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MATH 6370 Homework: Vector Iteration Methods and QR Decomposition, Assignments of Mathematical Methods for Numerical Analysis and Optimization

University of Houston (UH)Mathematical Methods for Numerical Analysis and Optimization

Solutions to problems 14 and 22 from chapter 6 of the textbook for math 6370, a graduate-level mathematics course. The solutions involve the vector iteration method for finding the eigenvalues of a matrix and the qr decomposition of a matrix using a householder projector. The document also includes additional material on the eigenvalues of a matrix and their associated eigenvectors.

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Uploaded on 08/19/2009

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Department of Mathematics Fall 2007
MATH 6370, Section 10583 MW 5:30-7:00PM
A. Caboussat
Answers Homework Chapter 8
HOMEWORK FROM THE TEXT
Problem 14 (Chapter 6 - Problem 14) 6 14
Following the steps and notations in class, the simple vector iteration methods leads to
yk=
n
X
i=1
ρiλk
ixi, Ayk=
n
X
i=1
ρiλk+1
ixi.
Therefore,
qkj =Pn
i=1 ρiλk+1
i(xi)j
Pn
i=1 ρiλk
i(xi)j
=
λ1ρ1(x1)j+Pn
i=2 ρiλi
λ1k+1
(xi)j
ρ1(x1)j+Pn
i=2 ρiλi
λ1k
(xi)j
.
In a first approximation, the term leading the error will be, when (x1)j6= 0,
qkj =λ1
ρ1(x1)j+Pn
i=2 ρiλi
λ1k+1
(xi)j
ρ1(x1)j
+ higher order terms,
meaning:
qkj =λ1 ρ1(x1)j+ρiλ2
λ1k+1
(xi)j!+ higher order terms .
Similarly:
rk=yT
kAyk
yT
kyk
=yT
kPn
i=1 ρiλk+1
ixi
Pn
i=1 ρ2
iλ2k
i|xi|2=Pn
i=1 ρ2
iλ2k+1
i|xi|2
Pn
i=1 ρ2
iλ2k
i|xi|2
and
rk=λ1
ρ1|x1|2+Pn
i=2 ρ2
iλi
λ12k+1
|xi|2
ρ1|x1|2Pn
i=2 ρ2
iλi
λ12k
|xi|2
=λ1 1 + ρ2
2λ2
λ12k+1
|x2|2+ higher order terms !.
Problem 22 (Chapter 6 - Problem 22) 6 22
The QR decomposition of the matrix
A=2ε
ε1
actually consists in one single multiplication oby a Householder projector matrix. The first step therefore reads
(by using the notations of the book):
P A =R=k r1
0r2
pf2

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Department of Mathematics Fall 2007 MATH 6370, Section 10583 MW 5:30-7:00PM A. Caboussat

Answers Homework Chapter 8

HOMEWORK FROM THE TEXT

Problem 14 (Chapter 6 - Problem 14) 6 14

Following the steps and notations in class, the simple vector iteration methods leads to

yk =

∑^ n

i=

ρiλki xi, Ayk =

∑^ n

i=

ρiλk i +1xi.

Therefore,

qkj =

∑n i=1 ρiλ

k+ ∑^ i^ (xi)j n i=1 ρiλ k i (xi)j

λ 1

ρ 1 (x 1 )j +

∑n i=2 ρi

λi λ 1

)k+ (xi)j

ρ 1 (x 1 )j +

∑n i=2 ρi

λi λ 1

)k (xi)j

In a first approximation, the term leading the error will be, when (x 1 )j 6 = 0,

qkj = λ 1

ρ 1 (x 1 )j +

∑n i=2 ρi

λi λ 1

)k+ (xi)j ρ 1 (x 1 )j

  • higher order terms,

meaning:

qkj = λ 1

ρ 1 (x 1 )j + ρi

λ 2 λ 1

)k+ (xi)j

  • higher order terms.

Similarly:

rk = yTk Ayk ykT yk

yTk

(∑n i=1 ρiλ

k+ i xi

∑n i=1 ρ

2 i λ

2 k i |xi|

∑n i=1 ρ 2 i λ

2 k+ i |xi|

2 ∑n i=1 ρ

2 i λ

2 k i |xi|

2

and

rk = λ 1

ρ 1 |x 1 |^2 +

∑n i=2 ρ 2 i

λi λ 1

) 2 k+ |xi|^2

ρ 1 |x 1 | 2 ∑n i=2 ρ

2 i

λi λ 1

) 2 k |xi| 2

= λ 1

1 + ρ^22

λ 2 λ 1

) 2 k+ |x 2 |^2 + higher order terms

Problem 22 (Chapter 6 - Problem 22) 6 22

The QR decomposition of the matrix

A =

2 ε ε 1

actually consists in one single multiplication oby a Householder projector matrix. The first step therefore reads (by using the notations of the book):

P A = R =

k r 1 0 r 2

with k = −

4 + ε^2 , P = I − βuuT^ , u = (2 +

4 + ε^2 , ε)T^ and β =

[√

4 + ε

4 + ε^2 + 2

)]− 1

Therefore

P = I − βuuT^ =

1 − β(2 +

4 + ε^2 )^2 −β(2 +

4 + ε^2 )ε −β(2 +

4 + ε^2 )ε 1 βε^2

Simplifications and multiplications of (ε, 1)T^ by P allows to obtain (r 1 , r 2 ). Q is the inverse of P.

ADDITIONAL HOMEWORK

Problem 1 49

Suppose that λ is an eigenvalue of A with associated eigenvector x, such that ||x||∞ = 1. Since Ax = λx, we have

∑^ n

j=

aij xj = λxj , i = 1,... , n.

If k is the integer such that |xk| = ||x||∞ = 1. Therefore

∑^ n

j=

akj xj = λxk

and

∑^ n

j=1,j 6 =k

akj xj = (λ − akk )xk.

Therefore:

|λ − akk| = |λ − akk| |xk| 6

∑^ n

j=1,j 6 =k

|akj | |xj | 6

∑^ n

j=1,j 6 =k

|akj | |xj | ,

since |xj | 6 |xk| = 1. Thus λ ∈ R.