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Computational Linear Algebra Syllabus

NUMERICAL ANALYSIS

Linear Systems of Equations and Matrix Computations

Module 1: Direct methods for solving linear system of equation

Simple Gaussian elimination method, gauss elimination method with partial pivoting, determinant evaluation, gauss Jordan method, L U decompositions Doolittle’s lu decomposition, Doolittle’s 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 eigenvalues of 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


Lecture Plan Modules Learning Units Hours per

Topics Total Hours

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. Doolittle’s LU Decomposition 2

1. Direct methods for solving linear system of equation.

7. Doolittle’s method with row interchange. 2

10

8. Iterative methods for the solution of systems equation

3

9. Jacobi iteration. 2

10. Gauss – Seidel method 2

2. Iterative methods for solving linear systems of equations.

11. Successive over relaxation method (sort method).

2

9

12. An introduction. 1

13. Eigenvalues and eigenvectors, 2

14. Similar matrices, 2

15. Hermitian matrices. 1

16. Gramm – Schmidt orthonormalization, 2

3. Eigenvalues and Eigenvectors

17. Vector and matrix norms. 1

9

18. Computation of eigenvalues 2

19. Computation of eigenvalues of a real symmetric matrix.

2

20. Determination of the eigenvalues of a real symmetric tridiagonal matrix,

2

21. Tridiagonalization of a real symmetric matrix

1

22. Jacobian iteration for finding eigenvalues of a real symmetric matrix

1

4. Computations of eigenvalues.

23. The Q R decomposition 1

11



24. The Q-R algorithm. 2


Numerical Analysis/ Direct methods for solving linear Lecture Notes

system of equation

1. DIRECT METHODS FOR SOLVING LINEAR SYSTEMS OF EQUATIONS

1.1. SIMPLE GAUSSIAN ELIMINATION METHOD Consider 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 the simple Gaussian elimination method for finding the solution. The method reduces the system 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

aaaaaa

)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

yy

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(

aam i

i −= i > 1.

Note A(2) is of the form

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


system of equation

M1/L1and L2/V1/May2004/2

A(2) =

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

nnn

n

n

n

aa

aaaaaaa

)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

Imm

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(

aam i

i −= 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


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

aaaaaaaa

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 get

M(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

aaaaaaaa

AAM

)1(1

)1(

1)1(

11)1(

)()(

2)2(

22)2(

1)1(

11)1(

)1()()(

...000.........

...0

......0

.........0

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

MMM

MM

M



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; and hence 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)



system of equation

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

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

A = αu of a given matrix A – as a product of a lower triangular matrix with an upper triangular matrix. 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

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

021112211

A ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

324

y

)2(2

)1(

210330

211

021112211

12

13

AARR

RR=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−= →

−

−

a(1)11 = 1 ≠ 0 m(1)

21 = -2 a(2)22 = -3 ≠ 0

m(1)31 = -1

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−=

101012001

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

16

4

324

yMyy ==⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=



system of equation

)3(31

)2(

210330

21123

AARR

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−→

+

a(3)33 = -3

M(2)31 = 1/3

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=

1310

010001

)2(M ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−==

36

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,

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

111

3

2

1

xxx

x

The determinant of the given matrix A is

a(1)11 a(2)

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

Now

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=−

101012001

)1(1M

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−

=−

1310

010001

)1(2M



system of equation

L = M(2) M(-1)

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

= ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

101012001

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

− 1310

010001

L = L(-1) = ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

− 1311

012001

u = A(n) = A(3) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

300330

211

Therefore A = lu i.e.,

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

021112211

= ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

− 1311

012001

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

300330

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. This

may 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) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

625675.0663257.0173257.0476625.0375623.0215512.0332147.0213472.0000003.0



system of equation

y(1) = a⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

285321.0127653.0235262.0

(1)11 = 0.000003 ≠ 0

3.71837000003.0215512.0

11)1(21

)1(

21)1( −=−=−=

aaM

3.57752000003.0173257.0

11)1(31

)1(

31)1( −=−=−=

aaM

M(1) = ; y⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

103.57752013.71837001

(2) = M(1) y(1) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

6.135865.16900

235262.0

A(2) = M(1) A(1) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−−

7.191818.1232700.238609.153340

332147.0213472.0000003.0

a(2)22 = - 15334.9 ≠ 0

803905.09.153348.12327

22)2(32

)2(

32)2( −=

−−

−=−=aaM

M(2) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

− 1803905.00010001

y(3) = M(2) y(2) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

20000.05.16900

235262.0

A(3) = M(2) A(2) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

50000.0000.238609.153340

332147.0213472.0000003.0

Thus the given system is equivalent to the upper triangular system

A(3)x = y(3)



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 of partial 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 r th diagonal position by a 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 GAUSS ELIMINATION METHOD WITH PARTIAL PIVOTING. We now illustrate this with a few examples:

Example:

x1 + x2 + 2 x3 = 4

2x1 – x2 + x3 = 2

x1 + 2x2 = 3

We have

Aavg = 324

021112211

−

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



system of equation

Aavg ⎯→⎯12R

342

021211112 −

Therefore we have

M(1) = and M⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

100001010

(1) A(1) = A(2) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −

021211112

M(1) A(1) = y(2) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

342

Next we have

R2 – ½ R1 2 -1 1 2

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

R3 – ½ R1 0 5/2 -1/2 3

Here

M(2) =

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

−

1021

0121

001 ; M(2) A(2) = A(3) =

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−

−

21

250

23

230

112

M2 y(2) = y(3) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

232

Now at the next stage the pivot is 25 since this is the entry with the largest absolute value

in 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



system of equation

0 3/2 3/2 3

M(3) = M⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

010100001

(3) A(3) = A(4) = ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−−

23

230

21

250

112

M(3) y(3) = y(4) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

322

Next we have

2 -1 1 2

A(4)avg ⎯⎯⎯ →⎯

− 23 53

RR 0 5/2 -1/2 2

0 0 9/5 9/5

Here

M(4) = ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

− 1530

010001

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

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

−

590021

250

112

M(4) y(4) = y(5) = ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

5922

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

25 x2 -

21 x3 = 2

59 x3 =

59

We now get the solution by back substitution:



system of equation

The 3rd equation gives,

x3 = 1

using this in second equation we get

25 x2 -

21 = 2

giving 25 x2 =

25

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 = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

625675.0663257.0173257.0476625.0375623.0215512.0332147.0213472.0000003.0

y = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

285321.0127653.0235262.0

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

A(1) = A A⎯→⎯ 12R (2) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

625675.0663257.0173257.0332147.0213472.0000003.0476625.0375623.0215512.0



system of equation

y(1) = y y⎯→⎯ 12R (2) = M⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

285321.0235262.0127653.0

(1) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

100001010

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

aa

= - 215512.0000003.0 = - 0.000014

m31 = - 11

31

aa

= - 215512.0173257.0 = - 0.803932

M(2) = ; y⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

10803932.001000014.0001

(2) = M(2) y(2) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

182697.0235260.0127653.0

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) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

332140.0213467.00242501.0361282.00476625.0375623.0215512.0

M(3) = y⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

010100001

(4) = M(3) y(3) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

235260.0182697.0127653.0

Now reduce the entry below 2nd diagonal as zero

A(4) A⎯⎯⎯ →⎯ + 2323 RMR 5 = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

188856.000242501.0361282.00476625.0375623.0215512.0



system of equation

M32 = - 361282.0213467.0 = - 0.590860

M(4) = y⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

− 159086.0010001

(5) = M(4) y(4) = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

127312.0182697.0127653.0

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 = 0.991291

which 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 GEM whereas we have come around this difficulty by using partial pivoting.


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)( 25 )(

59 ) = 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.

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.

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 = Lα Uα

is also a LU decomposition where α ≠ 0 is any scalar and Lα = α L and Uα = 1/α 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 uii 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 be considering three cases. First, we shall consider the decomposition Tridiagonal matrix; secondly the Doolittles’s method for a general matrix, and thirdly the Cholesky’s method for a symmetric matrix.



1 TRIDIAGONAL MATRIX

Let

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

=

−

−−

bcabc

abcabc

ab

A

i

iii

1

12

432

321

21

0........00....0

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

........................0....00....00....00

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

Let δi denote the determinant of the ith principal minor of A

ii

ii

i

bcabc

abcab

1

112

321

21

0....0....0

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

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

−

−−

=δ

Expanding by the last row we get,

δi = bi δi-1 – ci-1 ai δi-2 ; I = 2,3,4, ……..

……..(I)

δi = b1

We define δi = 1

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

1

21

1 −

−−

−

−=i

iiii

i

i acbδδ

δδ

setting ii

i k=−1δδ

this can be written as

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

1 IIkackb

i

iiii

−−+=



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

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

− 1........00........................0....0100........010............01

1

2

1

nw

ww

L ;

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

−

n

nn

uu

uu

U

0....00....00

....................0....00....0

1

32

21

α

αα

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 differ by 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 j<i and i-j ≥ 2.

Uii+1 = αi+1

Uii = ui ……………………. (V)

Uij = 0 if i>j or i<j and j-I ≥ 2.

Now A = LU is what is needed.

Therefore,

∑=

=n

kkjikij VIULA

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

Therefore

∑=

−− =n

kkikiii ULA

111

Using (III), (IV) AND (V) we get



Therefore

iiiiii ULa α== −−− 111

Therefore

αi = ai ………………….. (VII)

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

∑=

=n

kkiikii ULA

1

iiiiiiii ULUL += −− 11

Therefore

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

From (VI) we get further,

∑=

++ =n

kkikiii ULA

111

iiiiiiii ULUL 1111 ++++ +=

iii uWc =

Thus ………………… (IX) iii uWc =

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

ii

iii u

uac

b +=−

−

1

1

Therefore

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

1 Xu

acubi

iiii

−

−+=

Comparing (X) with (II) we get

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

XIkui

iii

−

==δδ



using this in (IX) we get

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

uc

wi

ii

i

ii δ

δ −==

From (VII) we get

αI = 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 δI are all nonzero. i.e. all the principal minors have nonzero determinant.

Example:

Let

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−−−−

−−

=

1300013900025200011200022

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 = 3

a2 = -2 a3 = 1 a4 = -2 a5 = 1

We have

δ0 = 1

δ1 = 2

δ2 = b2 δ1 – a2 c1 δ0 = 2-4 = -2

δ3 = b3 δ2 – a3 c2 δ1 = (-10) – (-2) (2) = -6

δ4 = b4 δ3 – a4 c3 δ2= (-3) (-6) – (-18) (-2) = -18

δ5 = b5 δ4 – a5 c4 δ3 = (-1) (-18) – (3) (-6)= 36.

Note δ1,δ2,δ3,δ4,δ5 are all nonzero. So we can apply the above method.

