Numerical Analysis 323

7/29/2019 Numerical Analysis 323

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Computational Linear Algebra Syllabus

NUMERICAL ANALYSIS

Linear Systems of Equations and Matrix Computations

Module1: Direct methods for solving linear system of equationSimple Gaussian elimination method, gauss elimination method with partial pivoting,determinant evaluation, gauss Jordan method, L U decompositions Doolittles lu

decomposition, Doolittles method with row interchange.

Module 2: Iterative methods for solving linear systems of equations

Iterative methods for the solution of systems equation, Jacobin iteration,gauss seidel

method, successive over relaxation method (sort method).

Module 3: Eigenvalues and Eigenvectors

An introduction, eigenvalues and eigenvectors, similar matrices,hermitian matrices,

gramm Schmidt orthonormalization,vector and matrix norms.Module 4: Computations of eigenvaues

Computation of eigenvalues of a real symmetric matrix, determination of the eigenvaluesof a real symmetric tridiagonal matrix, tridiagonalization of a real symmetric matrix,

Jacobin iteration for finding eigenvalues of a real symmetric matrix, the q r

decomposition, the Q-R algorithm.

Vittal Rao/IISc, Bangalore V1/1-4-04/1


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Lecture Plan

Modules Learning Units Hours perTopics

TotalHours

1. Simple Gaussian elimination method 1

2. Gauss elimination method with partial

pivoting.

2

3. Determinant evaluation 1

4. Gauss Jordan method 1

5. L U decompositions 1

6. Doolittles LU Decomposition 2

1. Direct

methods for

solving linearsystem of

equation.

7. Doolittles method with row interchange. 2

10

8. Iterative methods for the solution ofsystems equation

3

9.Jacobi iteration. 210.Gauss Seidel method 2

2. Iterativemethods for

solving linearsystems of

equations.

11.Successive over relaxation method (sortmethod).

2

9

12.An introduction. 113.Eigenvalues and eigenvectors, 214.Similar matrices, 215.Hermitian matrices. 116.Gramm Schmidt orthonormalization, 2

3. Eigenvaluesand Eigenvectors

17.Vector and matrix norms. 1

9

18.Computation of eigenvalues 219.Computation of eigenvalues of a real

symmetric matrix.2

20.Determination of the eigenvalues of a realsymmetric tridiagonal matrix,

2

21.Tridiagonalization of a real symmetricmatrix

1

22.Jacobian iteration for finding eigenvaluesof a real symmetric matrix

1

4. Computations

of eigenvalues.

23.The Q R decomposition 1

11



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24.The Q-R algorithm. 2



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Numerical Analysis/ Direct methods for solving linear Lecture Notes

system of equation

1.DIRECT METHODS FOR SOLVINGLINEAR SYSTEMS OF EQUATIONS

1.1.SIMPLE GAUSSIAN ELIMINATION METHODConsider a system of n equations in n unknowns,

a11x1 + a12x2 + . + a1nxn = y1

a21x1 + a22x2 + . + a2nxn = y2

an1x1 + an2x2 + . + annxn = yn

We shall assume that this system has a unique solution and proceed to describe thesimple Gaussian elimination method for finding the solution. The method reduces thesystem to an upper triangular system using elementary row operations (ERO).

Let A(1) denote the coefficient matrix A.

A(1) = Where a

nnnn

n

n

aaa

aaa

aaa

)1(2

)1(1

)1(

2)1(

22)1(

21)1(

1)1(

12)1(

11)1(

......

.......................

......................

.....

.....

(1)ij = aij

Let

y(1) = Where y

(1 )

1

(1 )

2

(1 )

n

y

y

y

M (1)

i = yi

We assume a(1)11 0

Then by ERO of type applied to A(1)

reduce all entries below a(1)

11 to zero. Let the

resulting matrix be denoted by A

(2)

.

A(1)

+ 11

)1(RmR ii

A(2)

Where ;11

)1(

1)1(

1)1(

a

am

ii = i > 1.

Note A(2)

is of the form

VittalRao/IISc, Bangalore M1/L1and L2/V1/May2004/1


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system of equation

M1/L1and L2/V1/May2004/2

A(2) =

nnn

n

n

n

aa

aa

aa

aaa

)2(2

)2(

3

)2(

32

)2(

2)2(

22)2(

1)1(

12)1(

11)1(

......0

.....................0

......0

......

Notice that the above row operations on A(1) can be effected by premultiplying A(1) by

M(1)

where

M(1) =

)1(

1

1

)1(

31

)1(

21

00001

n

n

m

Im

m

M

i.e.

M(1)

A(1)

= A(2)

Let

y(2) = M(1) y(1) i.e)2()1( 11 yy

RmR ii +

Then the system Ax = y is equivalent to

A(2)x = y(2)

Next we assume

a(2)22 0

and reduce all entries below this to zero by ERO

A(2) + 2

)2(ii mR

A(3) ; ;22

)2(

2)2(

2)2(

a

am

ii = i > 2.

Here

1 0 0 . . . 0

0 1 0 . . . 0

0 m(2)32

VittalRao/IISc, Bangalore

M(2) = 0 m(2)42 In-2


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system of equation

.

0 m(2)n2

and M(2) A(2) = A(3) ; M(2) y(2) = y(3) ;

and A(3) is of the form

=

nnn

n

n

n

aa

aa

aaa

aaa

A

)3(3

)3(

3)3(

33)3(

2)2(

23)2(

22)2(

1)1(

12)1(

11)1(

)3(

...00

...

...00

...0

......

MMMM

We next assume a

(3)

33

0 and proceed to make entries below this as zero. We thus getM(1)

, M(2)

, . , M(r)

where

1 0 . 0

0 1 . 0

0 1

m(r)

r+1r

M(r) = rxr m(r)r+2r In-r

.

.

.

m(r)nr

==

++

+

++

+++

+

nnr

nrr

nrr

rrr

rnr

rrr

n

n

rrr

aa

aa

aa

aa

aa

AAM

)1(1

)1(

1)1(

11)1(

)()(

2)2(

22)2(

1)1(

11)1(

)1()()(

...000

.........

...0

......0

.........0

............

MMM

MM

M



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system of equation

M(r) y(r) = y(r+1)

At each stage we assume a(r)rr 0.

Proceeding thus we get,

M(1), M(2), . , M(n-1) such that

M(n-1) M(n-2) . M(1) A(1) = A(n) ; M(n-1) M(n-2) . M(1) y(1) = y(n)

a(1)11 a(1)

12 . . . . . a(1)

1n

a(2)22 . . . . . a(2)

2n

where A(n) = .

.

a(n)nn

which is an upper triangular matrix and the given system is equivalent to

A(n)x = y(n)

and since this is an upper triangular, this can be solved by backward substitution; andhence the system can be solved easily

Note further that each M(r) is a lower triangular matrix with all diagonal entries as 1.

Thus let M(r) is 1 for every r. Now,

A(n) = M(n-1) . M(1) A(1)

Thus

det A(n)

= det M(n-1)

det M(n-2)

. det M(1)

det A(1)

det A(n) = det A(1) = det A since A = A(1)

Now A(n) is an upper triangular matrix and hence its determinant is

a(1)11 a(2)

22. a(n)nn. Thus det A is given by

det A = a(1)11 a(2)

22. a(n)nn

Thus the simple GEM can be used to solve the system Ax = y and also to evaluate det A

provided a(i)ii 0 for each i.

Further note that M(1), M(2), . , M(n-1) are lower triangular, and nonsingular as their det =

1 0. They are all therefore invertible and their inverses are all lower triangular, i.e. if L= M(n-1) M(n-2) . M(1) then L is lower triangular, and nonsingular and L-1 is also lower

triangular.

Now LA = LA(1) = M(n-1) M(n-2) . M(1) A(1) = A(n)



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system of equation

Therefore A = L(-1) A(n)

Now L(-1) is lower triangular which we denote by and A(n) is upper triangular which wedenote by u, and we thus get the so called u decomposition

A = u

of a given matrix A as a product of a lower triangular matrix with an upper triangularmatrix. This is another application of the simple GEM. REMEMBER IF AT ANY

STAGE WE GET a(1)ii = 0 WE CANNOT PROCEED FURTHER WITH THE SIMPLE

GSM.

