAPTER 5 • Eigenvalues and Eigen
v
r
t
c
n
d
r
l
e
n
e
8. The matrix from Exercise 2 has eig
e
and the others are less than one in
m
g
e
9.
2
17 3 det( )
12 8
AAI
.−.
⎡⎤
=,−λ=
−
⎢⎥
−. .
⎣⎦
2
25 25 4(1) 25
2
.± . − .±
==
o
e
s
v
ectors
r
e
nvalues 3, 4/5, and 3/5. Since one eigenvalue is grea
t
m
agnitude, the origin is a saddle point. The direction
o
25 10
−
.+=
225 25 15 2 and 5
22
..±.
==.
t
er than 1
o
f greatest
5.6 • Solutions 325
10.
2
34
det( )14 45 0
311
AAI
..
⎡⎤
=,−=−.+.=
⎢⎥
−. .
⎣⎦
2
14 14 4(45) 14 16 14 4 5 and 9
222
.± . − . .± . .±.
====..
11.
2
45
det( )17 72 0
413
AAI
..
⎡⎤
=,−=−.+.=
⎢⎥
−. .
⎣⎦
12.
2
56
det( )19 88 0
314
AAI
..
⎡⎤
=,−=−.+.=
⎢⎥
−. .
⎣⎦
The origin is a saddle point because one eigenvalue is greater than 1 and the other eigenvalue is less
than 1 in magnitude. The direction of greatest repulsion is through the origin and the eigenvector
1
v
found below. Solve 660 1 10
(11) ,
330 0 00
−. . −
⎡⎤⎡⎤
−. = :
⎢⎥⎢⎥
−. .
⎣⎦⎣⎦
x0 ∼AI so
12
,=xx
and
2
x is free. Take
326 CHAPTER 5 • Eigenvalues and Eigenvectors
13.
2
83
det( )23 132 0
415
AAI
..
⎡⎤
=,−=−.+.=
⎢⎥
−. .
⎣⎦
The origin is a repellor because both eigenvalues are greater than 1 in magnitude. The direction of
greatest repulsion is through the origin and the eigenvector
1
v found below. Solve
14.
2
17 6 det( )24 143 0
47
AAI
..
⎡⎤
=,−=−.+.=
⎢⎥
−. .
⎣⎦
The origin is a repellor because both eigenvalues are greater than 1 in magnitude. The direction of
greatest repulsion is through the origin and the eigenvector
1
v found below. Solve
4601150
(13) ,
460000
.. .
⎡⎤⎡⎤
−. = :
⎢⎥⎢⎥
−. −.
⎣⎦⎣⎦
x0 ∼AI so
12
15 ,=− .xx
and
2
x is free. Take
1
3.
2
−
⎡⎤
=⎢⎥
⎣⎦
v
15.
40 2
383.
⎢
⎢
⎣
..
⎡⎤
⎢⎥
=. . .
A
Given eigenvector
1
1
6
.
⎡
⎤
⎢
⎥
=.
v
and eigenvalues .5 and .2. To find the eigenvalue for
13
23 2
1020 10 20 2 2
For 5 3 3 3 0 0 1 3 0 3 . Set 3
−. . − =
⎡⎤⎡⎤ ⎡⎤
⎢⎥⎢⎥ ⎢⎥
=. : . . . , =− = − .
⎢⎥⎢⎥ ⎢⎥
v∼
xx
xx
λ
Given
0
(0 3 7),=,.,.x find weights such that
011 233
.=++xvvvccc
12 10 1001
.−
⎡⎤⎡⎤
5.6 • Solutions 327
16. [M]
90 01 09 1 0000
01 90 01 0 8900 To four decimal places
... .
⎡⎤ ⎡⎤
⎢⎥ ⎢⎥
=. . . ⋅ = . . ,
⎢⎥ ⎢⎥
2
ev = eig(A)
v = nulbasis(A –ev(2)
A
1
1
0
−
⎡⎤
⎢⎥
⎢⎥
⎢⎥
*eye(3))=
Note:
When working with stochastic matrices and starting with a probability vector (having nonnegative
entries whose sum is 1), it helps to scale
1
v to make its entries sum to 1. If
1
(91 209 19 209 99 209),=/,/,/v or
( 435 091 474).,.,.
to three decimal places, then the weight
1
c above
17. a. 016
38
A.