Therefore by (XI) we get


M1/L5/V1/May 2004/6

326;1

22;2

2

33

1

22

0

11 =

−−

==−=−

====δδ

δδ

δδ uuu

36

18

3

44 =

−−

==δδ

u ; and 218

36

4

55 −=

−==

δδu

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

122

1

11 −=

−==

ucw 222 −== aα

212

2

22 =

−−

==ucw 133 == aα

339

3

33 ===

ucw 244 −== aα

133

4

44 ===

ucw 155 == aα

Thus;

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛−

=

1100001300001200001100001

L ;

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

−−−

=

2000013000023000011000022

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 ui, the diagonal entries in U are small thus creating problems 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 be achieved as follows:

We seek

A = LU

where

VittalRao/IISc, Bangalore



l1

L = w1 l2

;

wn-1ln

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

−

n

nn

uu

uu

U

0....00....00

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

........00....0

1

32

21

α

αα

Lii = li

Now Li+1i = wi

Lij = 0 if j>i or j<i and i-j ≥ 2

Uii = ui

Uii+1 = αi+1

Uij = 0 if i>j; 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 α1−=

Therefore

ai = li-1 αI ………………. (VII`)

iii Ab = ki

n

kikUL∑

−

=1

iiiiiiii ULUL += −− 11 iiii ulW += − α1

`).......(..........1 VIIIulWb iiiii +=∴ − α

iii Ac 1+= ki

n

kki UL∑

−+=

11 iiii UL 1+= iiuw=

iii uwc = ……………….. (IX`)

From (VIII`) we get using (VII`) and (IX`)



iii

i

i

ii ul

la

uc

b +=−−

−

11

1 .

iiii

ii ululca

+=−−

−

11

1

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

1 Xppca

b ii

iii +=

−

−

where

pi = li ui

Comparing (X`) with (II) we get

1−==

i

iii kp δ

δ

therefore

1−=

i

iiiul δ

δ

we choose 1−

=i

iil δ

δ ………………. (XIV)

⎟⎠⎞⎜

⎝⎛=

−1sgn

i

iiu δ

δ

1−i

iδ

δ ………… (XV)

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

1−=

i

iil δ

δ ;

11

.sgn−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

i

i

i

iiu δ

δδδ

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

i

ii u

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

1−=

i

ii l

aα . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . .(XIII`)

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

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



We get;

δ0 = 1 δ1 = 2 δ2 = -2 δ3 = -6 δ4 = -18 δ5 = 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 δ1/δ0 = 2 ; δ2/δ1 = -1 ; δ3/δ2 = 3 ; δ4/δ3 = 3 ; δ5/δ4 = -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

222

1

11 −=

−==

uCw ; 2

12

2

22 =

−−

==uCw ;

333

9

3

33 ===

uCw ; 3

33

4

44 ===

uCw

From (XIII`) we get

222

1

22 −=

−==

laα ; 1

11

2

33 ===

laα ;

32

3

44

−==

laα ;

31

4

55 ==

laα

Thus, we have LU decomposition,



⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

−

−−−

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−−−−

−−

=

200003

13000

032300

0011000022

23000033300003200001200002

1300013900025200011200022

A

L U

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



DOOLITTLE’S 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 in which 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 determined as follows :

∑=

=n

kkk ula

11111

= l11 u11 Since l1k = 0 for k>1

= u11 Since l11 = 1.

.111 =∴u In general,

∑=

=n

kkjkj ula

111

= l11 u11 Since l1k = 0 for k>1

= u1j Since l11 = 1.



⇒ 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

kkjkj ula

111

= 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 we proceed to describe how one then determines the ith 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 known for 1 ≤ i ≤ n ; 1 ≤ k ≤ i-1.

Now

∑=

=n

kkjikij ula

1

Since l∑=

=i

kkjik ul

1ik = 0 for k>i

ijii

i

kkjik ulul += ∑

−

=

1

1



since lij

i

kkjik uul += ∑

−

=

1

1ii = 1.

∑−

=

−=⇒1

1

i

kkjikijij ulau

. . . . . ... . . . .(III)

Note that on the RHS we have aij 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 entries in 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 ith 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

kkijkji ula

1

Since u∑=

=i

kkijk ul

1ki = 0 if k>i

iiji

i

kkijk ulul += ∑

−

=

1

1

⎥⎦

⎤⎢⎣

⎡−=⇒ ∑

−

=

1

1

1 i

kkijkji

iiji ula

ul

…..(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-1 columns of L; and uki, 1 ≤ k ≤ i-1 and hence only entries in the first i-1 rows of U. Thus RHS in (IV) is completely known and hence lji, the entries in the ith column of L are completely determined by (IV).



Summarizing, Doolittle’s 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

kkjikijij ulau

; j = i, i+1, i+2, …….,n

(Note for j<i we have uij = 0)

⎥⎦

⎤⎢⎣

⎡−= ∑

−

=

1

1

1 i

kkijkji

iiji ula

ul

; j = i, i+1, i+2,…..,n

(Note for j<i we have ljj = 0)

We observe that the method fails if uii = 0 for some i.

Example:

Let

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

−−−

=

126064191444622

3112

A

Let us determine the Doolittle decomposition for this matrix.

First step:



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



= [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.



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

l44 = 1.

Thus.

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

−=

1213014200110001

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

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

−−

=

200021001530

3112

U

and

A = LU.

This gives us the LU decomposition by Doolittle’s method for the given A.

As we observed in the case of the LU decomposition of a tridiagonal matrix; it is not 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

kkjkij



11

1l

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

kkiikii ula

1

for l∑=

=n

kkiik ul

1ik = 0 if k>i

ii

i

kiikiik ulul∑

−

=

+=1

1

saypulaul i

i

kkiikiiiiii ,

1

1=−=∴ ∑

−

=

Choose ki

i

kikiiiii ulapl ∑

−

=

−==1

1



iiii ppu sgn−=

iforklulula ikkj

i

kikkj

n

kikij >=== ∑∑

=−

011

Q

ijiikj

i

kik ulul += ∑

−

=

1

1

⎟⎠

⎞⎜⎝

⎛−=⇒ ∑

−

=kj

i

kikijij ulau

1

1 lii

determining the ith row of U.

ki

n

kjkji ula ∑

=

=1

iifkuul kiki

i

kjk >== ∑

=

01

Q

iijiki

i

kjk ulul += ∑

−

=

1

1

⎟⎠

⎞⎜⎝

⎛−=⇒ ∑

−

=

1

1

i

kkijkjiji ulal uii ,

thus determining the ith column of L.

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



First Step:

2;22 1111111111 ==∴== ulaul

23;

21

21

11

1414

11

1313

11

1212 ==−====

lau

la

ulau

23;

21;

21;2 14131211 =−=== uuuu

22

2

11

2121 −=−==

ua

l

222

4

11

3131 ===

ual

232

6

11

4141 ===

ual

therefore

23

22

2

2

41

31

21

11

=

=

−=

=

l

l

l

l

Second Step:

1221222222 ulaul −=



( ) 32

122 =⎟⎠

⎞⎜⎝

⎛−−=

3;3 2222 ==∴ ul

( )13212323 ulau −= l22

( )3

53/2

126 =⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−−−=

[ ]14212424 ulau −= l22

( ) ( )3

13/2

324 −=⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−−−=

therefore

31;

35;3;0 24232221 −====∴ uuuu

( ) 2212313232 /uulal −=

( ) 3/2

12214 ⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−=

34=



( ) 2212414242 /uulal −=

( ) 3/2

1230 ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−=

3−=

therefore

3

34

3

0

42

32

22

12

=

=

=

=

l

l

l

l

Third Step:

23321331333333 ululaul −−=

( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠

⎞⎜⎝

⎛−−=

3534

212219

= 1

1;1 3333 ==∴ ul

( ) 33243214313434 / lululau −−=



( ) ( ) 1/3

1342

3224⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟

⎠

⎞⎜⎝

⎛−=

= 2

2,1;0;0 34333231 ====∴ uuuu

[ ] 33234213414343 / uululal −−=

( ) ( ) 1/3

532

1236 ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟

⎠

⎞⎜⎝

⎛−−−=

= 2

therefore

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

====

2100

43

33

23

13

llll

Fourth Step:

344324421441444444 ulululaul −−−=

( ) ( ) ( )( )223

132

32312 −⎟⎟⎠

⎞⎜⎜⎝

⎛−−−⎟

⎠

⎞⎜⎝

⎛−=

= -2



2;2 4444 −==∴ ul

2;0;0;0 44434241 −====∴ uuuu

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=

===

2

000

44

34

24

14

l

lll

Thus we get the LU decompositions,

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

−=

2232301342200320002

L ;

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

−

−

−

=

20002100

31

3530

23

21

212

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 replaced diagonals.



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.

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


DOOLITTLE’S 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 zero pivot 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 Doolittle’s procedure. The Doolittle’s method which is used 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

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−−

−−

=

32131451

32221213

A y =

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

−

138

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



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 any advantage 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 Doolittle’s method.

.133;

31;3

2;111

4141

11

3131

11

212111 ========

ual

ua

luall

Thus

L is of the form

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

***1

01*31

00132

0001

; and

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

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛ −−

*000**00***01213

Stage 2

We now calculate the second diagonal of U: By Doolittle’s method we have

12212222 ulau −= ( )381

322 −=⎟

⎠⎞

⎜⎝⎛−−=

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 2nd and 3rd row.

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−−−−

=

18

33

;

3213322214511213

NewyNewA



⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛ −−

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

*000**00

**3

1401213

;

1**101*3

20013

10001

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

102314 −=−⎟

⎠⎞

⎜⎝⎛−−=

14212424 ulau −= ( ) ( )321

311 −=−⎟

⎠⎞

⎜⎝⎛−−=

2nd column of L:

[ ] 2212313232 uulal ÷−= ( ) ( )3

141322 ÷⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−−=

74

−=

[ ] 1112414242 uulal ÷−= ( )( )[ ]3

14113 ÷−= = 0

Therefore new L has form

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

1*01

0174

32

00131

0001

New U has form

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛−−−−

*000**0032

310

3140

1213

This completes the 2nd stage of our computation.

Note: We had three choices of u22 to 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.



3rd Stage:

3rd diagonal of U:

233213313333 ululau −−= ( ) ⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−−−⎟

⎠⎞

⎜⎝⎛−=

310

742

322

710

=

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 ( )( ) ( ) ⎟⎠⎞

⎜⎝⎛−+−−=

3100212 = 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

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

−−−−

=

81

33

3222321314511213

NewYNewA

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛−−

−−

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

=

*000*40032

310

3140

1213

;

1*74

32

0101

00131

0001

NewUNewL

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

243214313434 ululau −−= ( )( ) ( ) 4320113 =⎟

⎠⎞

⎜⎝⎛ −−−−=

( ) 33234213414343 uululal ÷−−=



( ) 43

10742

322 ÷⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−−−⎟

⎠⎞

⎜⎝⎛−= = 5/14.

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛−−

−−

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

=∴

*000440032

310

3140

1213

;

1145

74

32

0101

00131

0001

NewUNewL

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

4th Stage

[ ]3443244214414444 ulululau −−−=

( ) ( )4145

32

741

323 ⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−−−⎟

⎠⎞

⎜⎝⎛−= = 13/7.

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

==

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

−−−−

==∴

81

33

3222321314511213

** YNewYANewA

New L = L* , New U = U*

,

713000440032

310

3140

1213

;

1145

74

32

0101

00131

0001

**

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛−−

−−

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

= 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



213331

221 =−=⇒=+ zzz

411 1331 −=−−=⇒−=+ zzzz

8145

74

32

4321 −=++− zzzz

( ) ( ) ( ) 841452

743

32

4 −=+−⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛ z

752

4 −=⇒ z

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

−=∴

7524

23

z

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

752

713

4 −=x therefore x4 = -4.