EXAMPLE:

Consider the system

x1 + x2 + 2x3 = 4

2x1 - x2 + x3 = 2

x1 + 2x2 = 3

Here

=

021

112

211

A

=

3

2

4

y

)2(2

)1(

210

330

211

021

112

21112

13

AARR

RR

=

=

a(1)11 = 1 0 m(1)

21 = -2 a(2)

22 = -3 0

m(1)31 = -1

=

101

012

001)1(

M )1()1()2()1(

1

6

4

3

2

4

yMyy ==

=



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system of equation

)3(3

1

)2(

210

330

21123

AA

RR

=

+

a(3)

33 = -3

M(2)31 = 1/3

=

13

10

010

001)2(

M

==

3

6

4)2()2()3(

yMy

Therefore the given system is equivalent to A(3)x = y(3)

x1 + x2 + 2x3 = 4

-3x2 - 3x3 = -6

- 3x3 = -3

Backward Substitution

x3 = 1

-3x2 - 3 = - 6 -3x2 = -3 x2 = 1

x1 + 1 + 2 = 4 x1 = 1

Thus the solution of the given system is,

=

=

1

1

1

3

2

1

x

x

x

x

The determinant of the given matrix A is

a(1)11 a(2)

22 a(3)

33 = (1) (-3) (-3) = 9.

Now

=

101

012

001)1(

1M

=

13

10

010

001)1(

2M



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system of equation

L = M(2) M(-1)

L-1 = (M(2) M(1))-1 = (M(1))-1 (M(2))-1

=

101

012001

13

10

010001

L = L(-1) =

13

11

012

001

u = A(n) = A(3) =

300

330

211

Therefore A = lu i.e.,

021

112

211

=

13

11

012

001

300

330

211

is the lu decomposition of the given matrix.

We observed that in order to apply simple GEM we need a(r)rr 0 for each stage r. Thismay not be satisfied always. So we have to modify the simple GEM in order to

overcome this situation. Further, even if the condition a(r)rr 0 is satisfied at each stage,simple GEM may not be a very accurate method to use. What do we mean by this?

Consider, as an example, the following system:

(0.000003) x1 + (0.213472) x2 + (0.332147) x3 = 0.235262

(0.215512) x1 + (0.375623) x2 + (0.476625) x3 = 0.127653

(0.173257) x1 + (0.663257) x2 + (0.625675) x3 = 0.285321

Let us do the computations to 6 significant digits.

Here,

A(1) =

625675.0663257.0173257.0

476625.0375623.0215512.0

332147.0213472.0000003.0



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system of equation

y(1) = a

285321.0

127653.0

235262.0(1)

11 = 0.000003 0

3.71837000003.0

215512.0

11)1(

21)1(

21)1( ===

a

aM

3.57752000003.0

173257.0

11)1(

31)1(

31)1( ===

a

aM

M(1) = ; y

103.57752

013.71837

001(2) = M(1) y(1) =

6.13586

5.16900

235262.0

A(2) = M(1) A(1) =

7.191818.123270

0.238609.153340

332147.0213472.0000003.0

a(2)22 = - 15334.9 0

803905.09.15334

8.12327

22)2(

32)2(

32)2( =

==

a

aM

M(2) =

1803905.00

010

001

y(3) = M(2) y(2) =

20000.0

5.16900

235262.0

A(3) = M(2) A(2) =

50000.000

0.238609.153340

332147.0213472.0000003.0

Thus the given system is equivalent to the upper triangular system

A(3)x = y(3)



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system of equation

Back substitution yields,

x1 = 0.40 00 00

x2 = 0.47 97 23

x3 = -1.33 33 3

This compares poorly with the correct answers (to 10 digits) given by

x1 = 0.67 41 21 46 9

x2 = 0.05 32 03 93 39.1

x3 = -0.99 12 89 42 52

Thus we see that the simple Gaussian Elimination method needs modification in order to

handle the situations that may lead to a(r)rr = 0 for some r or situations as arising in the

above example. In order to do this we introduce the idea of Partial Pivoting. The idea ofpartial pivoting is the following:

At the r th stage we shall be trying to reduce all the entries below the r th diagonal as zero.

Before we do this we look at the entries in the r th diagonal and below it and then pick

the one that has the largest absolute value and we bring it to the rth

diagonal position bya row interchange, and then reduce the entries below the r th diagonal as zero. When we

incorporate this idea at each stage of the Gaussian elimination process we get the GAUSSELIMINATION METHOD WITH PARTIAL PIVOTING. We now illustrate this with afew examples:

Example:

x1 + x2 + 2 x3 = 4

2x1 x2 + x3 = 2

x1 + 2x2 = 3

We have

Aavg =

3

2

4

021

112

211

1st Stage: The pivot has to be chosen as 2 as this is the largest absolute valued entry in the

first column. Therefore we do



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system of equation

Aavg 12R

3

4

2

021

211

112

Therefore we have

M(1) = and M

100

001

010

(1) A(1) = A(2) =

021

211

112

M(1) A(1) = y(2) =

3

4

2

Next we have

R2 R1 2 -1 1 2

A(2)avg 0 3/2 3/2 3

R3 R1 0 5/2 -1/2 3

Here

M(2) =

102

1

012

1001

; M(2)

A(2)

= A(3)

=

21

250

23

230

112

M2 y(2) = y(3) =

2

3

2

Now at the next stage the pivot is 2

5

since this is the entry with the largest absolute valuein the 1st column of the next sub matrix. So we have to do another row interchange.

Therefore

2 -1 1 2

A(3)avg 0 5/2 -1/2 2 23R



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system of equation

0 3/2 3/2 3

M(3) = M

010

100

001(3) A(3) = A(4) =

23

230

2

1

2

50

112

M(3) y(3) = y(4) =

3

2

2

Next we have

2 -1 1 2

A(4)

avg 23

5

3RR

0 5/2 -1/2 2

0 0 9/5 9/5

Here

M(4)

=

15

30

010

001

M(4) A(4) = A(5) =

5

900

2

1

2

50

112

M(4) y(4) = y(5) =

59

2

2

This completes the reduction and we have that the given system is equivalent to the

system

A(5)x = y(5)

i.e.

2x1 x2 + x3 = 2

2

5x2 -

2

1x3 = 2

5

9x3 =

5

9

We now get the solution by back substitution:



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system of equation

The 3rd equation gives,

x3 = 1

using this in second equation we get

2

5x2 -

2

1= 2

giving2

5x2 =

2

5

and hence x2 = 1.

Using the values of x1 and x2 in the first equation we get

2x1 1 + 1 = 2 giving x1 = 1

Thus we get the solution of the system as x1 = 1, x2 = 1, x3 = 1; the same as we had

obtained with the simple Gaussian elimination method earlier.

Example 2:

Let us now apply the Gaussian elimination method with partial pivoting to the following

example:

(0.000003)x1 + (0.213472)x2 + (0.332147) x3 = 0.235262

(0.215512)x1 + (0.375623)x2 + (0.476625) x3 = 0.127653

(0.173257)x1 + (0.663257)x2 + (0.625675) x3 = 0.285321,

the system to which we had earlier applied the simple GEM and had obtained solutions

which were for away from the correct solutions.

Note that

A =

625675.0663257.0173257.0

476625.0375623.0215512.0

332147.0213472.0000003.0

y =

285321.0

127653.0

235262.0

We observe that at the first stage we must choose 0.215512 as the pivot. So we have

A(1) = A A 12R (2) =

625675.0663257.0173257.0

332147.0213472.0000003.0

476625.0375623.0215512.0



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system of equation

y(1) = y y 12R (2) = M

285321.0

235262.0

127653.0(1) =

100

001

010

Next stage we make all entries below 1st diagonal as zero

R2 + m21R1 0.215512 0.375623 0.476625

A(2) A(3) 0 0.213467 0.332140

R3 + m21R1 0 0.361282 0.242501

Where

m21 = -11

21

a

a= -

215512.0

000003.0= - 0.000014

m31 = -11

31

a

a= -

215512.0

173257.0= - 0.803932

M(2) = ; y

10803932.0

01000014.0

001

(2) = M(2) y(2) =

182697.0

235260.0

127653.0

In the next stage we observe that we must choose 0.361282 as the pivot. Thus we have to

interchange 2nd and 3rd row. We get,

A(3) A 23R (4) =

332140.0213467.00

242501.0361282.00

476625.0375623.0215512.0

M(3) = y

010

100

001

(4) = M(3) y(3) =

235260.0

182697.0

127653.0

Now reduce the entry below 2nd diagonal as zero

A(4) A + 2323 RMR 5 =

188856.000

242501.0361282.00

476625.0375623.0215512.0



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system of equation

M32 = -361282.0

213467.0= - 0.590860

M(4) = y

159086.0

010

001

(5) = M(4) y(4) =

127312.0

182697.0

127653.0

Thus the given system is equivalent to

A(5) x = y(5)

which is an upper triangular system and can be solved by back substitution to get

x3 = 0.674122

x2 = 0.053205 ,

x1 = 0.991291which compares well with the 10 decimal accurate solution given at the end of page 9.

Notice that while we got very bad errors in the solutions while using simple GEMwhereas we have come around this difficulty by using partial pivoting.



