50
5.
110
220
001
ªº
«»
«»
«»
¬¼
A
Solution
(a) Block diagonal matrix;
() 0,1,3
O
A
. For
10
O
we solve
>@
111
110 0 1
2 2 0 0 1
001 0 0
ªº½ ½
°° ° °
«»
®¾ ® ¾
«»
°° ° °
«»
¬¼¯¿ ¯¿
Av 0 v v
6.
13 1
02 3
00 1
ªº
«»
«»
«»
¬¼
A
Solution
(a) Upper-triangular matrix; () 1,2,1
O
A. For
11
O
we solve
>@
111
03 1 0 1
0 1 3 0 0
00 2 0 0
ª º ½ ½
°° °°
«»
®¾ ®¾
«»
°° °°
«»
¬ ¼ ¯¿ ¯¿
AIv 0 v v
51
7.
312
44 6
21 1
ªº
«»
«»
«»
¬¼
A
Solution
(a)
() 1,2,3
O
A
. For
11
O
we solve
>@
111
212 0 0
4 3 6 0 2
21 2 0 1
ªº½ ½
°° °°
«»
®¾ ®¾
«»
°° °°
«»
¬¼¯¿ ¯¿
AIv 0 v v
52
8.
224
13 2
510 1
ªº
«»
«»
«»
¬¼
A
Solution
(a)() 1,4,1
O
A. For 11
O
we solve
>@
111
324 0 2
1 4 2 0 1
510 0 0 1
ªº½ ½
°° ° °
«»
®¾ ® ¾
«»
°° ° °
«»
¬¼¯¿ ¯¿
AIv 0 v v
D =
4.0000 0 0
0 1.0000 0
0 0 -1.0000
9.
22 20
132 2
0323
12 21
ªº
«»
«»
«»
«»
¬¼
A
Solution
(a)
( ) 0, 1, 1, 2
O
A
. For
10
O
we solve
V =
-0.8165 0.5774 0.0000 0.5000
0.4082 -0.5774 0.7071 -0.5000
10.
0000
0410
0520
000 1
ªº
«»
«»
«»
«»
¬¼
A
Solution
(a)
( ) 0, 1, 1, 3
O
A
. For
10
O
we solve
111
0000 0 1
0410 0 0
0520 0 0
000 1 0 0
ª º ½ ½
« » °° °°
°° °°
«»
®¾ ®¾
«»
°° °°
«»
°° °°
¬ ¼ ¯¿ ¯¿
Av 0 v v
54
In Problems 11–18, find the eigenvalues, eigenvectors, algebraic and geometric multiplicity of each
eigenvalue, and decide whether the matrix is defective or not. Then transform the matrix into either a diagonal or a
Jordan matrix, whichever applicable.
11.
11
66
ªº
«»
¬¼
A
Solution
>> A = [1 -1;6 6]; [V,D] = eig(A)
55
12.
31
11
ªº
«»
¬¼
A
Solution
>> A = [-3 -1;1 -1]; [V,D] = eig(A)
13.
10 0
01 1
50 2
ªº
«»
«»
«»
¬¼
A
Solution
>> A = [1 0 0;0 1 1;5 0 -2];
>> [V,D] = eig(A)
V =
0 0 0.0000
56
14.
010
001
111
ªº
«»
«»
«»
¬¼
A
Solution
>> A = [0 1 0;0 0 1;-1 -1 -1]; [V,D] = eig(A)
57
15.
211
321
110
ªº
«»
«»
«»
¬¼
A
Solution
Use the eig command to find eigenvalues and eigenvectors of matrix
A
.
>> A = [-2 -1 -1;3 2 1;1 1 0]; [V,D] = eig(A)
V =
16.
321
020
002
ªº
«»
«»
«»
¬¼
A
Solution
>> A = [3 2 1;0 2 0;0 0 2]; [V,D] = eig(A)
V =
58
17.
30 0
00 1
01 0
ªº
«»
«»
«»
¬¼
A
Solution
Use the eig command to find eigenvalues and eigenvectors of matrix
A
.
18.
100
211
002
ªº
«»
«»
«»
¬¼
A
Solution
>> A = [1 0 0;2 1 1;0 0 2];[V,D] = eig(A)
V =
19. Prove that if
22u
A
has a repeated eigenvalue
O
, then
A
must be defective.
Solution
If the eigenvalue is repeated, we have
AM( ) 2
O
. To determine the GM we must solve the eigenvalue problem
20. Prove that a singular matrix must have at least one zero eigenvalue.
Solution
The determinant of a matrix is the product of its eigenvalues. If the determinant is zero, then at least one of the
Review Problems
1. Prove that if
A
is
nnu
and
rank( ) nA
, then
1
A
does not exist.
Solution
2.Show that if
A
is lower triangular and one of its diagonal entries is zero, then
1
A
does not exist.
Solution
3. Prove that the product of two symmetric matrices is not necessarily symmetric.
Solution
4.If
A
is
mmu
and symmetric, and
B
is a general
mnu
matrix, show that
T
BAB
is
nnu
and symmetric.