311444 434343 =−−=⇒−=+⇒−=+ xxxxxx

therefore x3 = 3

232

310

314

432 =−− xxx

( ) ( ) 24323

310

314

2 =−⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛−x

22 =⇒ x

323 4321 =−−+ xxxx

134623 11 =⇒−+−+⇒ xx

Therefore the solution of the given system is



⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

=

4321

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 may then be uniquely determined as in Doolittle’s 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 crout’s method the diagonal elements of U are all chosen as unity. Apart from this, there is little distinction, as regards procedure or accuracy, between the two methods.

As already mentioned, Wilkinson’s 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 of U.

Stage 1: 1st row of U:



∑∑==

==n

kkk

n

kk uula

11

21

1111 11 kk ul =Q 1UL =Q

for k>1 since U is upper triangular 112u= 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 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 of U.

Stage 1: 1st row of U:

∑∑==

==n

kkk

n

kk uula

11

21

1111 11 kk ul =Q 1UL =Q

112u= for k>1 since U is upper triangular. 01 =kuQ

1111 au =∴

∑ ∑= =

==n

k

n

kkikkiki uuula

1 1111

iuu 111= 101 >= forkukQ

1111 au =

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



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

∑ ∑= =

===n

k

n

kkiikkikiikii uluula

1 1

2 Q

for k > i 01

2 == ∑=

ki

i

kki uu Q

ii

i

kki uu 2

1

1

2 += ∑−

=

∑−

=

−=∴1

1

22i

kkiiiii uau

∑−

=

−=∴1

1

2i

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

1st i-1 rows have already been obtained.

∑ ∑= =

==n

k

n

kkjkikjikij uuula

1 1 Now we need uij for j > i

∑=

=i

kkjkiuu

1 Because uki = 0 for k > i

∑−

=

+=1

1

i

kijiikjki uuuu

Therefore

ii

i

kkjkiijij uuuau ÷⎥

⎦

⎤⎢⎣

⎡−= ∑

−

=

1

1



⎪⎪

⎩

⎪⎪

⎨

⎧

÷⎥⎦

⎤⎢⎣

⎡−=

−=∴

∑

∑−

=

−

=

ij

i

kkjkiijij

i

kkiiiii

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

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−−

−

=

10131133133511111

A

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

1st row of U

⎪⎪

⎩

⎪⎪

⎨

⎧

=÷==÷=

−=÷=

==

11

11

111414

111313

111212

1111

uauuauuau

au

2nd row of U



( ) ( )( )( )( ) ( )( )( )⎪

⎪⎩

⎪⎪⎨

⎧

=÷−−=÷−=−=÷−−−=÷−=

=−=−=

2211312113

215

2214122424

2213122323

122

2222

uuuauuuuau

uau

3rd row of U

( ) ( )( ) ( )( )( )⎪⎩

⎪⎨⎧

=÷−−−=÷−−=

=−−=−−=

21211111113

33242314133434

232

132

3333

uuuuuauuuau

4th row of U

144110342

242

142

4444 =−−−=−−−= uuuau

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛−

−

=∴

1000210021201111

U

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

==∴

1221011100210001

1 LU and

A = LU

= LL1

= U1U

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, { }kx 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

if 1<M for some matrix norm.

We shall now consider some iterative schemes for solving systems of linear equations,

Ax = y …………….(3)

We write this system in detail as

11212111 ..... yxaxaxa nn =+++

22222121 ..... yxaxaxa nn =+++ . . . . . . . .(4) ...... ...... ......

nnnnnn yxaxaxa =+++ .....2211

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nnnn

n

n

aaa

aaaaaa

WehaveA

K

KKKK

K

K

21

22221

11211

. . . . . . . . . . . (5)

We denote by D, L, U the matrices


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

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

......00...............0...000......00......0

33

22

11

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

nna

aa

a

D

the diagonal part of A; and

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

0

0000000

121

3231

21

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

−nnn aaa

aaa

L

K

KKKKK

K

KK

KK

the lower triangular part of A; and

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

0...000...............

...00

......0

223

112

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

= n

n

uuuu

U

the upper triangular part of A.

Note that,

A = D + L + U ……………………… (9).

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

So that D-1 exists.

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


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

xx

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 again x1, 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 ˆ+= ……………… (14)

where

J = -D-1 (L + U) …………….(15)

and, we get

x(0) starting vector

,.......2,1;ˆ)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 1<

spJ . We shall see an easier condition below:


We have

1/a11

D-1 = 1/a22

1/ann

and therefore

( )

⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

−−−

−−−

−−−

=+−=

−

−

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

....0

....0

121

22

2

22

23

22

21

11

1

11

13

11

12

1

nn

nn

nn

n

nn

n

n

n

aa

aa

aa

aa

aa

aa

aa

aa

aa

ULDJ

Now therefore the ith Absolute row sum for J is

( )∑≠

+− ++++++==ij

iiiniiiiiiii

iji aaaaaa

aa

R /........ 1121

∴ If Ri <1 for i =1,2,3,…..,n

then

{ } 1,.....,max 1 <=∞ nRRJ

and we have convergence.

Now Ri < 1 means

iiiniiiiii aaaaaa <++++++ +− .......... 1121

i.e. in each row of A the sum of the absolute x values of the nondiagonal entries is dominated by the absolute value of the diagonal entry. i.e. A is ‘strictly row diagonally dominant’. Thus the Jacobi iteration scheme for the system (3) converges if A is strictly row diagonally dominant (Of course this condition may not be satisfied) and still Jacobi iteration scheme may converge if .1<

spJ



Example:

Consider the system

x1 + 2x2 – 2x3 = 1

x1 + x2 + x3 = 0 ………….(I)

2x1 + 2x2 + x3 = 0

Let us apply the Jacobi iteration scheme with the initial vector as

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛==

000

)0( θx ………….(II)

We ; ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −=

122111221

A⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

100010001

D

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −=+

022101220

UL ; ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

001

y

( )⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−−+−

=+−= −

022101220

1 ULDJ ; ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛== −

001

ˆ 1 yDy

Thus the Jacobi scheme (16) becomes

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

000

0x

( ) ( ) ,......2,1,ˆ1 =+= − kyJxx kk

( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛==+=+=∴

001

ˆˆˆ01 yyJyJxx θ



( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−−+−

=+=001

001

022101220

ˆ12 yJxx

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−=

21

1

001

21

0

( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−−

−=+=

001

21

1

022101

220ˆ23 yJxx

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

011

001

012

( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−−

−=+=

001

011

022101

220ˆ34 yJxx

( )3

011

001

012

x=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −=

∴ x(4) = x(5) = x(6) = ………. = x(3)

∴ x(k) = x(3) and x(k) converges to x(3)

∴ The solution is =x∞→k

lim ( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−==

011

3xx k

Can easily check that this is the exact solution.

Here, there is no convergence problem at all.



Example 2:

8x1 + 2x2 – 2x3 = 8

x1 - 8x2 + 3x3 = 19

2x1 + x2 + 9x3 = 30

Let us apply Jacobi iteration scheme starting with ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

000

0x

We have ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

900080008

D

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−=∴ −

9100

0810

0081

1D

( )⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−+

+−=+−= −

011111.022222.0375.00125.0

25.025.001 ULDJ

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−== −

33333.3375.2

1ˆ 1 yDy

Now the matrix is such that

4228 131211 =+=+= aaanda 131211 aaa +>∴

;4318 232122 =+=+= aaanda 232122 aaa +>∴

3129 323133 =+=+= aaanda 323133 aaa +>∴

Thus we have strict row diagonally dominant matrix A. Hence the Jacobi iteration scheme will converge. The scheme is,



⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

000

0x

yJxx kk ˆ)1()( += − yx k ˆ011111.022222.0375.00125.025.025.00

)1( +⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−= −

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−==

33333.3375.2

1ˆ1 yx

We continue the iteration until the components of x(k) and x(k+1) differ by at most, say; 3x10-5 = ( ) ( ) 51 103.. −

∞

+ ≤−∈ xxxei kk we get ( ) ( ) .33333.301 =−∞

xx So we

continue

( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+=

37500.300000.1

42708.2ˆ12 yJxx ( ) ( ) ≥∈=−

∞42708.112 xx

( ) ( ) ;90509.280599.0

09375.2ˆ23

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+= yJxx ( ) ( ) ≥∈=−

∞46991.023 xx

( ) ( ) ;95761.202387.1

92777.1ˆ34

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+= yJxx ( ) ( ) ≥∈=−

∞21788.034 xx

( ) ( ) ;01870.302492.1

99537.1ˆ45

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+= yJxx ( ) ( ) ≥∈=−

∞06760.045 xx

( ) ( ) ;00380.399356.0

01091.2ˆ56

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+= yJxx ( ) ( ) ≥∈=−

∞03136.056 xx



( ) ( ) ;99686.299721.0

99934.1ˆ67

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+= yJxx ( ) ( ) ≥∈=−

∞01157.067 xx

( ) ( ) ;99984.200126.1

99852.1ˆ78

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+= yJxx ( ) ( ) ≥∈=−

∞00405.078 xx

( ) ( ) ;00047.300025.1

00027.2ˆ89

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+= yJxx ( ) ( ) ≥∈=−

∞00176.089 xx

( ) ( ) ;99997.299979.0

00018.2ˆ910

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+= yJxx ( ) ( ) ≥∈=−

∞00050.0910 xx

( ) ( ) ;99994.299999.0

99994.1ˆ1011

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+= yJxx ( ) ( ) ≥∈=−

∞00024.01011 xx

( ) ( ) ;00001.300003.1

99998.1ˆ1112

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+= yJxx ( ) ( ) ≥∈=−

∞00008.01112 xx

( ) ( ) ;00001.300000.1

00001.2ˆ1213

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+= yJxx ( ) ( ) =∈=−

∞00003.01213 xx

∴ SOLUTION IS 2 = x1 ; -1 = x2, 3.00001 = x3 (Exact solution is x1 = 1, x2 = -2, x3 =3).


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

Gauss – Seidel Method

Once again we consider the system

Ax = y …………….. (I)

In the Jacobi scheme we used the values of x(k)2, x(k)

3, ….., x(k)n obtained in the k the

iteration, in place of x2, x3, ….., xn in the first equation,

11212111 ..... yxaxaxa nn =+++

to calculate x(k+1)1 from

( ) ( ) ( ) ( )113132121

111 ..... yxaxaxaxa n

kn

kkk +−−−=+

Similarly, in the ith equation we used the values, x(k)1, x(k)

2, …., x(k)i-1, x(k)

i+1, …., x(k)n, in

place of x1, x2, ….., xi-1, xi+1, ….., xn to calculate x(k+1)i from

( ) ( ) ( ) ( )112211

1 ...... −−+ −−−−= i

kii

ki

kii

kii xaxaxaxa

( ) ( ) ......11 ink

inik

ii yxaxa +−−− ++

What Gauss – Seidel suggest is that having obtained x(k+1)1from the first equation use this

value for x1 in the second equation to calculate x(k+1)2 from

( ) ( ) ( ) ( )223231

1212

122 ...... yxaxaxaxa n

kn

kkk +−−−−= ++

and use these values of x(k+1)1, x(k+1)

2, in the 3rd equation to calculate x(k+1)3, and so on.