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Numerical analysis/Direct methods for solving linear system of equation Lecture notes

VittalRao/IISc, Bangalore M1/L3/V1/May 2004/1

DETERMINANT EVALUATION

Notice that even in the partial pivoting method we get Matrices

M(k), M(k-1) . M(1) such that

M(k), M(k-1) . M(1) A is upper triangular and therefore

det M(k), det M(k-1) . det M(1) det A = Product of the diagonal entries in the final upper

triangular matrix.

Now det M(i) = 1 if it refers to the process of nullifying entries below a diagonal to zero;

and

det M(i) = 1 if it refers to a row interchange necessary for a partial pivoting.

Therefore det M(k) . det M(1) = (-1)m where m is the number of row inverses effected in

the reduction.

Therefore det A = (-1)

m

product of the diagonals in the final upper triangular matrix.In our example 1 above, we had M(1), M(2), M(3), M(4) of which M(1) and M(3) referred to

row interchanges. Thus therefore there were to row interchanges and hence

det A = (-1)2 (2)(2

5)(

5

9) = 9.

In example 2 also we had M(1), M(3) as row interchange matrices and

therefore det A = (-1)2 (0.215512) (0.361282) (0.188856) = 0.013608

LU decomposition:

Notice that the M matrices corresponding to row interchanges are no longer lower

triangular. (See M(1) & M(3) in the two examples.) Thus,

M(k) M(k-1) . . . . . M(1)

is not a lower triangular matrix in general and hence using partial pivoting we cannot get

LU decomposition in general.


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Numerical Analysis/Direct methods for solving linear system of equation Lecture notes


GAUSS JORDAN METHOD

This is just the method of reducing Aavg to (AR / yR ) where AR = In is the Row Reduced

Echelon Form of A (in the case A is nonsingular). We could also do the reduction here by

partial pivoting.

Remark:

In case in the reduction process at some stage if we get arr = ar+1r = . . . . = ar+1n = 0, then

even partial pivoting does not being any nonzero entry to rth

diagonal because there is no

nonzero entry available. In such a case A is singular matrix and we proceed to the RRE

form to get the general solution of the system. As observed earlier, in the case A is

singular, Gauss-Jordan Method leads to AR = In and the product of corresponding M(i)

give us A-1

.


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Numerical Analysis / Direct methods for solving linear system of equation Lecture notes


LU decompositions

We shall now consider the LU decomposition of matrices. Suppose A is an nxn matrix.

If L and U are lower and upper triangular nxn matrices respectively such that A = LU.

We say that this is a LU decomposition of A. Note that LU decomposition is not unique.

For example if A = LU is a decomposition then A = L U is also a LU decompositionwhere 0 is any scalar and L = L and U = 1/ U.

Suppose we have a LU decomposition A = LU. Then, the system, Ax = y, can be solved

as follows:

Set Ux = z (1)

Then the system Ax = y can be written as,

LUx = y,

i.e.,

Lz = y ..(2)Now (2) is a triangular system infact lower triangular and hence we can solve it by

forward substitution to get z.

Substituting this z in (1) we get an upper triangular system for x and this can be solved by

back substitution.

Further if A = LU is a LU decomposition then det. A can be calculated as det. A = det. L .det. U = l11 l22 .lnn u11u22 ..unn

Where lii are the diagonal entries of L and u ii are the diagonal entries of U.

Also A-1 can be obtained from an LU decomposition as A-1 = U-1 L-1.

Thus an LU decomposition helps to break a system into Triangular system; to find the

determinant; and to find the inverse of a matrix.

We shall now give methods to find LU decomposition of a matrix. Basically, we shall beconsidering three cases. First, we shall consider the decomposition Tridiagonal matrix;

secondly the Doolittless method for a general matrix, and thirdly the Choleskys method

for a symmetric matrix.


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1 TRIDIAGONAL MATRIX

Let

=

bc

abc

abc

abc

ab

A

i

iii

1

12

432

321

21

0........0

0....0

........................

........................

0....0

0....0

0....00

be an nxn tridiagonal matrix. We seek a LU decomposition for this. First we shall givesome preliminaries.

Let i denote the determinant of the ith

principal minor of A

ii

ii

i

bc

abc

abc

ab

1

112

321

21

0....0

....0

....................

....................

0....

0....0

=

Expanding by the last row we get,

i = bii-1 ci-1 aii-2 ; I = 2,3,4, ..

..(I)

i = b1

We define i = 1

From (I) assuming that i are all nonzero we get

1

21

1

=i

i

iii

i

iacb

setting ii

ik=

1

this can be written as

).(..............................1

1 IIk

ackb

i

iiii

+=


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Now we seek a decomposition of the form A = LU where,

=

1........00

........................

0....010

0........01

0............01

1

2

1

nw

w

w

L ;

=

n

nn

u

u

u

u

U

0....00

....00

....................

0....0

0....0

1

32

21

i.e. we need the lower triangular and upper triangular parts also to be tridiagonal

triangular.

Note that if A = (Aij) then because A is tridiagonal, Aij is nonzero only when i and j differby 1. i.e. only Ai-1i, Aii, Aii+1 are nonzero. In fact,

Ai-1i = ai

Aii = bi .. (III)

Ai+1i = ci

In the case of L and U we have

Li + 1i = wi

Lii = 1 .. (IV)

Lij = 0 if j>i or jj or i


23/161



Therefore

iiiiiiULa == 111

Therefore

i = ai .. (VII)

This straight away gives us the off diagonal entries of U. From (VI) we also get

=

=n

k

kiikii ULA1

iiiiiiiiULUL += 11

Therefore

)......(..........1 VIIIuwb iiii +=

From (VI) we get further,

=

++ =n

k

kikiii ULA1

11

iiiiiiii ULUL 1111 ++++ +=

iii uWc =

Thus (IX)iii uWc =

Using (IX) in (VIII) we get (also using I = ai)

i

i

ii

i uu

acb +=

1

1

Therefore

)..(....................1

1 Xu

acub

i

ii

ii

+=

Comparing (X) with (II) we get

).(....................1

XIkui

iii

==


24/161



using this in (IX) we get

).......(....................1 XIIc

u

cw

i

ii

i

i

i

==

From (VII) we get

I = ai .(XIII)

(XI), (XII) and (XIII) completely determine the matrices L and U and hence we get the

LU decomposition.

Note : We can apply this method only when I are all nonzero. i.e. all the principalminors have nonzero determinant.

Example:

Let

=

13000

13900

02520

00112

00022

A

Let us now find the LU decomposition as above.

We have

b1 = 2 b2 = 1 b3 = 5 b4 = -3 b5 = -1

c1 = -2 c2 = -2 c3 = 9 c4 = 3a2 = -2 a3 = 1 a4 = -2 a5 = 1

We have

0 = 1

1 = 2

2 = b2 1 a2 c10 = 2-4 = -2

3 = b32 a3 c21 = (-10) (-2) (2) = -6

4 = b43 a4 c3 2= (-3) (-6) (-18) (-2) = -18

5 = b54 a5 c43 = (-1) (-18) (3) (-6)= 36.

Note 1,2,3,4,5 are all nonzero. So we can apply the above method.

Therefore by (XI) we get


25/161


M1/L5/V1/May 2004/6

32

6;1

2

2;2

2

33

1

22

0

11 =

===

====

uuu

36

183

44 =

==

u ; and 2

1836

4

55 ===

u

From (XII) we get From (XIII) we get

12

2

1

11 =

==

u

cw 222 == a

2

1

2

2

22 =

==

u

cw 133 == a

33

9

3

33 ===

u

cw 244 == a

13

3

4

44 ===

u

cw 155 == a

Thus;

=

11000

01300

00120

00011

00001

L ;

=

20000

13000

02300

00110

00022

U

In the above method we had made all the diagonal entries of L as 1. This will facilitate

solving the triangular system LZ = y (equation (2)) in page 17. However by choosing

these diagonals as 1 it may be that the u i, the diagonal entries in U are small thus creatingproblems in backward substitution for the system Ux = z (equation (1) on page 17). In

order to avoid this situation Wilkinson suggests that in any triangular decomposition

choose the diagonal entries of L and U to be of the same magnitude. This can beachieved as follows:

We seek

A = LU

where

VittalRao/IISc, Bangalore


26/161



l1

L = w1 l2

;

wn-1ln

=

n

nn

u

u

u

u

U

0....00

....00

....................

