Math 2270 Spring 2004
Homework 26 Solutions
Section 8.3) 2, 3, 5, 6, 8, 9, 12
(2,3) We are told that A is an orthogonal matrix, so it preserves both length and angle.
Therefore, when A is 2 x 2, the image of the unit circle is the unit circle. The length of
semimajor and semiminor axes are both one, so by Fact 8.3.2 σ1 = σ2 = 1. Algebraically, we
remember that AT A = I for an orthogonal matrix. The eigenvalues of the identity matrix
are λ1 = · · · = λn = 1, so the singular values are all equal to one as well.
(5,8)
T
A A=
"
p q
−q p
#"
p −q
q p
#
"
=
#
p2 + q 2
0
2
0
p + q2
√ 2
p + q2.
The eigenvalues of AT A are λ1,2 = p2 + q 2 , so the singular values
are
σ
=
1,2
√ 2
2
Geometrically, A represents a rotation-dilation
dilation factor p + q . Therefore, the
√ 2 with
2
unit circle will map to a circle of radius p + q . The eigenspace is all of ℜ2 , so ~v1 = ~e1
and ~v2 = ~e2 . Then,
"
#
1
1
p
~u1 = A~v1 = √ 2
σ1
p + q2 q
1
1
~u2 = A~v2 = √ 2
σ2
p + q2
"
−q
p
#
Putting together all the pieces we have
#" √
#"
#
"
√
√
p2 + q 2 √ 0
1 0
p/√p2 + q 2 −q/√ p2 + q 2
T
A = UΣV =
0
q/ p2 + q 2 p/ p2 + q 2
p2 + q 2
0 1
(6,9) The characteristic equation of AT A is λ2 − 25λ = λ(λ − 25), so the eigenvalues are
λ1 = 25 and λ2 = 0. Singular values are σ1 = 5 and σ2 = 0. Eigenspaces are
We let ~v1 =
"
E25 = ker
"
20 −10
−10
5
E0 = ker
"
5 10
10 20
#
#
"
1
2
#
2
−1
#
1
= sp √
5
= sp
= sp
"
"
√ #
√ #
1/√5
2/ √5
and ~v2 =
. Then,
−1/ 5
2/ 5
1
1
~u1 = A~v1 =
σ1
5
"
1 2
2 4
#"
1
1
= sp √
5
"
"
1
2
#!
2
−1
#!
√ # " √ #
1/√5
1/√5
=
2/ 5
2/ 5
Math 2270 Spring 2004
Homework 26 Solutions
The second singular value is zero, so to find ~u2 , we can use Gram-Schmidt
the #basis
" on √
2/ √5
will
{~u1 , ~e1 }, or we can try to find a unit vector that is orthogonal to ~u1 . ~u2 =
−1/ 5
work. Putting together all the pieces we have
"
#"
√
√ #"
√ #
√
1/√5 2/ √5
5 0
1/√5 2/ √5
T
A = UΣV =
0 0
2/ 5 −1/ 5
2/ 5 −1/ 5
(12)
AT A =
"
0 1 1
1 1 0
#


"
#
0 1
2 1


 1 1 =
1 2
1 0
The characteristic equation is λ2 − 4λ√+ 3 = (λ − 3)(λ − 1). The eigenvalues are λ1 = 3 and
λ2 = 1. The singular values are σ1 = 3 and σ2 = 1. The eigenspaces are
E3 = ker
E1 = ker
Then,
"
"
1 −1
−1 1
−1 −1
−1 −1
#
#
1
= sp √
2
1
= sp √
2
"
"
1
1
1
−1
#!
#!
√ 


1/√6
0 1 " √ #
1
1
 1/√2

~u1 = A~v1 = √ 
=
 1 1 
 2/ 6 
√
1/ 2
σ1
3 1 0
1/ 6
√ 



√ #
0 1 "
−1/ 2
1
1/ √2



0√ 
=
~u2 = A~v2 =  1 1 

−1/ 2
σ2
1 0
1/ 2

We find ~u3 using Gram-Schmidt on the basis {~u1 , ~u2, ~e1 }. We have
√ 
√ 




1/√6
1
−1/ 2
−1 
1 
1  
~e1 − (~u1 · ~e1 )~u1 − (~u2 · ~e1 )~u2


0√ 
=
~u3 =
 0  − √  2/ 6  − √ 

√
||~e1 − (~u1 · ~e1 )~u1 − (~u2 · ~e1 )~u2 ||
length
6
2
0
1/ 2
1/ 6
√ 




1/ √3
1 − 1/6 − 1/2
1/3
1 
1 




=
 0 − 2/6 − 0  =
 −1/3  =  −1/ 3 
√
length
length
0 − 1/6 + 1/2
1/3
1/ 3

2
Math 2270 Spring 2004
Homework 26 Solutions
Putting together all the pieces we have
√
√
√  √


√ #
1/√6 −1/ 2 1/ √3
3 0 " √
1/ √2


 1/√2
T
A = UΣV =  2/√6
0√ −1/√ 3   0 1 
1/ 2 −1/ 2
0 0
1/ 6 1/ 2 1/ 3
3