Thus in the equation use x(k+1)1, ….., x(k+1)

i-1 to calculate x(k+1)i from

( ) ( ) ( ) ( )1

112

121

11

1 ...... −+

−+++ −−−−= i

kii

ki

kii

kii xaxaxaxa

( ) ( ) ( )

ink

inik

iiik

ii yxaxaxa +−−− ++++ .....2211

In matrix notation we can write this as, ( ) ( ) ( ) yUxLxDx kkk +−−= ++ 11

which can be rewritten as,

( ) ( ) ( ) yUxxLD kk +−=+ +1 , and hence

Thus we get the Gauss – Seidel iteration scheme as,

( ) ( ) ( ) yLDUxLDx kk 111 −−+ +++−=

x(0) initial guess ( ) ( ) yGxx kk ˆ1 +=+ ……..(II)



where,

G = -(D+L)-1U

is the Gauss – Seidel iteration matrix, and

( ) yLDy 1ˆ −+=

The scheme converges if and only if .1<sp

G Of course, the scheme will converge if

1<G in some matrix norm. But some matrix norm, 1≥G does not mean that the

scheme will diverge. The acid test for convergence is .1<sp

G

We shall now consider some examples.

Example 3:

Let us consider the system

x1 + 2x2 – 2x3 = 1

x1 + x2 + x3 = 0

2x1 + 2x2 + x3 = 0

considered on page 5; and for which the Jacobi scheme gave the exact solution in the 3rd iteration. (see page 6). We shall now try to apply the Gauss – Seidel scheme for this system. We have,

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −=

122111221

A ; ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

001

y

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=+

122011001

LD ; ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

−=−

000100

220u

( )⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−=+ −

120011001

1LD

Thus,



( )⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−=+−= −

200320

220

000100

220

120011001

1uLDG

Thus, Gauss – Seidel iteration matrix is,

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

−=

200320

220G

Since G is triangular we get its eigenvalues immediately, as its diagonal entries. Thus λ1 = 0, λ2 = 2, λ3 = 2 are the three eigenvalues. Therefore,

12 >=sp

G

Hence the Gauss – Seidel scheme for this system will not converge. Thus for this system the Jacobi scheme converges so rapidly giving the exact solution in the third iteration itself whereas the Gauss – Seidel scheme does not converge.

Example 4:

Consider the system

021

21

0

121

21

321

321

321

=+−−

=++

=−−

xxx

xxx

xxx

Let us apply the Gauss – Seidel scheme to this system. We have,

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

−−

=

121

21

11121

211

A ; ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

001

y

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−−

=+

121

21

011001

LD ; ( )⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−=+ −

1210

011001

1LD ,



⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−=−000100

21

210

u .

Thus,

( )⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−=+−= −

000100

21

210

1210

011001

1 uLDG

..(*)..........

210023

210

21

210

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

−−=∴G

is the Gauss – Seidel matrix for this sytem.

The Gauss – Seidel scheme is ( ) ( )

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

+=+

000

ˆ

0

1

x

yGxx kk

where

( ) andyLDy ;01

1

001

1210

011001

ˆ 1

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−=+= −

where G is given (*).



Notice that G is upper triangular and hence we readily get the eigenvalues of G as its diagonal entries. Thus the eigenvalues of G are, λ1 = 0, λ2 = -1/2, λ3 = -1/2. Hence

121<=

spG . Hence the Gauss – Seidel scheme will converge.

Let us now carry out a few steps of the Gauss – Seidel iteration, since we have now been assured of convergence. (We shall first do some exact calculations).

( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=+=

01

1

01

1

000

ˆ01 GyGxx

( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+=

01

1

01

1ˆ12 GyGxx

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

−−=01

1

01

1

210023

210

21

210

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞

⎜⎝⎛ −−

−

=

0211

211

( ) ( ) ( )⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+−−

+−

=+=

02

12

112

12

11

ˆ 2

2

23 yGxx

If we continue this process we get



( )

( )

( )

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞

⎜⎝⎛ −+−+−−

−+−+−

= −

−

−

−

02

1.....21

211

21.....2

12

11

1

1

2

1

1

2

k

k

k

k

kx

Clearly,

( ) ( )⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−+−−

+−+−

→

0.....2

12

12

11

.....21

21

21

211

32

432

kx

and by summing up the geometric series we get,

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛→

03232

kx

which is the exact solution.

Of course, here ‘we’ knew ‘a priori’ that the sequence is going to sum up neatly for each component and so we did exact calculation. If we had not noticed this we still would have carried out the computations as follows:

( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+=01

1ˆ01 yGxx as before

( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+=

05.0

5.0ˆ12 yGxx

( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+=

0625.0

625.0ˆ23 yGxx

( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+=

06875.0

6875.0ˆ34 yGxx



( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+=

065625.0

65625.0ˆ45 yGxx ; ( ) ( ) 03125.045 =−

∞xx

( ) ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=+=

0671875.0

671875.0ˆ56 yGxx ; ( ) ( ) 025625.056 =−

∞xx

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

0664062.0

664062.07x ; ( ) ( ) 007813.067 =−

∞xx

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

0667969.0

667969.08x ; ( ) ( ) 003907.078 =−

∞xx

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

0666016.0

666016.09x ; ( ) ( ) 001953.089 =−

∞xx

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

0666504.0

666504.010x ; ( ) ( ) 000488.0910 =−

∞xx

(Since now error is < 10-3 we may stop here and take x(10) as our solution for the system. Or we may improve our accuracy by doing more iterations, to get,

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

0666748.0

666748.011x ; ; ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

0666626.0

666626.012x ( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

0666687.0

666687.013x

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

0666656.0

666656.014x ( ) ( ) 41314 10000031.0 −

∞<=− xx

and hence we can take x(14) as our solution within error 10-4.

Let us now try to apply the Jacobi scheme for this system. We have



⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

−−

=

121

21

11121

211

A ; and therefore,

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−=

021

21

10121

210

J

We have the characteristic polynomial of J as

⎟⎠⎞

⎜⎝⎛ +−⎟⎠⎞

⎜⎝⎛ +=

−−++−−

=− 122

1

21

21

112

12

12 λλλ

λλ

λλ JI

Thus the eigenvalues of J are

415

21;4

1521;

21

321 ii −=+=−= λλλ

2416

415

41;

21

321 ==+===∴ λλλ

2=∴sp

J which is >1. 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 which the 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



Here

,

21001210

01210012

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−−

−−−

=A is a symmetric tridiagonal matrix.

The Jacobi matrix for this scheme is

⎟⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

=

02100

210

210

0210

21

00210

J

The characteristic equation is,

16λ 4 - 12λ 2 + 1 = 0 ………………(CJ)

Set λ 2 = α

Therefore

16α2 - 12α + 1 = 0 ………………(CJ1)

∴ λ 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:



( )⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

−=+

2100021000210002

LD

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=−

0000100001000010

U

( )

⎟⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

=+ −

21

41

81

161

021

41

81

0021

41

00021

1LD

( )⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

=+−= −

0000100001000010

21

41

81

161

021

41

81

0021

41

00021

1ULDG

⎟⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

=

41

81

1610

21

41

810

021

410

00210

The characteristic equation of G is

0=−GIλ , which becomes in this case

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



This can be factored as

( ) 011216 22 =+− λλλ

Thus the eigenvalues of G are roots of

λ2 = 0 ; and

16λ2 - 12λ + 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 <=sp

G

Thus the Gauss – Seidel scheme also converges. Observe that

spspJG 2= ; spsp

JG <

Thus the Gauss – Seidel scheme converges faster than the Jacobi scheme.

In many class of problems where both schemes converge it is the Gauss – Seidel scheme that converges faster. We shall not go into any further details of this aspect.


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 ω ≠ 0 and multiply both sides of (I) to get an equivalent system,

ωAx = ωy ………………(II)

Now

( )ULDA ++=

We write (II) as

(ωD + ωL + ωU)x = ωy,

i.e.

(ωD + ωL) = - ωUx + ωy

i.e.

(D + ωL)x + (ω-1) Dx = - ωUx +ωy

i.e.

(D + ωL)x = - [(ω – 1)D + ωU]x + ωy

i.e.

x = - (D + ωL)-1 [(ω-1)D + ωU]x + ω [D + ωL]-1y.

We thus get the SOR scheme as

( ) ( )

( ) ;

ˆ0

1

θω

=

+=+

x

yxMx kk

initial guess ……………(III)

where,

( ) ( )[ ]uDLDM ωωωω +−+−= − 11

and

( ) yLDy ωω 1ˆ −+=

Mω is the SOR matrix for the system.



Notice that if ω = 1 we get the Gauss – Seidel scheme. The strategy is to choose ω such that ,1<isM

spω and is al small as possible so that the scheme converges as rapidly as

possible. This is easier said than achieved. How does one choose ω? It can be shown that convergence cannot be achieved if ω ≥ 2. (We assume ω > 0). ‘Usually’ ω is chosen

between 1 and 2. Of course, one must analyse sp

Mω as a function of ω and find that

value ω0 of ω for which this is minimum and work with this value of ω0.

Let us consider an example of this aspect.

Example 6:

Consider the system given in example 5.

For that system,

Mω = - (D +ω L)-1 [(ω-1) D +ωU]

⎟⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

+−+−+−−

+−+−−

+−−

−

=

23243243

23232

22

411

81

21

21

161

41

41

81

81

21

411

81

21

21

41

41

021

411

21

21

00211

ωωωωωωωωωω

ωωωωωωωω

ωωωωω

ωω

and the characteristic equation is

( ) ( ) ( )ωλωλωλωλω MC.................0112116 24224 =++−−+−

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

16(ω-1)4 = 0 ⇒ ω = 1,

i.e. in the Gauss – Seidel case. So let us take ω ≠ 1; so λ = 0 is not a root. So we can divide the above equation (CMω) by ω4λ2 to get

( ) ( ) 01112116 2

22

2

2

=++−

−⎥⎦

⎤⎢⎣

⎡ +−λωλω

λωλω

Setting



( )λωλωµ 2

22 1+−= we get

011216 24 =+− µµ

which is the same as (CJ). Thus

.8090.0;3090.0 ±±=µ

Now

( ) 22

21 µλωλω

=+−

= 0.0955 or 0.6545 ……….(*)

Thus, this can be simplified as

( ) ( ) 21

2222 1411

21

⎭⎬⎫

⎩⎨⎧ −−±−−= ωωµµωωωµλ

as the eigenvalues of Mω.

With ω = 1.2 and using the two values of µ2 in (*) we get,

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

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

Thus

sp

M ω when ω = 1.2 is 0.4545

which is less that sp

J = 0.8090 and sp

G = 0.6545 computed in Examples. Thus for

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

We can show that in this example when ω = ω0 = 1.2596, the spectral radius 0ωM is

smaller than ωM for any other ω. We have

2596.1M = 0.2596

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

Note:

We had sp

M 2.1 = 0.4545



And

spM 2596.1 = 0.2596

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

radius spM ω .