........0

0....0

1

32

21

Lii = li

Now Li+1i = wi

Lij = 0 if j>i or jj; or j>i and j-i 2

Now (VII), (VIII) and (IX) change as follows:

iii Aa 1=

ki

n

kki UL

= 11 iiii UL 111

= iil

1

=

Therefore

ai = li-1I. (VII`)

iii Ab = kin

k

ikUL

=1

iiiiiiiiULUL += 11 iiii ulW += 1

`).......(..........1 VIIIulWb iiiii +=

iii Ac 1+= kin

k

ki UL += 1 1 iiii UL 1+= iiuw=

iii uwc = .. (IX`)From (VIII`) we get using (VII`) and (IX`)


27/161



ii

i

i

i

i

i ull

a

u

cb +=

11

1 .

ii

ii

ii ululca +=

11

1

`)..(....................1

1 Xpp

cab i

i

ii

i +=

where

pi = li ui

Comparing (X`) with (II) we get

1

==i

iii kp

therefore

1

=i

iiiul

we choose1

=i

iil

. (XIV)

=

1

sgni

iiu

1i

i

(XV)

Thus li and ui have same magnitude. These then can be need to get wi and i from (VII`)and (IX`). We get finally,

1

=i

iil

;

11

.sgn

=

i

i

i

i

iu

. . . . . . . . . . . .(XI`)

i

ii u

Cw = . . . . . . . . . . . . .. . . . . . . . . (XII`)

1

=i

ii l

a . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . .(XIII`)

These are the generalizations of formulae (XI), (XII) and (XIII).

Let us apply this to our example matrix (on page 21).


28/161



We get;

0 = 1 1 = 2 2 = -2 3 = -6 4 = -18 5 = 36

b1 = 2 b2 = 1 b3 = 5 b4 = -3 b5 = -1

c1 = -2 c2 = -2 c3 = 9 c4 = 3

a1 = -2 a3 = 1 a4 = -2 a5 = 1

We get 1/0 = 2 ; 2/1 = -1 ; 3/2 = 3 ; 4/3 = 3 ; 5/4 = -2

Thus from (XI`) we get

l1 = 2 u1 = 2

l2 = 1 u2 = -1

l3 = 3 u3 = 3

l4 = 3 u4 = 3

l5 = 2 u5 = -2

From (XII`) we get

22

2

1

11 =

==

u

Cw ; 2

1

2

2

22 =

==

u

Cw ;

333

9

3

33 ===

u

Cw ; 3

3

3

4

44 ===

u

Cw

From (XIII`) we get

22

2

1

22 =

==

l

a ; 1

1

1

2

33 ===

l

a ;

3

2

3

44

==

l

a ;

3

1

4

55 ==

l

a

Thus, we have LU decomposition,


29/161



=

=

20000

3

13000

03

2300

00110

00022

23000

033300

00320

00012

00002

13000

13900

02520

00112

00022

A

L U

in which the L and U have corresponding diagonal elements having the same magnitude.


30/161



DOOLITTLES LU DECOMPOSITION

We shall now consider the LU decomposition of a general matrix. The method we

describe is due to Doolittle.

Let A = (aij). We seek as in the case of a tridiagonal matrix, an LU decomposition inwhich the diagonal entries lii of L are all 1. Let L = (lii) ; U = (uij). Since L is a lower

triangular matrix, we have

lij = 0 if j > i ; and by our choice, lij =1.

Similarly, since U is an upper triangular matrix, we have

uij = 0 if i > j.

We determine L and U as follows : The 1st

row of U and 1st

column of L are determinedas follows :

=

=n

k

kkula1

1111

= l11 u11 Since l1k= 0 for k>1

= u11 Since l11 = 1.

.111 =u In general,

=

=n

k

kjkj ula1

11

= l11 u11 Since l1k= 0 for k>1

= u1j Since l11 = 1.


31/161



u1j = a1j . . . . . . . . . (I)Thus the first row of U is the same as the first row of A. The first column of L is

determined as follows:

=

=n

k

kjkj ula1

11

= lj1 u11 Since uk1 = 0 if k>1

lj1 = aj1/u11 . . . . . . . . . (II)Note : u11 is already obtained from (I).

Thus (I) and (II) determine respectively the first row of U and first column of L. The

other rows of U and columns of L are determined recursively as given below: Suppose

we have determined the first i-1 rows of U and the first i-1 columns of L. Now weproceed to describe how one then determines the i th row of U and ith column of L. Since

first i-1 rows of U have been determined, this means, ukj ; are all known for 1 k i-1 ; 1

j

n. Similarly, since first i-1 columns are known for L, this means, lik are all knownfor 1 i n ; 1 k i-1.

Now

=

=n

k

kjikij ula1

Since l==

i

kkjik ul1 ik= 0 for k>i

ijii

i

k

kjik ulul +=

=

1

1


32/161



since lij

i

k

kjik uul +=

=

1

1ii = 1.

=

=1

1

i

k

kjikijij ulau . . . . . ... . . . .(III)

Note that on the RHS we have a ij which is known from the given matrix. Also the sum

on the RHS involves lik for 1 k i-1 which are all known because they involve entriesin the first i-1 columns of L ; and they also involve ukj ; 1 k i-1 which are also known

since they involve only the entries in the first i-1 rows of U. Thus (III) determines the i

th

row of U in terms of the known given matrix and quantities determined upto the previous

stage. Now we describe how to get the ith column of L :

=

=n

k

kijkji ula1

Since u==i

k

kijk ul1

ki = 0 if k>i

iiji

i

k

kijk ulul +=

=

1

1

=

=

1

1

1 i

k

kijkji

ii

ji ulau

l..(IV)

Once again we note the RHS involves uii, which has been determined using (III); aij

which is from the given matrix; ljk; 1 k i-1 and hence only entries in the first i-1columns of L; and uki, 1 k i-1 and hence only entries in the first i-1 rows of U. ThusRHS in (IV) is completely known and hence lji, the entries in the i

th column of L are

completely determined by (IV).


33/161



Summarizing, Doolittles procedure is as follows:

lii = 1; 1st

row U = 1st

row of A ; Step 1 determining 1st

row of U and

lj1 = aj1/u11 1st

column of L.

For i 2; we determine

=

=1

1

i

k

kjikijij ulau ; j = i, i+1, i+2, .,n

(Note for j


34/161



1st row of U : Same as 1st row of A.

u11 = 2 ; u12 = 1 ; u13 = -1 ; u14 = 3

1st column of L:

l11 = 1;

l21 = a21/u11 = -2/2 = -1.

l31 = a31/u11 = 4/2 = 2.

l41 = a41/u11 = 6/2 = 3.

Second step:

2nd row of U : u12 = 0 (Because upper triangular)

u22 = a22 l21 u12 = 2 (-1) (1) = 3.

u23 = a23 l21 u13 = 6 (-1) (-1) = 5.

u24 = a24 l21 u14 = - 4 (-1) (3) = -1.

2nd

column of L : l12 = 0 (Because lower triangular)

l22 = 1.

l32 = (a32 l31 u12) /u22

= [14 (2)(1)]/3 = 4.

L42 = (a42 l41 u12) /u22


35/161



= [0 (3)(1)]/3 = -1.

Third Step:

3rd

row of U: u31 = 0 Because upper triangular

u32 = 0

u33 = a33 l31 u13 l32 u23

= 19 (2) (-1) (4)(5) = 1.

u34 = a34 l31 u14 l32 u24

= 4 (2) (3) (4)(-1) = 2.

3rd column of L : l13 = 0 Because lower triangular

l23 = 0

l33 =1

l43 = (a43 l41 u13 l42 u23)/ u33

= [-6 (3) (-1) (-1) (5)]/1

= 2.

Fourth Step:

4th row of U: u41 = 0

u42 = 0 Because upper triangular

u43 = 0

u44 = a44 l41 u14 l42 u24 l43 u34

= 12 (3) (3) (-1) (-1) (2) (2)

= -2.


36/161



4th column of L : l14 = 0 = l24 = l34 Because lower triangular

l44 = 1.

Thus.

=

1213

0142

0011

0001

L ; . . . . . . . . . . .(V)

=

2000

2100

1530

3112

U

and

A = LU.

This gives us the LU decomposition by Doolittles method for the given A.

As we observed in the case of the LU decomposition of a tridiagonal matrix; it isnot advisable to choose the lii as 1; but to choose in such a way that the diagonal entries

of L and the corresponding diagonal entries of U are of the same magnitude. We describe

this procedure as follows:

Once again 1st

row and 1st

column of U & L respectively is our first concern:

Step 1: a11 = l11 u11

Choose ( ) 1111111111 .sgn; aaual ==

Next 1011111

1 >====

forkaslulula kj

n

k

kjkij


37/161



11

1

l

au jij

Thus note that u1j have been scaled now as compared to what we did earlier.