Solution
5. Find the rank of
21 0 3
49 27
34 15
ªº
«»
«»
«»
¬¼
A
Solution
6. Determine
a
such that rank( ) 3 A, where
1210
23 4 1
,
350 2
26 1
a parameter
a
ªº
«»
«»
«»
«»
¬¼
A
Solution
Using EROs we find
12 10
ªº
7.Find
a
such that the following homogeneous system has a non-trivial ( zx0
) solution:
1
2
3
21 1 0
23 0
252 0
x
ax
x
½
ªº½
°°°°
«»
®¾®¾
«»
°°°°
«»
¬¼¯¿
¯¿
Solution
The system has a non-trivial solution if and only if
0 A
. Thus,
8. Find the value(s) of
a
for which the following system only has a trivial solution:
1
2
3
143 0
221 0
24 0
x
x
ax
½
ªº½
°°°°
«»
®¾®¾
«»
°°°°
«»
¬¼¯¿
¯¿
Solution
The homogeneous system has a trivial solution if and only if the coefficient matrix is non-singular. Since
9. Using Gauss elimination solve
1
2
3
4
20 2 1 1
01 3 2 5
, , ,
21 4 3 1
2114 7
x
x
x
x
½
ªº ½
°°
«» °°
°° °°
«»
®¾ ®¾
«»
°° °°
«»
°° °°
¬¼ ¯¿
¯¿
Ax b A x b
61
Solution
Using EROs, the augmented matrix is transformed into
202 11
012 20
ªº
«»
«»
10. Solve the system in Problem 9 using the inverse of the coefficient matrix.
Solution
We find
1
A
as
8 11 2 9 4 5.5 1 4.5
14 24 6 20 7 12 3 10
ªºª º
«»« »
11. Find the inverse of the rotation matrix
cos sin 0
sin cos 0
001
TT
TT
ªº
«»
«»
«»
¬¼
R
Solution
Noting
1 R
, we find
cos sin 0
TT
ªº
12. Solve the following system using
(a) Cramer’s rule,
(b) the “\” operator.
1
2
3
4
10 12 2
210 2 4
, , ,
031 1 2
10 2 3 2
x
x
x
x
½
ªº ½
°°
«» °°
°° °°
«»
®¾ ®¾
«»
°° °°
«»
°° °°
¬¼ ¯¿
¯¿
Ax b A x b
62
Solution
(a)
>> A = [1 0 -1 2;2 1 0 -2;0 3 1 -1;-1 0 2 3]; b = [-2;4;2;-2]; d = det(A);
>> A1=A; A1(:,1)=b; d1=det(A1);
>> A2=A; A2(:,2)=b; d2=det(A2);
13. Solve using Cramer’s rule:
1
2
3
4
110 3 2
021 1 2
, , ,
1031 4
112 0 1
x
x
x
x
½
ªº ½
°°
«» °°
°° °°
«»
®¾ ®¾
«»
°° °°
«»
°° °°
¬¼ ¯¿
¯¿
Ax b A x b
Solution
Since
33 0D zA
we proceed to calculate 12 3 4
0, 33, 33, 33DD D D
, hence
14. Show that any matrix with distinct eigenvalues is non-defective.
Solution
Since the eigenvalues are assumed distinct, each has
AM 1
. But
GM AMd
so that each eigenvalue also has
15.(a) Find all eigenvalues and eigenvectors of
120
030
112
ªº
«»
«»
«»
¬¼
A
(b) Repeat in MATLAB.
Solution
(a)
A
is block lower triangular, hence its eigenvalues are those of the upper-left corner block and the single block
63
16. Find the eigenvalues and the algebraic and geometric multiplicity of each, and decide whether the matrix is
defective.
100 0
23 0 0
10 11
020 1
ªº
«»
«»
«»
«»
¬¼
A
Solution
A
is block lower triangular, comprised of two
22u
blocks, one block is lower triangular, the other upper
triangular. Therefore
1, 1, 1, 3
O
. For the two eigenvalues that occur only once, the AM and GM are both 1. Let
us inspect 1
O
as follows.
17. Prove that the eigenvalues of a matrix are preserved under a similarity transformation, that is, if
1 SAS B
,
then the eigenvalues of
A
and
B
are the same. Hint: Show that if
O
is an eigenvalue of
B
, it is also an
eigenvalue of
A
.
Solution
Let
O
be an eigenvalue of
B
so that
0
O
I-B
. Substitute for
B
:
18. Find the modal matrix and use it to transform
A
into a diagonal or a Jordan matrix:
14 1 2
03 1 4
00 1 0
00 2 3
ªº
«»
«»
«»
«»
¬¼
A
Solution
>> A = [1 4 -1 2;0 3 1 4;0 0 1 0;0 0 2 -3];
>> [V,J] = jordan(A) % [V,D]=eig(A) returns a singular modal matrix V
19. Prove that if
nnu
A
has eigenvalues 1, … , n
OO
, then
1
A
exists if
0
i
O
z
for
1, … ,in
.
Solution
20. Prove that if
k const
and
nnu
A
has eigenvalues 1, … , n
OO
with corresponding eigenvectors
1, … , n
vv
, then
the eigenvalues of
kA
are
1, … , n
kk
OO
with eigenvectors
1, … , n
vv
.
Solution
Multiply both sides of
O
Av v by
k
to get
kk
O
Av v
. This is the eigenvalue problem for
kA
so that the