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 α is called an eigenvalue of A if there exists a nonzero nx1 vector x such that

Ax = αx

Example:

Let

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

=7816438449

A

1−=αLet

Consider

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

021

x . We have

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

=021

1021

021

7816438449

Ax

( ) xx α=−= 1

1−=∴α is such that there exists a nonzero vector x such that Ax = αx. Thus α is an eigenvalue of A.

Similarly, if we take α = 3, we find that ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

021

x

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

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

Let α be an eigenvalue of A. Let,

{ }.: xAxCx n αω α =∈=

Then : (i) ωα is nonempty. αωθ ∈= nxQ



α

α

ωαααω

∈+⇒+=+⇒==⇒∈

yxyxyxAyAyxAxyxii )()(,,)(

(iii)

For any constant )(; xxAx κακακκ == )()( xxA κακ =⇒

αωκ ∈⇒ x

Thus ωα is a subspace of Cn. This is called the characteristic subspace or the eigensubspace corresponding to the eigenvalue α.

Example:

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

We want to find all x such that

Ax = -x

i.e., (A+I)x = θ.

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

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

=+=7816448448

IAM

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

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛ −−

⎯⎯ →⎯⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −−−

−→

000000

21

211

000000448

112

13

81

2

RRR

RRM

Thus, 321 21

21 xxx +=

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



⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛ +

201

21

021

212

121

32

3

2

32

xxxx

xx

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

201

021

21 AA

where A1 and A2 are arbitrary constants.

Thus ω-1 consists of all vectors of the form

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

201

021

21 AA .

Note: The vectors form a basis for ω⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

201

,021

-1 and therefore

dim ω-1 = 2.

What is ω3 the eigensubspace corresponding to the eigenvalue 3 for the above matrix We need to find all solutions of Ax = 3x,

i.e., Ax – 3x = θ

i.e., Nx = θ Where

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

=−=48164084412

3 IAN

Again we use row reduction



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−

−

⎯⎯ →⎯

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

−

−+

−

−

→00034

380

4412

34

380

34

380

441243

12

13

32

34

RRRR

RR

N

321 4412 xxx +=∴

32 34

38 xx = 23 2 xx =∴

2221 128412 xxxx =+=∴

12 xx =∴

12312 22; xxxxx ===∴

∴ The general solution is

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

211

21

1

1

1

xx

xx

Thus ω3 consists of all vectors of the form

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

211

κ

Where κ is an arbitrary constant.

Note: The vector forms a basis for ω⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

211

3 and hence

dim. ω3 = 1.

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



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

.;)( θθα ≠=−⇒ andxxIA

The system θα =− xIA )( has at least one nonzero

solution.

nullity (A - αI) ≥ 1

rank (A - αI) < n

(A - αI) is singular

det. (A - αI) = 0

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

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

This (A - αI) is singular

rank (A - αI) < n

nullity (A - αI) ≥ 1

The system θα =− xIA )( has nonzero solution.

α is an eigenvalue of A.

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

Combining the two we get,

α is an eigenvalue of A

det. (A - αI) = 0

det. (αI - A) = 0

Now let C(λ) = det. (λI - A) Thus we see that,

“The eigenvalues of a matrix A are precisely the roots of C(λ) = det. (λI - A)”.



( )

nnnn

n

n

aaa

aaaaaa

C

−−−

−−−−−−

=

λ

λλ

λ

K

KKKK

KKKK

K

K

21

22221

11211

( ) ( ) Aaa nnnn

n .det1111 −++++−= − KK λλ

Thus ; C(λ) is a polynomial of degree n. Note the ‘leading’ coefficient of C(λ) is 1. We say C(λ) is a ‘monic’ polynomial of degree n. This is called CHARACTERISTIC POLYNOMIAL of A. The roots of the characteristic polynomial are the eigenvalues of A. The equation C(λ) = 0 is called the characteristic equation.

Sum of the roots of C(λ) = Sum of the eigenvalues of A = a11 + . . . . . . + ann , and this is called the TRACE of A.

Product of the roots of C(λ) = Product of the eigenvalues of A = det. A. In our example in page 1 we have

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

=7816438449

A

( )7816

438449

).(det−−−−−−+

=−=∴λ

λλ

λλ AIC

781431441

321

−−+−−+−−+

⎯⎯⎯ →⎯ ++

λλλλ

λCCC



( )781

431441

1−−

−−−−

+=λ

λλ

340

010441

)1(12

13 −−+

−−+=→

−

− λλλ

RR

RR

( )( )( )311 −++= λλλ

( ) ( )31 2 −+= λλ

Thus the characteristic polynomial is

( ) ( )31)( 2 −+= λλλC

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 −= λλ)( . . . . . . . . . . . . .(1)

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

( ) ( ) ( ) kak

aaC λλλλλλλ −−−= KK2121)( . . . . . . . .(2)

Where λ1, λ2, . . . . . ., λk are the distinct roots; these distinct roots are the distinct eigenvalues of A and the multiplicities of these roots are called the algebraic multiplicities of these eigenvalues of A. Thus when C(λ) is as in (2), the distinct eigenvalues are λ1, λ2, . . . . . ., λk and 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 −+= λλλC )



Thus the distinct eigenvalues of this matrix are λ1 = -1 ; and λ2 = 3 and their algebraic multiplicities are respectively a1 = 2 ; a2 = 1.

If λi is an eigenvalues of A the characteristic subspace corresponding to λi is iλω and is

defined as

{ }xAxx iiλωλ == :

The dimension of iλω is called the GEOMETRIC MULTIPLICITY of the eigenvalue λi

and is denoted by gi.

Again for the matrix on page 1, we have found on pages 3 and 4 respectively that, dim ω-

1 = 2 ; and dim. ω3 = 1. Thus the geometric multiplicities of the eigenvalues λ1 = -1 and λ2 = 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(λ) as in (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 α1, α2, . . . . . . . ., αs be a distinct scalars, (i.e., αi ≠ αj if i ≠ j ). Consider,

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

)(1121

1121

siiiiiii

siiip

αααααααααααλαλαλαλαλ

λ−−−−−

−−−−−=

+−

+−

KK

KK

)()(

1 ji

j

sjij

αααλ

−

−∏=≠≤≤ for i = 1,2, . . . . . . ., s . . . . . . .. (4)

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

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

( ) 1=iip α



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

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

We call these the Lagrange Interpolation polynomials. If p(λ) is any polynomial of degree ≤ s-1 then it can be written as a linear combination of p1(λ),p2(λ), . . ., ps(λ) as follows:

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

( ) (∑=

=s

iii pp

1λα )

With this preliminary, we now proceed to study the properties of the eigenvalues and eigenvectors of an nxn matrix A.

Let λ1, . . . . , λk be the distinct eigenvalues of A. Let φ1, φ2, . . . , φk be eigenvectors corresponding to these eigenvalues respectively ; i.e., φi are nonzero vectors such that

Aφi = λiφi . . . . . . . . . . . .(6)

From (6) it follows that

iiiiiiii AAAAA φλφλφλφφ 22 )()( ====

iiiiiiii AAAAA φλφλφλφφ 32223 )()( ====

and by induction we get

iim

imA φλφ = for any integer m ≥ 0 . . . . . . . . . . .(7)

(We interpret A0 as I).

Now let,

s

saaap λλλ +++= KK10)(be any polynomial. We define p(A) as the matrix,

s

s AaAaIaAp +++= KK10)(


Now

is

si AaAaIaAp φφ )()( 10 +++= KK

is

sii AaAaa φφφ +++= KK10

by (6) iis

siii aaa φλφλφ +++= KK10

iis

si aaa φλλ )( 10 +++= KK

.)( iip φλ=

Thus,

envalues λ

Np

a

N

Now are the eigenvectors, φ1, φ2, . . . . , φk corresponding to the distinct eig

If λi is any eigenvalue of A and φi is an eigenvector corresponding to λi then

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


1, λ2, . . . . , λk of A, linearly independent ? In order to establish this linear independence, we must show that

0212211 ====⇒=+++ KnKK CCCCCC KK θφφφ . . . (8)

ow if in (4) & (5) we take s = k ; αi = λi then we get the Lagrange Interpolation olynomials as

( ))()(

1 ji

j

kji

ij

pλλλλ

λ−

−∏=≠≤≤ ; i = 1,2,….., k …………(9)

nd

( ) ijjip δλ = if j ≠ i …………(10)

ow,

nkkCCC θφφφ =+++ ....2211

For 1≤ i ≤ k,



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

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

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

;1; kiC ii ≤≤=⇒ θφ by (10)

kiCi ≤≤=⇒ 1;0 since φi are nonzero vectors

Thus

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

Eigen vectors corresponding to distinct eigenvalues of A are linearly independent.



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 nxn matrix 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 the set 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(λ) and CB (λ) be the characteristic polynomials of A and B respectively. We have,

( ) BPPIAIC A1−−=−= λλλ

BPPPP 11 −− −= λ

( )PBIP −= − λ1

PBIP −= − λ1



1sin 1 =−= − PPceBIλ

= CB (λ )

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

k BPPBPPBPPA 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(λ) = a0 + a1λ + ….. + akλ be any polynomial.

Then

( ) kk AaAaIaAp +++= .....10

PBPaPBPaBPPaIa kk

1212

110 ..... −−− ++++=

[ ]PBaBaBaIaP kk++++= − .....2

2101

( )PBpP 1−=

Thus

( ) ( ) nn OPBpPOAp =⇔= −1

( ) nOBp =⇔

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

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

p(A) = On, i.e.



A (A) = {p(λ) : 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 ”. Thus similar matrices have the same set of annihilating polynomials.

We shall discuss more about annihilating polynomials later.

We now investigate the following question? Given an nxn matrix A when is it similar to a “simple matrix”? What are simple matrices? The simplest matrix we know is the zero matrix On. Now A ∼ On . There is a nonsingular matrix P such that A = P-1 On P = On.

∴ “ THE ONLY MATRIX SIMILAR TO On IS ITSELF ”.

The next simple matrix we know is the identity matrix In. Now A ∼ In there is a nonsingular P such that A = P-1 In P A = In.

Thus “THE ONLY MATRIX SIMILAR TO In IS ITSELF ”.

The next class of simple matrices are the DIAGONAL MATRICES. So we now ask the question “ Which type of nxn matrices are similar to diagonal matrices”?

Suppose now A is an nxn matrix; and A is similar to a diagonal matrix,

λ1

D = λ2

λn

(λI not necessarily distinct).

Then there exists a nonsingular matrix P such that

P-1 A P = D

AP = PD ………..(1)

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nnninn

ni

ni

PPPP

PPPPPPPP

LetP

..........

..........

..........

21

222221

111211

MMMMMM



⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nnnn

n

n

aaa

aaaaaa

A

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

.....

.....

21

22221

11211

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

ni

i

i

i

P

PP

LetPM2

1

denote the ith column of P.

Now the ith column of AP is

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

+++

++++++

ninninin

ninii

ninii

PaPaPa

PaPaPaPaPaPa

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

.....

.....

2211

2222121

1212111

which is equal to APi.

Thus the ith column of A P, the L.H.S. of (1), is A Pi.