Similarly,

11

11 u

al jj

These determine the first row of U and first column of L. Suppose we have determined

the first I-1 rows of U and first I-1 columns of L. We determine now the ith row of U and

ith

column of L as follows:

=

=n

k

kiikii ula1

for l=

=n

k

kiik ul1

ik= 0 if k>i

ii

i

k

iikiik ulul

=

+=1

1

saypulaul i

i

k

kiikiiiiii ,1

1

==

=

Choose ki

i

k

ikiiiii ulapl

=

==1

1


38/161



iiii ppu sgn=

iforklulula ikkj

i

k

ikkj

n

k

ikij >=== =

011

Q

ijiikj

i

k

ik ulul +=

=

1

1

=

=

kj

i

k

ikijij ulau1

1

lii

determining the ith row of U.

ki

n

k

jkji ula =

=1

iifkuul kiki

i

k

jk >== = 01Q

iijiki

i

k

jk ulul +=

=

1

1

=

=

1

1

i

k

kijkjiji ulal uii ,

thus determining the ith column of L.

Let us now apply this to matrix A in the example in page 30.


39/161



First Step:

2;22 1111111111 ==== ulaul

2

3;

2

1

2

1

11

1414

11

1313

11

1212 ======

l

au

l

au

l

au

2

3;

2

1;

2

1;2 14131211 ==== uuuu

22

2

11

21

21 ===u

al

222

4

11

3131 ===

u

al

232

6

11

4141 ===

ual

therefore

23

22

2

2

41

31

21

11

=

=

=

=

l

l

l

l

Second Step:

1221222222 ulaul =


40/161



( ) 32

122 =

=

3;3 2222 == ul

( )13212323 ulau = l22

( )3

53/2

126 =

=

[ ]14212424 ulau = l22

( ) ( )3

13/2

324 =

=

therefore

3

1;

3

5;3;0 24232221 ==== uuuu

( ) 2212313232 /uulal =

( ) 3/2

12214

=

34=


41/161



( ) 2212414242 /uulal =

( ) 3/2

1230

=

3=

therefore

3

34

3

0

42

32

22

12

=

=

=

=

l

l

l

l

Third Step:

23321331333333 ululaul =

( ) ( )

=

3

534

2

12219

= 1

1;1 3333 == ul

( ) 33243214313434 /lululau =


42/161



( ) ( ) 1/3

134

2

3224

=

= 2

2,1;0;0 34333231 === uuuu

[ ] 33234213414343 /uululal =

( ) ( ) 1/3

532

1236

=

= 2

therefore

=

=

=

=

2

1

0

0

43

33

23

13

l

l

l

l

Fourth Step:

344324421441444444 ulululaul =

( ) ( ) ( )( )223

13

2

32312

=

= -2


43/161



2;2 4444 == ul

2;0;0;0 44434241 ==== uuuu

=

=

=

=

2

0

0

0

44

34

24

14

l

l

l

l

Thus we get the LU decompositions,

=

22323

013422

0032

0002

L ;

=

2000

21003

1

3

530

2

3

2

1

2

12

U

in which iiii ul = , i.e. the corresponding diagonal entries of L and U have the same

magnitude.

Note: Compare this with the L and U of page 32. What is the difference.

The U in page 36 can be obtained from the U of page 32 by

(1)replacing the numbers in the diagonal of that U and keeping the same sign.Thus the first diagonal 2 is replaced by 2 ; 2nd diagonal 3 is replaced by 3 ,

third diagonal1 by 1 and 4th diagonal 2 by - 2 . These then give the diagonals

of the U in page 36.

(2)Divide each entry to the right of a diagonal in the U of page 32 by these replaceddiagonals.


44/161



Thus 1st row changes to 1st row of U in page 36

2nd row changes to 2nd row of U in page 36

3rd row changes to 3rd row of U in page 36

4th

row changes to 4th

row of U in page 36

This gives the U of page 36 from that of page 32.

The L in page 36 can be obtained from the L of page 32 as follows:

(1) Replace the diagonals in L by magnitude of the diagonals in U of page 36.(2) Multiply each entry below the diagonal of L by this new diagonal entry.

We get the L of page 32 changing to the L of page 36.


45/161

Numerical Analysis/ Direct methods for solving linear system of equation Lecture notes


DOOLITTLES METHOD WITH ROW INTERCHANGES

We have seen that Doolittle factorization of a matrix A may fail the moment at stage i we

encounter a uii which is zero. This occurrence corresponds to the occurrence of zeropivot at the ith stage of simple Gaussian elimination method. Just as we avoided this

problem in the Gaussian elimination method by introducing partial pivoting we can adopt

this procedure in the modified Doolittles procedure. The Doolittles method which isused to factorize A as LU is used from the point of view of reducing the system

Ax = y

To two triangular systems

Lz = y

Ux = z

as already mentioned in page 17.

Thus instead of actually looking for a factorization A = LU we shall be looking for a

system,

A*x = y*

and for which A* has LU decomposition.

We illustrate this by the following example: The basic idea is at each stage calculate all

the uii that one can get by the permutation of rows of the matrix and choose that matrix

which gives the max. absolute value for uii.

As an example consider the system

Ax = y

where

=

3213

1451

3222

1213

A y =

1

3

8

3

We want LU decomposition for some matrix that is obtained from A by row interchanges.

We keep lii = 1.Stage 1:

1st diagonal of U. By Doolittle decomposition,

u11 = a11 = 3


46/161



If we interchange 2nd or 3rd or 4th rows with 1st row and then find the u11 for the new

matrix we get respectively u11 = 2 or 1 or 3. Thus interchange of rows does not give anyadvantage at this stage as we have already got 3 without row interchange for u11.

So we keep the matrix as it is and calculate 1st row of U, by Doolittles method.

.13

3;

3

1;

32;1

11

4141

11

3131

11

212111 ========

u

al

u

al

u

all

Thus

L is of the form

***1

01*3

1

0013

20001

; and

U is of the form ; A and Y remaining unchanged.

*000

**00

***0

1213

Stage 2

We now calculate the second diagonal of U: By Doolittles method we have

12212222 ulau = ( )3

81

3

22 =

=

Suppose we interchange 2nd row with 3rd row of A and calculate u22 : our new a22 is 5.

But note that the L gets in the 1st column 2nd and 3rd row interchanged. Therefore new l21

is1/3.

Suppose instead of above we interchange 2nd row with 4th row of A:

New a22 = 1 and new l21 = 1 and therefore new u22 = 1 (1) (1) = 0

Of these 14/3 has largest absolute value. So we prefer this. Therefore we interchange 2ndand 3rd row.

=

=

1

8

33

;

3213

3222

14511213

NewyNewA


47/161



=

=

*000

**00

**3

140

1213

;

1**1

01*3

2

0013

1

0001

NewUNewL

Now we do the Doolittle calculation for this new matrix to get 2nd row of U and 2nd

column of L.

13212323 ulau = ( ) ( )3

102

3

14 =

=

14212424 ulau = ( ) ( )3

21

3

11 =

=

2

nd

column of L:

[ ] 2212313232 uulal = ( ) ( )3

141

3

22

=

7

4=

[ ] 1112414242 uulal = ( )( )[ ]3

14113 = = 0

Therefore new L has form

1*01

017

4

3

2

001

3

10001

New U has form

*000

**003

2

3

10

3

140

1213

This completes the 2nd stage of our computation.

Note: We had three choices of u22to be calculated, namely 8/3, 14/3, 0 before we chose

14/3. It appears that we are doing more work than Doolittle. But this is not really so.For, observe, that the rejected u22 namely 8/3 and 0 when divided by the chosen u22

namely 14/3 give the entries of L below the second diagonal.


48/161



3rd Stage:

3rd diagonal of U:

233213313333 ululau = ( )

=

3

10

7

42

3

22

7

10=

Suppose we interchange 3rd row and 4th row of new A obtained in 2nd stage. We get new

a33 = 2.

But in L also the second column gets 3rd

and 4th

row interchanges

Therefore new l31 = 1 and new l32 = 0

Therefore new u33 = a33 l31 u13 l32 u23 ( )( ) ( )

+=

3

100212 = 4.

Of these two choices of u33 we have 4 has the larges magnitude. So we interchange 3rd

and 4th rows of the matrix of 2nd stage to get

=

=

8

1

3

3

3222

3213

1451

1213

NewYNewA

=

=

*000

*4003

2

3

10

3

140

1213

;

1*7

4

3

20101

0013

10001

NewUNewL

Now for this set up we calculate the 3rd stage entries as in Doolittles method:

243214313434 ululau = ( )( ) ( ) 43

20113 =

=

( ) 33234213414343 uululal =


49/161



( ) 43

10

7

42

3

22

= = 5/14.

=

=

*000

44003

2

3

10

3

140

1213

;

114

5

7

4

3

20101

0013

10001

NewUNewL

Note: The rejected u33 divided by chosen u33 gives l43.

4th Stage

[ ]3443244214414444 ulululau =

( ) ( )414

5

3

2

7

41

3

23

= = 13/7.