=⎢⎥
..
⎣⎦
b.
2
16
det 8 48 0.
38
−.
⎡⎤
=−.−.=
⎢⎥
..−
⎣⎦
λλλ
λ
The eigenvalues of A are given by
The numbers of juveniles and adults are increasing because the largest eigenvalue is greater than
1. The eventual growth rate of each age class is 1.2, which is 20% per year.
To find the eventual relative population sizes, solve
(12) :−. =x0AI
(4 3)
xx
=/
328 CHAPTER 5 • Eigenvalues and Eigenvectors
c. [M] Suppose that the initial populations are given by
0
(15 10).=,x The Study Guide describes
18. a.
0042
60 0
07595
A
.
⎡⎤
⎢⎥
=.
⎢⎥
⎢⎥
..
⎣⎦
b.
0 0774 0 4063
0 0774 0 4063
1 1048
i
i
.+.
⎡⎤
⎢⎥
.−.
⎢⎥
⎢⎥
.
ev = eig(A)=
5.7 SOLUTIONS
1. From the “eigendata” (eigenvalues and corresponding eigenvectors) given, the eigenfunctions for the
differential equation
A′=xx
are
4
1
t
ev
and
2
2
.v
t
e
The general solution of
A′=xx
has the form
42
12
31
11
tt
cece
−−
⎡⎤ ⎡⎤
+
⎢⎥ ⎢⎥
⎣⎦ ⎣⎦
The initial condition 6
(0) 1
−
⎡⎤
=⎢⎥
⎣⎦
x determines
1
c and
2
:c
4(0) 2(0)
12
316
111
−−−
⎡⎤ ⎡⎤ ⎡⎤
+=
⎢⎥ ⎢⎥ ⎢⎥
cece
5.7 • Solutions 329
2. From the eigendata given, the eigenfunctions for the differential equation
A′=xx
are
3
1
t
e
−
v
and
1
2.
−
v
t
e
The general solution of
A′=xx
has the form
The initial condition 2
(0) 3
⎡⎤
=⎢⎥
⎣⎦
x determines
1
c and
2
c:
3(0) 1(0)
12
112
113
ce ce
−−
−
⎡⎤ ⎡⎤ ⎡⎤
+=
⎢⎥ ⎢⎥ ⎢⎥
⎣⎦ ⎣⎦ ⎣⎦
3.
2
23
det( )1( 1)( 1) 0.
12
⎡⎤
=,−=−=−+=
⎢⎥
−−
⎣⎦
AAI Eigenvalues: 1 and
1.−
For λ = 1: 130 130
,
1 30 000
⎡⎤⎡⎤
⎢⎥⎢⎥
−−
⎣⎦⎣⎦
∼ so
12
3xx=− with
2
x free. Take
2
1x= and
1
3.
1
−
⎡⎤
=⎢⎥
⎣⎦
v
12
313 1052
(0) 112 0192
−− −/
⎡⎤⎡ ⎤
=
⎡⎤
⎢⎥⎢ ⎥
⎣⎦ /
⎣⎦⎣ ⎦
vvx ∼
4.
2
25
det( ) 2 3( 1)( 3) 0.
14
AAI
−−
⎡⎤
=,−=−−=+−=
⎢⎥
⎣⎦ Eigenvalues:
1−
and 3.
For λ = 3: 550 110
,
110 000
−−
⎡⎤⎡⎤
⎢⎥⎢⎥
⎣⎦⎣⎦
∼ so
12
xx=− with
2
x free. Take
2
1x= and
1
1.
1
−
⎡⎤
=⎢⎥
⎣⎦
v
330 CHAPTER 5 • Eigenvalues and Eigenvectors
For the initial condition 3
(0) ,
2
⎡⎤
=⎢⎥
⎣⎦
x find
1
c and
2
c such that
11 2 2
(0)cc+=vvx
:
5. 71
,
33
−
⎡⎤
=⎢⎥
⎣⎦
A det
2
( ) 10 24 ( 4)( 6) 0.−=−+=− −=AI
Eigenvalues: 4 and 6.