Now the ith column of P D is

ii

ni

i

i

i

ini

ii

ii

P

P

PP

P

PP

λλ

λ

λλ

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

MM2

1

2

1

Thus the ith column of P D, the R.H.S. of (1), is λI Pi. Since L.H.S. = R.H.S. by (1) we have

APi = λi Pi ; i = 1, 2, …., n ……………..(2)

Note that since P is nonsingular no column of P can be zero vector. Thus none of the column vectors Pi are zero. Thus we conclude that,

“IF A IS SIMILAR TO A DIAGONAL MATRIX D THEN THE DIAGONAL ENTRIES OF D MUST BE THE EIGENVALUES OF A AND IF P-1AP = D THEN THE ith COLUMN VECTOR MUST BE AN EIGENVECTOR CORRESPON DING TO THE EIGENVALUE WHICH IS THE ith DIAGONAL ENTRY OF D”.

Note:



The n columns of P must be linearly independent since P is nonsingular and thus these n columns give us n linearly independent eigenvectors of A

Thus the above result can be restated as follows: A is similar to a diagonal matrix D and P-1 A P = D A has n linearly independent eigenvectors; taking these as the columns of P we get P-1 A P we get D where the ith diagonal entry of D is the eigenvalue corresponding to the ith eigenvector.

Conversely, it is now obvious that if A has n linearly independent eigenvectors then A is similar to a diagonal matrix D and if P is the matrix whose ith column is the eigenvector, then D is P-1 A P and ith diagonal of D is the eigenvalue corresponding to the ith eigenvector.

When does then a matrix have n linearly independent eigenvectors’. It can be shown that a matrix A has n linearly independent eigenvectors the algebraic multiplicity of each eigenvalue of A is equal to its geometric multiplicity. Thus

A IS SIMILAR TO A DIAGONAL MATRIX

FOR EVERY EIGENVALUE OF A, ALGEBRAIC MULTIPLICITY IS EQUAL TO ITS GEOMETRIC MULTPLICITY”.

RECALL; if ( ) ( ) ( ) ( ) kak

aaC λλλλλλλ −−−= .....2121 where λ 1, λ 2, ….., λ k are

the distinct eigenvalues of A, then ai is called the algebraic multiplicity of the eigenvalue λ i. Further, let

{ }xAxx ii λω == :

be the eigensubspace corresponding to λ i. Then gi = dim ωi is called the geometric multiplicity of λ i.

Therefore, we have,

“ If A is an nxn matrix with C(λ ) = (λ -λ 1)a1…… (λ -λ k) ak where λ 1, ….., λ k are the

district eigenvalues of A, then A is similar to a diagonal matrix ai = gi (=dimωi) ; 1≤ i ≤ k”.

Example:

Let us now consider

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

=7816438449

A



On page 6, we found the characteristic polynomial of A as

C(λ ) = (λ +1)2 (λ - 3)

Thus λ 1 = -1 ; a1 = 2

λ 2 = 3 ; a2 = 1

On pages 3 and 4 we found,

ω1 = eigensubspace corresponding to λ = -1

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛==

201

021

: 21 AAxx

ω2 = eigensubspace corresponding to λ = 3

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛==

211

: kxx

Thus dim ω1 = 2 ∴ g1 = 2

dim ω2 = 2 ∴ g2 = 1

Thus a1 = 2 = g1 and ence A must be similar.

a2 = 1 = g2

to a diagonal matrix. How do we get P such that P-1AP is a diagonal matrix? Recall the columns of P must be linearly independent eigenvectors. From ω1 we get two linearly

eigenvectors, namely, and ; and from ω⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

021

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

201

2 we get third as . ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

211

Thus if we take these as columns and write,

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

220102111

P



Then ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−−−

−

=−

1122

1122

1011P ; and it can be verified that

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−−−

−

=−

220102111

7816438449

1122

1122

1011 APP

a diagonal matrix. ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

−=

300010001

Thus we can conclude that A is similar to a diagonal matrix, i.e., P-1 AP = D A has n linearly independent eigenvectors namely the n columns of AP.

Conversely, A has n linearly independent eigenvectors P-1 AP is a diagonal matrix where the columns of P are taken to be the n linearly eigenvectors.

We shall now see a class of matrices for which it is easy to decide whether they are similar to a diagonal matrix; and in which case the P-1 is easy to compute. But we shall first introduce some preliminaries.

If ; are any two vectors in C

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nx

xx

xM2

1

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

ny

yy

yM2

1

n, we define the INNER PRODUCT OF

x with y (which is denoted by (x,y)) as,

( ) nn yxyxyxyx +++= K2211, ∑=

=n

iii yx

1

Example 1:

If ; ; then, ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−+=1

2 ii

x⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=i

iy 11

( )( ) ( )( )iiiiyx 1121.),( −+−++=



( )( ) ( )( ) iiiii 51112 +=−−++++=

Whereas ( ) ( )( ) ( )( ) iiiiixy 511211),( −=−++−+=

We now observe some of the properties of the inner product, below:

(1) For any vector x in Cn, we have

( ) ,,1

2

1∑∑==

==n

ii

n

iii xxxxx

Which is real ≥ 0. Further,

( ) ∑=

=⇔=n

iixxx

1

2 00,

nix i ≤≤=⇔ 1;0

nx θ=⇔

Thus,

(x,x) is real and ≥ 0 and = 0 x = θn

(2) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛== ∑∑

==

n

iii

n

iii xyyxyx

11,

( )xy ,=

Thus,

( ) ( )xyyx ,, =

(3) For any complex number α, we have,

( ) ( ) ∑∑==

==n

iii

n

iii yxyxyx

11

, ααα

( )yx ,α=

Thus



)

(αx,y) = α (x,y) for any complex number α.

We note,

( ) ( xyyx ,, αα = by (2)

( ) ( ) ( )yxxyxy ,,, ααα ===

(4) ( ) ( )∑=

+=+n

iiii zyxzyx

1, ∑ ∑

= =

+=n

i

n

iiiii zyzx

1 1

( ) ( zxyx ,, += )

Thus

(x + y, z) = (x,z) + (y,z) and similarly

(x, y + z) = (x, y) + (x, z)

We say that two vectors x and y are ORTHOGONAL if (x, y) = 0.

Example :

(1) If x = ,0

1;

1

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−iy

ii

then,

( ) ( ) ( ) ( )( )011, iiiyx −++−=

= -1 + 1 = 0

Thus x and y are orthogonal.

(2) If ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

1

1,

1ay

iix



then

( ) iiayx −+−= 1,

∴ x, y orthogonal

( )

( )

ia

iiii

ia

iai

+=⇔

−=+−=⎟⎠⎞

⎜⎝⎛ +

=⇔

=++−⇔

1

11101


Vittal rao/IISc.Bangalore M3/L4/V1/May 2004/1

HERMITIAN MATRICES

Let A = (aij); be an nxn matrix. We define the Hermitian conjugate of A, denoted by A* as ; A* = (a*ij) where a*

ij = aji.

A* is the conjugate of the transpose of A.

Example 1:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=iii

A1

Transpose of ⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

iii

A1

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=∴ii

iA

1*

Example 2:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=2

1i

iA

Transpose of ⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

21i

iA

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=∴2

1*

ii

A

Observe that in Example 1. A* ≠ A. Whereas in Example 2, A* = A.

DEFINITION: An nxn matrix A is said to be HERMITIAN if

A* = A.

We now state some properties of Hermitian matrices.

(1) If A = (aij) ; A* = (a*ij) and A = A* then aii = a*

ii = aii

Thus the DIAGONAL ENTRIES OF A HERMITIAN MATRIX ARE REAL.



(2) Let be any two vectors in C

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nn y

yy

y

x

xx

xMM

2

1

2

1

; n.

Let

( )( )

( )

( )( )

( ) ⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nn Ay

AyAy

Ay

Ax

AxAx

AxMM

2

1

2

1

;

We have

( ) ( ) .;11

∑∑==

==n

iijij

n

jjiji yaAyxaAx

Now

( ) ( )∑=

=n

iii yAxyAx

1,

∑ ∑= =

⎟⎟⎠

⎞⎜⎜⎝

⎛=

n

ii

n

jjij yxa

1 1

∑ ∑= =

⎟⎠

⎞⎜⎝

⎛=

n

j

n

iiijj yax

1 1

∑ ∑= =

⎟⎠

⎞⎜⎝

⎛=

n

j

n

iiijj yax

1 1

∑ ∑= =

⎟⎠

⎞⎜⎝

⎛=

n

j

n

iijij yax

1 1)sin( *AceAaa jiij ==Q



( )∑=

=n

jjj Ayx

1

= (x, Ay)

Thus IF A IS HERMITIAN THEN

(Ax, y) = (x, Ay)

FOR ANY TWO VECTORS x, y.

(3) Let λ be any eigenvalue of A. Then there is an x ∈ Cn, x ≠ θn such that

Ax = λx. Now,

( ) ( ) (

( )

)

( )xx

xxAxx

xAxxxxx

,

,),(

,,,

λ

λ

λλ

=

==

==

A is Hermitian.

( )( ) ( ) nxxxButxx θλλ ≠≠=−∴ Q0,.0,

λλλλ =∴=−∴ 0 ∴ λ is real.

THUS THE EIGENVALUES OF A HERMITIAN MATRIX ARE ALL REAL.

(4) Let λ, µ be two different eigenvalues of A and x, y corresponding eigenvectors. We have,

Ax = λx and Ay = µy and λ, µ are real by (3). Now,

( ) ( yxyx ,, )λλ =



( )( )( )

( )( ) .,

,

,)2(,

,

isrealyxyx

yxbyAyx

yAx

µµµ

µ

Q==

===

( )( ) µλµλ ≠=−∴ Butyx .0,

∴ (x,y) = 0 x and y are orthogonal.

THUS IF A IS A HERMITIAL MATRIX THEN THE EIGENVECTORS CORRESPONDING TO DISTINCT EIGENVALUES ARE ORTHOGONAL.



Gramm – Schmidt OrthonormalizationWe shall now discuss the Gramm – Schmidt Orthonormalization process:

Let U1, U2, …., Uk be k linearly independent vectors in Cn. The Gramm – Schmidt process is the method to get an orthonormal set kφφφ ,.....,, 21 show that the subspace ω

spanned by U1, ….., Uk is the same as the subspace spanned by kφφ ,.....,1 thus providing an orthonormal basis for ω.

The process goes as follows:

Let ;11 U=ψ

( )1

, 111

1

1

11 === φ

ψψψ

ψψ

φ Note

Next, let,

( ) 11222 φφψ UU −=

Note that

( )12φψ

( ) ( )( )112212 ,, φφφφ UU −=

( ) ( )( 112212 , )φφφφ UU −=

( ) ( ) ( ) 1, 111212 =−= φφφφ QUU

.12 φψ ⊥∴

Let

;2

22 ψ

ψφ = clearly ( ) 0,,1,1 2112 === φφφφ

Also

x = α1 U1 + α2 U2 then

( )( )1122211 , φφψαψα Ux ++=⇔

( )[ ]112222111 , φφφψαφψα Ux ++=⇔



2211 φβφβ +=⇔ x where

( )122111 ,φαψαβ U+=

222 ψαβ =

Thus xε subspace spanned by U1, U2

xε subspace spanned by φ1, φ2.