==

==

8

1

3

3

3222

3213

1451

1213

** YNewYANewA

New L = L* , New U = U*

,

7

13000

44003

2

3

10

3

140

1213

;

114

5

7

4

3

20101

0013

10001

**

=

= UL

and A* = L*U*

The given system Ax=y is equivalent to the system

A*x=y*

and hence can be split into the triangular systems,

L*z = y*

U*x = z

Now L*z = y* gives by forward substitution:

Z1 =3


50/161



21333

1221 ===+ zzz

411 1331 ===+ zzzz

814

5

7

4

3

24321 =++ zzzz

( ) ( ) ( ) 8414

52

7

43

3

24 =+

+

z

7

52

4 = z

=

7

524

23

z

Therefore U*x = z gives by back-substitution;

7

52

7

134 =x therefore x4 = -4.

311444 434343 ===+=+ xxxxxx

therefore x3 = 3

23

2

3

10

3

14432 = xxx

( ) ( ) 243

23

3

10

3

142 =

x

22 = x

323 4321 =+ xxxx

134623 11 =++ xx

Therefore the solution of the given system is


51/161



=

4

3

2

1

x

Some Remarks:

The factorization of a matrix A as the product of lower and upper triangular matrices is by

no means unique. In fact, the diagonal elements of one or the other factor can be chosen

arbitrarily; all the remaining elements of the upper and lower triangular matrices maythen be uniquely determined as in Doolittles method; which is the case when we choose

all the diagonal entries of L as 1. The name of Crout is often associated with triangular

decomposition methods, and in crouts method the diagonal elements of U are all chosenas unity. Apart from this, there is little distinction, as regards procedure or accuracy,

between the two methods.

As already mentioned, Wilkinsons suggestion is to get a LU decomposition in which

niul iiii = 1; .

We finally look at the cholesky decomposition for a symmetric matrix:

Let A be a symmetric matrix.

Let A = LU be a LU decomposition

Then A1

= U1

L1

U1

is also lower triangular

L1 is upper triangular

Therefore U1L1 is a decomposition of A1 as product of lower and upper triangular

matrices. But A1 = A since A is symmetric.

Therefore LU = U1L1

We ask the question whether we can choose L as U1; so that

A = U1U (or same as LL1)

Now therefore determining U automatically gets L = U1

We now do the Doolittle method for this. Note that it is enough to determine the rows ofU.

Stage 1: 1st row of U:


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==

==n

k

kk

n

k

k uula1

12

1

1

111 11 kk ul =Q 1UL =Q

for k>1 since U is upper triangular112

u= 01 =kuQ

1111 au =

We finally look at the cholesky decomposition for a symmetric matrix:

Let A be a symmetric matrix.

Let A = LU be a LU decomposition

Then A1 = U1 L1 U1 is also lower triangular

L1 is upper triangular

Therefore U1L

1is a decomposition of A

1as product of lower and upper triangular

matrices. But A1

= A since A is symmetric.Therefore LU = U1L1

We ask the question whether we can choose L as U1; so that

A = U1U (or same as LL1)

Now therefore determining U automatically gets L = U1

We now do the Doolittle method for this. Note that it is enough to determine the rows ofU.

Stage 1: 1st row of U:

==

==n

k

kk

n

k

k uula1

12

1

1

111 11 kk ul =Q 1UL =Q

112

u= for k>1 since U is upper triangular.01 =kuQ

1111 au =

= =

==n

k

n

k

kikkiki uuula1 1

111

iuu 111= 101 >= forkukQ

1111 au =

1111 /uau ii = determines first row of U. and hence first column of L.


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Having determined the 1st i-1 rows of U; we determine the ith row of U as follows:

= =

===n

k

n

k

kiikkikiikiiuluula

1 1

2Q

for k > i01

2 == =

ki

i

k

ki uu Q

ii

i

k

ki uu2

1

1

2 +=

=

=

=1

1

22i

k

kiiiii uau

=

=1

1

2i

k

kiiiii uau ; Note: uki are known for k i -1,

1st i-1 rows have already been obtained.

= =

==n

k

n

k

kjkikjikij uuula1 1

Now we need uij for j > i

==i

k

kjkiuu1

Because uki = 0 for k > i

=

+=1

1

i

k

ijiikjki uuuu

Therefore

ii

i

k

kjkiijij uuuau

=

=

1

1


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VittalRao/IISc, Bangalore M1/L7/V1/May

2004/10

=

=

=

=

ij

i

k

kjkiijij

i

k

kiiiii

uuuau

uau

1

1

1

1

2

determines the ith

row of U in terms of the previous rows. Thus we get U and L is U1.

This is called CHOLESKY decomposition.

Example:

Let

=

10131

1331

3351

1111

A

This is a symmetric matrix. Let us find the Cholesky decomposition.

1st row of U

==

==

==

==

1

1

1

1

111414

111313

111212

1111

uau

uau

uau

au

2nd row of U


55/161


VittalRao/IISc, Bangalore M1/L7/V1/May

2004/11

( ) ( )( )( )

( ) ( )( )( )

===

===

===

22113

12113

215

2214122424

2213122323

122

2222

uuuau

uuuau

uau

3rd row of U

( ) ( )( ) ( )( )( )

===

===

2121111

1113

33242314133434

232

132

3333

uuuuuau

uuau

4th row of U

144110342

242

142

4444 === uuuau

=

1000

2100

2120

1111

U

==

1221

0111

0021

0001

1 LU and

A = LU

= LL1

= U1U


56/161

Numerical Analysis / Iterative methods for solving linear systems of equations Lecture notes

ITERATIVE METHODS FOR THE SOLUTION OF SYSTEMS EQUATION

In general an iterative scheme is as follows:

We have an nxn matrix M and we want to get the solution of the systems

x = Mx + y ..(1)

We obtain the solution x as the limit of a sequence of vectors, { }k

x which are obtained as

follows:

We start with any initial vector x(0)

, and calculate x(k)

from,

x(k) = Mx(k-1) + y .(2)

for k = 1,2,3, .. successively.

A necessary and sufficient condition for the sequence of vectors x(k) to converge to

solution x of (1) is that the spectral radius spM

of the iterating matrix M is less than 1 or

if1


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Numerical Analysis / Iterative methods for solving linear systems of equations Lecture notes

)6......(....................

......00

...............

0...00

0......0

0......0

33

22

11

=

nna

a

a

a

D

the diagonal part of A; and

)7....(..............................

0

00

00

000

121

3231

21

=

nnn aaa

aa

a

L

K

KKKKK

K

KK

KK

the lower triangular part of A; and

)8(........................................

0...000

...............

...00

......0

223

112

= n

n

uu

uu

U

the upper triangular part of A.

Note that,

A = D + L + U (9).

We assume that ; i = 1, 2, , n (10)0iia

So that D-1

exists.

We now describe two important iterative schemes, below, for solving the system (3).



58/161

Numerical Analysis/Iterative methods for solving linear system of equation Lecture notes

JACOBI ITERATION

We write the system as in (4) as

11313212111 ..... yxaxaxaxa nn +=

22323121222 ..... yxaxaxaxa nn+=

. . . . . . . .(11)


...... ...... ......

nnnnnnnnn yxaxaxaxa += 112211 ..... We start with an initial vector,

( )

( )

( )

( )

)12........(....................

0

20

10

0

=

nx

x

x

xM

and substitute this vector for x on the RHS of (11) and calculate x1,x2, .., xn and this

vector is called x(1)

. We now substitute this vector on the RHS of (11) to calculate againx1, x2, .., xn and call this new vector as x

(2) and continue this procedure to calculate the

sequence x(k)

. We can describe this briefly as follows:

The equation (11) can be written as,

Dx = - (L + U) x + y . (13)

which we can write as

x = -D-1 (L+U) x +D-1 y,

giving

yJxx += (14)

where

J = -D-1

(L + U) .(15)

and, we get

x(0)

starting vector

,.......2,1;)1()( =+= kyJxx kk .(16)

as the iterating scheme. This is similar to (2) with the iterating matrix M as

J = -D-1 (L + U); J is called the Jacobi Iteration Matrix. The scheme will converge to the

solution x of our system if 11. Thus the Jacobi scheme for this system will not converge.

Thus, in example 3 we had a system for which the Jacobi scheme converged but Gauss

Seidel scheme did not converge; where in example 4 above we have a system for whichthe Jacobi scheme does not converge, but the Gauss Seidel scheme converges. Thus,

these two examples demonstrate that, in general, it is not correct to say that one scheme

is better than the other.

Let us now consider another example.

Example 5:

2x1 x2 =y1

-x1 + 2x2 x3 = y2

-x2 + 2x3 x4 =y3

-x3 + 2x4 = y4



73/161

Numerical Analysis/ Iterative methods for solving linear system of equation Lecture notes

Here

,

2100

1210

0121

0012

=A is a symmetric tridiagonal matrix.