For λ = 4: 310 1130
,
310 0 00
−−/
⎡⎤⎡ ⎤
⎢⎥⎢ ⎥
−
⎣⎦⎣ ⎦
∼ so
12
(1 3)xx=/ with
2
x free. Take
2
3x= and
1
1.
3
⎡
⎤
=
⎢
⎥
⎣
⎦
v
⎡
⎢
⎣
6. 12
,
34
−
⎡⎤
=⎢⎥
−
⎣⎦
A det
2
( ) 3 2( 1)( 2) 0.−=++=+ +=AI
Eigenvalues:
1−
and
2.−
For λ = –2: 320 1230
,
320 0 00
−−/
⎡⎤⎡ ⎤
⎢⎥⎢ ⎥
−
⎣⎦⎣ ⎦
∼ so
12
(2 3)xx=/ with
2
x free. Take
2
3x= and
1
2.
3
⎡
⎤
=
⎢
⎥
⎣
⎦
v
5.7 • Solutions 331
7. From Exercise 5, 71
,
33
−
⎡
⎤
=
⎢
⎥
⎣
⎦
A with eigenvectors
1
1
3
⎡
⎤
=
⎢
⎥
⎣
⎦
v and
2
1
1
⎡
⎤
=
⎢
⎥
⎣
⎦
v corresponding to
eigenvalues 4 and 6 respectively. To decouple the equation
,′=xxA
set
12
11
[]
31
⎡⎤
==
⎢⎥
⎣⎦
vvP and let
8. From Exercise 6, 12
,
34
−
⎡
⎤
=
⎢
⎥
−
⎣
⎦
A with eigenvectors
1
2
3
⎡
⎤
=
⎢
⎥
⎣
⎦
v and
2
1
1
⎡
⎤
=
⎢
⎥
⎣
⎦
v corresponding to
21
1
() () ()
dPAPPDPPPD
dt
−
== =yy yy
Since P has constant entries,
()
() (),=yy
dd
dt dt
PP
so that left-multiplying the equality
9. 32
.
11
−
⎡⎤
=⎢⎥
−−
⎣⎦
A An eigenvalue of A is
2i−+
with corresponding eigenvector 1.
1
−
⎡⎤
=⎢⎥
⎣⎦
vi The
complex eigenfunctions
t
e
λ
v and
λ
v
t
e form a basis for the set of all complex solutions to
.′=xxA
The general complex solution is
332 CHAPTER 5 • Eigenvalues and Eigenvectors
where
1
c and
2
c are arbitrary complex numbers. To build the general real solution, rewrite
(2 )it
e
−+
v
as:
(2 ) 2 2
2
2
11
(cos sin )
11
cos cos sin sin
cos sin
−+ − −
−
−−
⎡⎤ ⎡⎤
== +
⎢⎥ ⎢⎥
⎣⎦ ⎣⎦
⎡⎤
−+−
=⎢⎥
+
⎣⎦
v
it t it t
t
ii
eeeetit
ti ti ti t
e
ti t
10. 31
.
21
⎡⎤
=⎢⎥
−
⎣⎦
A An eigenvalue of A is
2i+
with corresponding eigenvector 1.
2
+
⎡⎤
=⎢⎥
−
⎣⎦
vi The complex
eigenfunctions
t
ev and
t
ev
form a basis for the set of all complex solutions to
.A′=xx
The
general complex solution is
where
1
c and
2
c are arbitrary complex numbers. To build the general real solution, rewrite
(2 )it
e
+
v
as:
(2 ) 2 2
11
(cos sin )
22
it t it t
ii
eeeetit
+
++
⎡⎤ ⎡⎤
== +
⎢⎥ ⎢⎥
−−
⎣⎦ ⎣⎦
v
The general real solution has the form
11. 39
.
23
−−
⎡⎤
=⎢⎥
⎣⎦
A An eigenvalue of A is 3i with corresponding eigenvector 33.
2
−+
⎡⎤
=⎢⎥
⎣⎦
vi The