Thus φ1, φ2 is an orthonormal basis for the subspace [U1,U2].

Having defined φ1, φ2,….., φi-1 we define φi as follows:

( )∑−

=

−=1

1

,i

piiiii UU φφψ Clearly ( ) 0, =pi φψ 1≤ p ≤ i-1

and i

ii ψ

ψφ =

Obviously ( ) 110,1 −≤≤== ijforand jii φφφ

and xε [U1, U2, ….., Ui] xε [φ1, ….., φi]

and thus φ1, φ2, ….., φi is an orthonormal basis for [U1, ….., Uk]. Thus at the kth stage we get an orthonormal basis φ1, …., φk for [U1, ….., Uk].

Example:

Let

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

0132

;

01

11

;

0111

321 UUU

be l.i. Vectors in R4. Let us find an orthonormal basis for the subspace ω spanned by U1, U2, U3 using the Gramm – Schmidt process.

;

0111

11

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

== Uψ ( )⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

==

0111

31

, 11

11 ψψ

ψφ



⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

=

∴

03

13

13

1

1φ

( ) 11222 , φφψ UU −=

⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−=

03

13

13

1

31

31

31

01

11

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−=

0313

13

1

01

11

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−=

03

43

23

2

and 3

629

1694

94

2 =++=ψ

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−==∴

06

26

16

1

03

43

23

2

623

2

22 ψ

ψφ



Thus

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−=

06

26

16

1

2φ

Finally,

( ) ( ) 22311333 ,, φφφφψ UUU −−=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

06

26

16

1

63

03

13

13

1

36

0132

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

012

12

1

0222

0132

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛ −

=

002

12

1

21

21

41

41

3 ==+=ψ

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛ −

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛ −

==∴

00

21

21

002

12

1

23

33 ψ

ψφ



Thus the required orthonormal basis for ω, the subspace spanned by U1,U2, U3 is φ1, φ2, φ3, where

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛−

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

00

21

21

;

06

26

16

1

;

03

13

13

1

321 φφφ

Note that these φi are mutually orthogonal and have, each, ‘length’ one.

We now get back to Hermitian matrices. We had seen that the eigenvalues of a Hermitian matrix are all real; and that the eigenvectors corresponding to district eigenvalues are mutually orthogonal. We can further show the following: (We shall not give a proof here, but illustrate with an example).

Let A be any nxn Hermitian matrix. Let

( ) ( ) ( ) ( ) kak

aaC λλλλλλλ −−−= .....2121 be its characteristic polynomial,

where λ1, λ2, ….., λk are its distinct eigenvalues and a1, ….., ak are their algebraic multiplicities. If ωi is the characteristic subspace corresponding to the eigen value λi ; that is,

{ }xAxx ii λω == :

then it can be shown that dim is ωi = ai.

We then choose any basis for ωi and orthonormalize it by G-S process and get an orthonormal basis for ωi. If we now take all these orthonormal basis vectors for ω1, . . ., ωk and write them as the columns of a matrix P then

P*AP

Will be a diagonal matrix.

Example :

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

−=

312132

226A

Notice

A* = A1 = A1 = A.

Thus the matrix A is Hermitian.



Characteristic Polynomial of A:

312132226

−−−

−−=−

λλ

λλ AI

( )

3121320222

21 2

−−−−−

⎯⎯⎯ →⎯ +

λλλλ

RR

( )312

132021

2−−

−−=λ

λλ

( )350

170021

212

13

2

2 −−−=→

−

+ λλλ

RR

RR

( ) ( )( )[ ]5372 −−−−= λλλ

( )[ ]16102 2 +−−= λλλ

( )( )( 822 −−−= )λλλ

( ) ( 82 2 −−= λλ ) Thus

( ) ( ) ( )82 2 −−= λλλC

∴ λ1 = 2 a1 = 2



λ2 = 8 a2 = 1 The characteristic subspaces:

{ }xAxx 2:1 ==ω

( ){ }θ=−= xIAx 2:

i.e. We have to solve

(A – 2I) x = θ

i.e. ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

−

000

112112

224

3

2

1

xxx

2x1 – x2 + x3 = 0

x3 = - 2x1 + x2

arbitraryxxxx

xx

x 21

21

2

1

,;2 ⎟

⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+−=∴

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+−==∴ scalarsxx βα

βαβα

ω ,;2

:1

∴ A basis for ωi is

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

110

;2

01

21 UU

We now orthonormalize this:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−==

201

11 Uψ 51 =ψ 1

11 ψ

ψφ =



⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

=∴

52

05

1

1φ

( ) 11222 , φφψ UU −=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

52

05

1

52

110

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

54

052

110

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=

5115

2

530

2530

2511

254

2 ==++=ψ

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

==∴

301

305

302

5115

2

305

2

22 ψ

ψφ

∴ φ1, φ2 is an orthonormal basis for ω1.

{ }xAxx 8:2 ==ω

( ){ }θ=−= xIAx 8:

So we have to solve

(A-8I) x = θ i.e.



⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−−−

−−

000

512152

222

3

2

1

xxx

This yields x1 = -2x2 = 2x3 and therefore the general solution is

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−11

2

2

21γ

γ

γγ

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=∴11

2: 3UBasis

∴ Orthonormalize: only one step:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−==11

2

33 Uψ

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−==

61

61

62

11

2

61

3

33 ψ

ψφ

∴ If

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

−=

61

301

52

61

3050

62

302

51

P

Then

P* = P1 and

;800020002

1*

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛== APPAPP adiagonal matrix.



VECTOR AND MATRIX NORMS

Consider the space,

,,; 212

12

⎭⎬⎫

⎩⎨⎧

∈⎟⎟⎠

⎞⎜⎜⎝

⎛== Rxx

xx

xR

our ‘usual’ two-dimensional plane. If x = is any vector in this space we

defineits‘usual’ ‘length’ or ‘norm’ as

⎟⎟⎠

⎞⎜⎜⎝

⎛

2

1

xx

22

12 xxx +=

We observe that

(i) 0≥x for every vector x in R2

0≥x if and only if x is θ;

(ii) xx αα = for any scalar α; for any vector x.

(iii) yxyx +≤+ for any two vectors x and y. (The inequality (iii) is usually referred to as the triangle inequality).

We now generalize this idea to define the concept of a norm on Cn or Rn.

The norm on a vector space V is a rule which associates with each vector x in V, a real number x satisfying,

(i) 0≥x for every x ε V and

0≥x if and only if x = θ;

(ii) xx αα = for every scalar α and every vector x in V,

(iii) yxyx +≤+ for every x, y in V.

Examples of Vector Norms on Cn and Rn

Let be any vector x in C

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nx

xx

xM2

1

n (or Rn)



We can define various norms as follows:(1)

[ ] 21

1

221

222

212

..... ⎥⎦

⎤⎢⎣

⎡=+++= ∑

=

n

iin xxxxx

(2) ∑=

=+++=n

iin xxxxx

1211

....

In general for 1 ≤ p < ∞ we can define,

(3) pn

i

pip

xx

1

1 ⎭⎬⎫

⎩⎨⎧

= ∑=

If we set p = 2 in (3) we get 2

x as in (1) and if we set p = 1 in (3) we get 1

x as in (2).

(4) { }nxxxx ,.....,,.max 21=∞

All these can be verified to satisfy the conditions (i), (ii) and (iii) required of a norm. Thus these give several types of norms on Cn and Rn.

Example:

(1) Let in R⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−=

12

1x 3

Then

( ) 621141

4121

2

1

=++=

=++=

x

x

{ }

( ) 41

41

4444

18121

21,2,1.max

=++=

==∞

x

x



(2) Let in C⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

iix2

13

Then

( ) 6141

4121

21

2

1

=++=

=++=

x

x

{ }

( ) 31

31

10121

21,2,1.max

3333

=++=

==∞

x

x

Consider a sequence of vectors in C{ } 1)(

=∞

kkx n (or Rn)

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nk

k

k

k

x

xx

x

)(

2)(

1)(

)(

M

Suppose )(2

1

nn

n

orRC

x

xx

x ∈

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=M

DEFINITION:

We say that the sequence { })( kx of vectors CONVERGES to the vector x if the

sequence of numbers, { }1)( kx converges to the number x1; { }2

)( kx converges to x2, ….

and { }nkx )( converges to xn i.e.

As k → ∞; x(k)i → xi for every i=1, 2, …., n.



Example:

Let

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

+

−=

11

21

2

)(

k

k

ki

x k be a sequence of vectors in R3.

Let . Here ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

010

x 11)( 01 x

kx k =→=

22)( 121 x

kx k =→−=

323)( 0

11 x

kx k =→

+=

forxx iik →∴ )( I=1,2,3.

xx k →∴ )(

If { })(kx is a sequence of vectors such that in some norm, the sequence of real numbers, xx k −)( converges to the real number 0 then we say that the sequence of vectors

converges to x with respect to this norm. We then write,

xx k ⎯→⎯)(

For example consider the sequence,

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

+

−=

11

21

1

2

)(

k

k

kx k in R3 as before and,

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

010

x

We have



( )

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

+

−=−

11

2

1

2k

k

kxx k

Now

01

12121

)( →+

++=−kkk

xx k

xx k ⎯→⎯∴ 1)(

Similarly

021

1,2,1.max 2)( →=

⎭⎬⎫

⎩⎨⎧

+=−

∞ kkkkxx k

xx k ⎯⎯ →⎯∴ ∞)(

( ) 01

121 21

22222

)( →⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+++=−

kkkxx k

xx k ⎯→⎯∴ 2)(

Also,

( ) 01

1211

2

)( →⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

++⎟

⎠⎞

⎜⎝⎛+=−

p

p

p

pp

k

kkkxx

xx pk ⎯⎯→⎯∴ )( ∞≤≤∀ pp 1;

It can be shown that

“ IF A SEQUENCE { })(kx OF VECTORS IN Cn (or Rn) CONVERGES TO A VECTOR x IN Cn (or Rn) WITH RESPECT TO ONE VECTOR NORM THEN THE SEQUENCE CONVERGES TO x WITH RESPECT TO ALL VECTOR NORMS AND ALSO THE



SEQUENCE CONVERGES TO x ACCORDING TO DEFINITION ON PAGE 40. CONVERSELY IF A SEQUENCE CONVERGES TO x AS PER DEFINITION ON PAGE 40 THEN IT CONVERGES WITH RESPECT TO ALL VECTOR NORMS”.

Thus when we want to check the convergence of a sequence of vectors we can choose that norm which is convenient to that sequence.

MATRIX NORMS

Let M be the set of all nxn matrices (real or complex). A matrix norm is a rule, which associates a real number A with each matrix A and satisfying,

(i) 0≥A for all matrices A

0=A if and only if A = On,

(ii) AA αα = for every scalar α and every matrix A,

(iii) BABA +≤+ for all matrices A and B,

(iv) BAAB ≤ for all matrices A and B.

Before we give examples of matrix norms we shall see a method of getting a matrix norm starting with a vector norm.