The Jacobi matrix for this scheme is

=

0

2

100

2

10

2

10

02

10

2

1

002

10

J

The characteristic equation is,

164 - 122 + 1 = 0 (CJ)

Set 2 =

Therefore

162 - 12 + 1 = 0 (CJ1)

is the square root of the roots of (CJ1).

Thus the eigenvalues of J are 0.3090; 0.8090.

Hence

8090.0=sp

J ; and the Jacobi scheme will converge.

The Gauss Seidel matrix for the system is found as follows:



74/161


( )

=+

2100

0210

0021

0002

LD

=

0000

1000

0100

0010

U

( )

=+

2

1

4

1

8

1

16

1

021

41

81

002

1

4

1

0002

1

1LD

( )

=+=

0000

1000

0100

0010

21

41

81

161

02

1

4

1

8

1

002

1

4

1

0002

1

1ULDG

=

4

1

8

1

16

10

2

1

4

1

8

10

02

1

4

10

002

10

The characteristic equation of G is

0=GI , which becomes in this case

)(....................01216 234 GC=+



75/161


This can be factored as

( ) 011216 22 =+ Thus the eigenvalues of G are roots of

2 = 0 ; and

162 - 12 + 1 = 0 .(CG1)

Thus one of the eigenvalues of G is 0 (repeated twice), and two eigenvalues of G are

roots of (CG1). Notice that roots of (CG1) are same as those of (CJ1). Thus nonzero

eigenvalues of G are squares of eigenvalues of J. the nonzero eigenvalues of G are,

0.0955, 0.6545.

Thus,

16545.0


76/161

Numerical Analysis ( Iterative methods for solving linear systems of equations) Lecture notes


SUCCESSIVE OVERRELAXATION METHOD (SOR METHOD)

We shall now consider SOR method for the system

Ax = y ..(I)

We take a parameter 0 and multiply both sides of (I) to get an equivalent system,

Ax = y (II)

Now

( )ULDA ++= We write (II) as

(D + L + U)x = y,

i.e.

(D + L) = - Ux + y

i.e.

(D + L)x + (-1) Dx = - Ux +y

i.e.

(D + L)x = - [( 1)D + U]x + y

i.e.

x = - (D + L)-1 [(-1)D + U]x + [D + L]-1y.

We thus get the SOR scheme as

( ) ( )

( ) ;

0

1

=

+=+

x

yxMx kk

initial guess (III)

where,

( ) ( )[ ]uDLDM ++=

11

and

( ) yLDy 1 +=

M is the SOR matrix for the system.


77/161



Notice that if = 1 we get the Gauss Seidel scheme. The strategy is to choose suchthat ,1 0). Usually is chosen

between 1 and 2. Of course, one must analysesp

M as a function of and find that

value 0 of for which this is minimum and work with this value of0.

Let us consider an example of this aspect.

Example 6:

Consider the system given in example 5.

For that system,

M = - (D + L)-1

[(-1) D +U]

+++

++

+

=

23243243

23232

22

4

11

8

1

2

1

2

1

16

1

4

1

4

1

8

1

8

12

1

4

11

8

1

2

1

2

1

4

1

4

1

02

1

4

11

2

1

2

1

002

11

and the characteristic equation is

( ) ( ) ( ) MC.................011211624224 =+++

Thus the eigenvalues of M are roots of the above equation. Now when is = 0 a root?If = 0 we get from (CM),

16(-1)4 = 0 = 1,

i.e. in the Gauss Seidel case. So let us take 1; so = 0 is not a root. So we candivide the above equation (CM) by

42 to get

( ) ( )01

112

116 2

22

2

2

=++

+

Setting


78/161



( )

2

2

2 1+= we get

011216 24 =+

which is the same as (CJ). Thus

.8090.0;3090.0 =

Now

( ) 22

21

=

+= 0.0955 or 0.6545 .(*)

Thus, this can be simplified as

( ) ( )2

1

22221

4

11

2

1

=

as the eigenvalues of M.

With = 1.2 and using the two values of2 in (*) we get,

= 0.4545, 0.0880, -0.1312 i (0.1509).

as the eigenvalues. The modulus of the complex roots is 0.2

Thus

spM when = 1.2 is 0.4545

which is less thatsp

J = 0.8090 andsp

G = 0.6545 computed in Examples. Thus for

this system, SOR with = 1.2 is faster than Jacobi and Gauss Seidel scheme.

We can show that in this example when = 0 = 1.2596, the spectral radius 0M is

smaller than M for any other . We have

2596.1M = 0.2596

Thus the SOR scheme with = 1.2596 will be the method which converges fastest.

Note:

We hadsp

M 2.1 = 0.4545


79/161



And

spM 2596.1 = 0.2596

Thus a small change in the value of brings about a significant change in the spectral

radiussp

M .


80/161

Numerical analysis /Eigenvalues and Eigenvectors Lecture notes

Vittal rao/IISc.Bangalore M3/L2/V1/May2004/1

EIGENVALUES AND EIGENVECTORS

Let A be an nxn matrix. A scalar is called an eigenvalue of A if there exists a nonzeronx1 vector x such that

Ax = xExample:

Let

=

7816

438

449

A

1=Let Consider

=

0

2

1

x. We have

=

=

=

0

2

1

1

0

2

1

0

2

1

7816

438

449

Ax

( ) xx == 1

1= is such that there exists a nonzero vector x such that Ax = x. Thus is aneigenvalue of A.

Similarly, if we take = 3, we find that

=

0

2

1

x

Ax = x. Thus, = 3 is also an eigenvalue of A.

Let be an eigenvalue of A. Then any nonzero x such that Ax = x is called aneigenvector of A.

Let be an eigenvalue of A. Let,

.: xAxCx n ==

Then : (i) is nonempty. = nxQ


81/161



+

+=+==

yx

yxyxAyAyxAxyxii )()(,,)(

(iii)

For any constant )(; xxAx == )()( xxA =

x

Thus is a subspace of Cn. This is called the characteristic subspace or the

eigensubspace corresponding to the eigenvalue .

Example:

Consider the A in the example on page 1. We have sum = -1 is an eigenvalue. What is-1, the eigensubspace corresponding to 1?

We want to find all x such that

Ax = -x

i.e., (A+I)x = .

i.e., we want to find all solutions of the homogeneous system Mx = ; where

=+=

7816

448

448

IAM

We now can use our row reduction to find the general solution of the system.

000

0002

1

2

11

000

000

4481

12

13

8

1

2

RRR

RR

M

Thus, 32121

21 xxx +=

Thus the general solution of (A+I) x = is


82/161



+

=

+

2

0

1

2

1

0

2

1

2

12

1

2

1

32

3

2

32

xx

x

x

xx

+

=

2

0

1

0

2

1

21 AA

where A1 and A2 are arbitrary constants.

Thus -1 consists of all vectors of the form

+

20

1

02

1

21 AA .

Note: The vectors form a basis for

2

0

1

,

0

2

1

-1 and therefore

dim -1 = 2.

What is 3 the eigensubspace corresponding to the eigenvalue 3 for the above matrix

We need to find all solutions of Ax = 3x,

i.e., Ax 3x =

i.e., Nx =

Where

==

4816

408

4412

3IAN

Again we use row reduction


83/161



+

000 34

3

80

4412

3

4

3

80

3

4

3

80

4412

43

12

13

3

2

3

4

RR

RR

RR

N

321 4412 xxx +=

323

4

3

8xx = 23 2xx =

2221 128412 xxxx =+=

12 xx =

12312 22; xxxxx === The general solution is

=

2

1

1

2

1

1

1

1

x

x

x

x

Thus 3 consists of all vectors of the form

2

1

1

Where is an arbitrary constant.

Note: The vector forms a basis for

2

1

1

3 and hence

dim. 3 = 1.

Now When can a scalar be an eigenvalue of a matrix A? We shall now investigate thisquestion. Suppose is an eigenvalue of A.


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This There is a nonzero vector x such that Ax = x.

.;)( = andxxIA

The system = xIA )( has at least one nonzerosolution.

nullity (A - I) 1

rank(A - I) < n

(A - I) is singular

det. (A - I) = 0

Thus, is an eigenvalue of A det. (A - I) = 0.

Conversely, is a scalar such that det. (A - I) = 0.

This (A - I) is singular

rank(A - I) < n

nullity (A - I) 1

The system = xIA )( has nonzero solution. is an eigenvalue of A.

Thus, is a scalar such that det. (A - I) = 0 is an eigenvalue.

Combining the two we get,

is an eigenvalue of A

det. (A - I) = 0

det. (I - A) = 0

Now let C() = det. (I - A)

Thus we see that,

The eigenvalues of a matrix A are precisely the roots of C() = det. (I - A).


85/161



( )

nnnn

n

n

aaa

aaa

aaa

C

=

K

KKKK

KKKK

K

K

21

22221

11211

( ) ( ) Aaa nnnnn

.det1111 ++++=KK

Thus ; C() is a polynomial of degree n. Note the leading coefficient of C() is 1. Wesay C() is a monic polynomial of degree n. This is called CHARACTERISTICPOLYNOMIAL of A. The roots of the characteristic polynomial are the eigenvalues of

A. The equation C() = 0 is called the characteristic equation.

Sum of the roots of C() = Sum of the eigenvalues of A = a11 + . . . . . . + ann ,

and this is called the TRACE of A.

Product of the roots of C() = Product of the eigenvalues of A = det. A.

In our example in page 1 we have

=7816

438

449

A

( )7816

438

449

).(det

+

==

AIC

781

431

441

321

+

+

+

++

CCC


86/161



( )781

431

441

1

+=

340

010

441

)1(12

13

+

+=

RR

RR

( )( )( )311 ++=

( ) ( )31 2 += Thus the characteristic polynomial is

( ) ( )31)( 2 += C The eigenvalues are 1 (repeated twice) and 3.

Sum of eigenvalues = (-1) + (-1) + 3 = 1

Trace A = Sum of diagonal entries.

Product of eigenvalues = (-1) (-1) (3) = 3 = det. A.

Thus, if A is an nxn matrix, we define the CHARACTERISTIC POLYNOMIAL as,

AIC = )( . . . . . . . . . . . . .(1)

and observe that this is a monic polynomial of degree n. When we factorize this as,

( ) ( ) ( ) kakaa

C = KK21 21)( . . . . . . . .(2)Where 1, 2, . . . . . ., k are the distinct roots; these distinct roots are the distincteigenvalues of A and the multiplicities of these roots are called the algebraic

multiplicities of these eigenvalues of A. Thus when C() is as in (2), the distincteigenvalues are 1, 2, . . . . . ., kand the algebraic multiplicities of these eigenvalues are

respectively, a1, a2, . . . . . , ak.For the matrix in Example in page 1 we have found the characteristic polynomial on page

6 as

( ) ( 31)( 2 += C )


87/161



Thus the distinct eigenvalues of this matrix are 1 = -1 ; and 2 = 3 and their algebraicmultiplicities are respectively a1 = 2 ; a2 = 1.

Ifi is an eigenvalues of A the characteristic subspace corresponding to i is i and is

defined as

{ }xAxx ii == :

The dimension of i is called the GEOMETRIC MULTIPLICITY of the eigenvalue i

and is denoted by gi.

Again for the matrix on page 1, we have found on pages 3 and 4 respectively that, dim -1 = 2 ; and dim. 3 = 1. Thus the geometric multiplicities of the eigenvalues 1 = -1 and2 = 3 are respectively g1 = 2 ; g2 = 1. Notice that in this example a1 = g1 = 2 ; and a2 = g2

= 1. In general this may not be so. It can be shown that for any matrix A having C() asin (2),

1 gi ai ; 1 i k . . . . . . . . . . . .(3)

i.e., for any eigenvalue of A,

1 geometric multiplicity algebraic multiplicity.

We shall study the properties of the eigenvalues and eigenvectors of a matrix. We shall

start with a preliminary remark on Lagrange Interpolation polynomials :

Let 1, 2, . . . . . . . ., s be a distinct scalars, (i.e., ij if i j ). Consider,

)())(())((

)())(())(()(

1121

1121

siiiiiii

sii

ip

=

+

+

KK

KK

)(

)(

1ji

j

sjij

=

for i = 1,2, . . . . . . ., s . . . . . . .. (4)

Then pi() are all polynomials of degree s-1.

Further notice that ( ) ( ) ( ) 0)(111 ====== + siiiiii pppp KK

( ) 1=iip


88/161



Thus pi()are all polynomials of degree s-1 such that,

ijjip = if j i . . . . . . . . . . (5)

We call these the Lagrange Interpolation polynomials. If p() is any polynomial ofdegree s-1 then it can be written as a linear combination ofp1(),p2(), . . ., ps()as follows:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ss ppppppp +++= L2211 . . . . (6)

( ) (=

=s

i

ii pp1

)

With this preliminary, we now proceed to study the properties of the eigenvalues and

eigenvectors of an nxn matrix A.

Let 1, . . . . , k be the distinct eigenvalues of A. Let 1, 2, . . . , k be eigenvectorscorresponding to these eigenvalues respectively ; i.e., i are nonzero vectors such that

Ai = ii . . . . . . . . . . . .(6)From (6) it follows that

iiiiiiii AAAAA 22

)()( ====

iiiiiiii AAAAA 32223 )()( ====

and by induction we get

iim

i

mA = for any integer m 0 . . . . . . . . . . .(7)(We interpret A0 as I).

Now let,s

saaap +++= KK10)(

be any polynomial. We define p(A) as the matrix,

s

s AaAaIaAp +++= KK10)(


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Now

i

s

si AaAaIaAp )()( 10 +++= KK

issii AaAaa +++= KK10

by (6)iis

siii aaa +++= KK10

iis

si aaa )( 10 +++= KK

.)( iip = Thus,

Now are the eigenvectors, 1, 2, . . . . , kcorresponding to the distinct eigenvalues1, 2, . . . . , kof A, linearly independent ?

Ifi is any eigenvalue of A and i is an eigenvector corresponding to i then

for any polynomial p() we have .)()( iii pAp =

In order to establish this linear independence, we must show that

0212211 =====+++ KnKK CCCCCC KK . . . (8)Now if in (4) & (5) we take s = k ; i = i then we get the Lagrange Interpolationpolynomials as

( ))(

)(

1ji

j

kji

ij

p

=

; i = 1,2,.., k (9)

and

ijjip = if j i (10)Now,

nkkCCC =+++ ....2211

For 1 i k,


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( )[ ] ( ) nnikki ApCCCAp ==+++ ....2211

( ) ( ) nkikii ApCApCApC =+++ ....)( 2211

( ) ( ) ,....)( 222111 nkkikii pCpCpC =+++ (by property I on page 10)

;1; kiC ii = by (10)

kiCi = 1;0 since i are nonzero vectors

Thus

0........ 212211 =====+++ nnkk CCCCCC proving (8). Thus we have

Eigen vectors corresponding to distinct eigenvalues of A are linearlyindependent.


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SIMILAR MATRICES

We shall now introduce the idea of similar matrices and study the properties of similar

matrices.

DEFINITION

An nxn matrix A is said to be similar to a nxn matrix B if there exists a nonsingular nxnmatrix P such that,

P-1 A P = B

We then write,

A B

Properties of Similar Matrices

(1) Since I-1

A I = A it follows that A A

(2) A B P, nonsingular show that., P-1 A P = B

A = P B P-1

A = Q-1 B P, where Q = P-1 is nonsingular

nonsingular Q show that Q-1 B Q = A

B A

Thus

A B B A

(3) Similarly, we can show that

A B, B C A C.

(4) Properties (1), (2) and (3) above show that similarity is an equivalence relation on theset of all nxn matrices.

(5) Let A and B be similar matrices. Then there exists a nonsingular matrix P such that

A = P-1 B P

Now, let CA() and CB () be the characteristic polynomials of A and B respectively. Wehave,

( ) BPPIAICA1==

BPPPP 11 =

( )PBIP = 1

PBIP = 1


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1sin 1 == PPceBI

= CB ()

Thus SIMILAR MATRICES HAVE THE SAME CHARACTERISTIC

POLYNOMIALS .

(6) Let A and B be similar matrices. Then there exists a nonsingular matrix P such that

A = P-1 B P

Now for any positive integer k, we have

( )( ) ( )4444 34444 21

ktimes

kBPPBPPBPPA

111 ..... =

= P-1

Bk

P

Therefore, Ak

= On P-1

Bk

P = On

Bk= On

Thus if A and B are similar matrices then Ak= On Bk= On .

Now let p() = a0 + a1 + .. + ak be any polynomial.

Then

( ) kkAaAaIaAp +++= .....10

PBPaPBPaBPPaIak

k

121

2

1

10 ..... ++++=

PBaBaBaIaP kk++++=

.....22101

( )PBpP 1=

Thus

( ) ( ) nn OPBpPOAp ==1

( ) nOBp =

Thus IF A and B ARE SIMILAR MATRICES THEN FOR ANY POLYNOMIAL p ();p (A) = On p (B) = On .

(7) Let A be any matrix. By A(A) we denote the set of all polynomials p() such that

p(A) = On, i.e.


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A (A) = {p() : p(A) = On}

Now from (6) it follows that,

IF A AND B ARE SIMILAR MATRICES THEN A(A) = A (B) .

Then set A (A) is called the set ANNIHILATING POLYNOMIALS OF A

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Numerical Analysis 323