Suppose . is a vector norm. Then, consider x

Ax (where A is an nxn matrix); for x ≠

θn. This given us an idea to by what proportion the matrix A has distorted the length of x. Suppose we take the maximum distortion as we vary x over all vectors. We get

nx θ≠

max

xAx

a real number. We define

=Anx θ≠

max

xAx

We can show this is a matrix norm and this matrix norm is called the matrix norm subordinate to the vector norm . We can also show that



=Anx θ≠

max

xAx

= 1max

=x Ax

For example,

=1

A 1max

1=x 1

Ax

=A 1max

2=x 2

Ax

=A 1max

=∞

x ∞Ax

=A 1max

=p

x pAx

How hard on easy is it to compute these matrix norms? We shall give some idea of computing

1A ,

∞A and

2A for a matrix A.

Let

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nnnn

n

n

aaa

aaaaaa

A

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

.....

.....

21

22221

11211

The sum of the absolute values of the entries in the ith column is called the absolute column sum and is denoted by Ci. We have

∑=

=++++=n

iin aaaaaC

1113121111 .....

∑=

=++++=n

iin aaaaaC

1223222122 .....

….. ….. ….. ….. ….. ….. …..



∑=

=n

iijj aC

1 ; 1 ≤ j ≤ n

Thus we have n absolute column sums, C2 , C2, ….., Cn.

Let

{ }nCCCC ,.....,,.max 21=

This is called the maximum absolute column sum. We can show that,

{ }nCCCA ,.....,.max 11==

nj ≤≤=1

max⎥⎦

⎤⎢⎣

⎡ ∑=

n

iija

1

For example, if

,423

101321

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

−=A

then

and C = max. {5, 4, 8} = 8 8413

;4202;5311

3

2

1

=++==++=

=++=

CCC

81=∴ A

Similarly we denote by Ri the sum of the absolute values of the entries in the ith

row ∑=

=+++=n

jjn aaaaR

11112111 .....

∑=

=+++=n

jjn aaaaR

12222212 .....

….. ….. ….. ….. ….. ….



∑=

=+++=n

jijiniii aaaaR

121 ..... and define R, the maximum absolute row

sum as, R = max {R1, ….., Rn}

It can be show that,

{ }nRRRA ,.....,max 1==∞

ni ≤≤=1

max

⎭⎬⎫

⎩⎨⎧∑=

n

jija

1

For example, for the matrix

,423

101321

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

−=A we have

R1 = 1 + 2 + 3 = 6;

R2 = 1 + 0 + 1 =2; and R = max {6, 2, 9}= 9

R3 = 3 + 2 + 4 = 9

9=∴∞

A

The computation of 1A and ∞

A for a matrix are thus fairly easy. However, the

computation of 2

A is not very easy; but somewhat easier in the case of the Hermitian matrix.

Let A be any nxn matrix; and

( ) ( ) ( ) kak

aC λλλλλ −−= L11 , be its characteristic polynomial, where

λ1, λ2, ….., λk are the district characteristic values of A.

Let

{ }kP λλλ ,,,.max 21 K=

This is called the spectral radius of A and is also denoted by spA



It can be show that for a Hermitian matrix A,

2A = P = sp

A

For example, for the matrix,

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

−=

312132

226A

which is Hermitian we found on page 33, the district eigenvalues as λ1 = 2; λ2 = 8

{ 8,2.max== }∴ PAxp = 8

82

==∴sp

AA

If A is any general nxn matrix (not Hermitian) then let B = A* A. Then B* = A* A = B, and hence B is Hermitian and its eigenvalues are real and in fact its eigenvalues are nonnegative. Let the eigenvalues (district) of B be µ1, µ2, ….., µr. Then let

µ = max {µ1, µ2, ….., µr}

We can show that

{ }nA µµµ ,.....,max 12==

If follows from the matrix norm definition subordinate to a vector norm, that

=Anx θ≠

max

xAx

∴ For any x in Cn or Rn , we have, if x ≠ θn

≤x

Axnx θ≠

maxA

xAx

=

and therefore

xAAx ≤ for all x ≠ θn

But this is obvious for x = θn

Thus if A is a matrix norm subordinate to the vector norm x then



xAAx ≤

for every vector x in Cn (or Rn).

Numerical analysis/ Computations of eigenvalues Lecture notes


COMPUTATION OF EIGEN VALUES

In this section we shall discuss some standard methods for computing the eigenvalues of an nxn matrix. We shall also briefly discuss some methods for computing the eigenvectors corresponding to the eigenvalues.

We shall first discuss some results regarding the general location of the eigenvalues.

Let A = (aij) be an nxn matrix; and let λ1, λ2, ….., λn be its eigenvalues (including multiplicities). We defined

{ }nxpAP λλλ ,.....,,max 21==

Thus if we draw a circle of radius P about the origin in the complex plane, then all the eigenvalues of A will lie on or inside this closed disc. Thus we have

(A) If A is an nxn matrix then all the eigenvalues of A lie in the closed disc { }P≤λλ : in the complex plane.

This result give us a disc inside which all the eigenvalues of A are located. However, to locate this circle we need P and to find P we need the eigenvalues. Thus this result is not practically useful. However, from a theoretical point of view, this suggests the possibility of locating all the eigenvalues in some disc. We shall now look for other discs which can be easily located and inside which the eigenvalues can all be trapped.

Let A be any matrix norm. Then it can be shown that AP ≤ . Thus if we draw a disc

of radius A and origin as center then this disc will be at least as big as the disc given in (A) above and hence will trap all the eigenvalues. Thus, the idea is to use a matrix norm, which is easy to compute. For example we can use

∞A or

1A which are easily

computed as MARS or MACS respectively. Thus we have,(B) If A is an nxn matrix then all its eigenvalues are trapped in the closed disc { }

∞≤ Aλλ : or the disc

{ }1

: A≤λλ . (The idea is to use ∞

A if it is smaller than 1

A if it is smaller than

∞A ).

COROLLORY

(C) If A is Hermitian, all its eigenvalues are real and hence all the eigenvalues lie in the intervals,

{ }pp ≤≤− λλ : by (A)



{ }{ }

11:

:

AA

AA

≤≤−

≤≤−∞∞

λλ

λλ by (B).

Example 1:

Let ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

−=

021321211

A

Here ‘Row sums’ are P1 = 4 P2 = 6 P3 = 3

6==∴∞

MARSA

Thus the eigenvalues are all in the disc ; { }6: ≤λλ

The ‘Column sums’ are C1 = 3, C2 = 5, C3 = 5.

51

==∴ MACSA

∴ The eigenvalues are all in the disc, { },5: ≤λλ

In this example ;651

=<=∞

AA and hence we use 1

A and get the smaller disc

{ ,5: ≤λλ } inside which all eigenvalues are located.

The above results locate all the eigenvalues in one disc. The next set of results try to isolate these eigenvalues to some extent in smaller discs. These results are due to GERSCHGORIN.

Let A = (aij) be an nxn matrix.

The diagonal entries are

;111 a=ξ ;222 a=ξ ….., ;nnn a=ξ

Now let Pi denote the sum of the absolute values of the off-diagonal entries of A in the ith row.

iniiiiiii aaaaaP ++++++= +− .......... 1121



Now consider the discs:

{ }11111 ::;: PradiusPCentreG ≤−ξλλξ

{ }222212 ::;: PradiusPCentreG ≤−ξλλξ

….. ….. ….. ….. ….. ….. ….

and in general

{ }iiiii PradiusPCentreG ≤−ξλλξ ::;:

Thus we get n discs G1, G2, ….., Gn. These are called the GERSCHGORIN DISCS of the matrix A.

The first result of Gerschgorin is the following:

(D) Every eigenvalue of A lies in one of the Gerschgorin discs.

Example 2:

Let ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

−=

513140011

A

The Gerschgorin discs are found as follows:

ξ1 = (1,0) ; ξ2 = (4,0) ; ξ3 = (-5,0)

P1 = 1 ; P2 = 1 ; P3 = 4

G1 : Centre (1,0) radius 1

G2 : Centre (4,0) radius 1

G3 : Centre (-5,0) radius 4.



G1 (1,0) G2 (4,0)

G3 (-5,0)

Thus every eigenvalue of A must lie in one of these three discs.

Example 3:

Let ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

2035.15.0101

1410A

(It can be shown that the eigenvalues are exactly λ1 = 8, λ2 = 12, λ3 = 20).

Now for this matrix we have,

ξ1 = (10,0) ξ2 = (10,0) ξ3 = 20

P1 = 5 P2 = 1.5 P3 = 4.5

Thus we have the three Gerschgorin discs

{ }{ }{ }5.420:

5.110:

510:

3

2

1

≤−=

≤−=

≤−=

λλ

λλ

λλ

G

G

G



G1 G3

Thus all the eigenvalues of A are in these discs. But notice that our exact eigenvalues are 8,12 and 20. Thus no eigenvalue lies in G2; and one eigenvalue lie in G3 (namely 20) and two lie in G1 (namely 8 and 12).

Example 4:

Let ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

501021101

A

Now,

ξ1 = (1,0) ξ2 = (2,0) ξ3 = (5,0) P1 = 1 P2 = 1 P3 = 1

The Gerschgorin discs are

{ }{ }{ }15:

12:

11:

3

2

1

≤−=

≤−=

≤−=

λλ

λλ

λλ

G

G

G



G1 (1,0) G2 (2,0) G3 (5,0)

Thus every eigenvalue of A must lie in one of these three discs.

In example 2, all the Gerschgorin discs were isolated; and in examples 3 and 4 some discs intersected and others were isolated. The next Gerschgoin result is to identify the location of the eigenvalues in such cases.

(E) If m of the Gerschgorin discs intersect to form a common connected region and the remaining discs are isolated from this region then exactly m eigenvalues lie in this common – region. In particular if Gerschgorin disc is isolated from all the rest then exactly one eigenvalue lies in this disc.

Thus in example 2 we have all three isolated discs and thus each disc will trap exactly one eigenvalue.

In example 3; G1and G2 intersected to form the connected (shaded) region and this is isolated from G3. Thus the shaded region has two eigenvalues and G3 has one eigenvalue.

In example 4, G1and G2 intersected to form a connected region (shaded portion) and this is isolated from G3. Thus the shaded portion has two eigenvalues and G3 has one eigenvalue.

REMARK:

In the case of Hermitian matrices, since all the eigenvalues are real, the Gerschgorin discs, { } { iiiiii PPaG ≤−=≤−= ξλλλλ :: } can be replaced by the Gerschgorin intervals,

{ } { }iiiiiii PPPG +≤≤−=≤−= ξλξλξλλ ::

Example 5:



Let ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−−

−=

2101051111

A

Note A is Hermitian. (In fact A is real symmetric)

Here; ξ1 = (1,0) P1 = 2

ξ2 = (5,0) P2 = 1

ξ3 = (-1/2,0) P3 = 1 Thus the Gerschgorin intervals are

G1 : -1≤ λ ≤ 3

G2 : 4 ≤ λ ≤ 6

G3 : -3/2 ≤ λ ≤ ½

-2 -1 0 1 2 3 4 5 6

G1 G2

G3

Note that G1 and G3 intersect and give a connected region, -3/2 ≤ λ ≤ 3; and this is isolated from G2 : 4 ≤ λ ≤ 6. Thus there will be two eigenvalues in –3/2 ≤ λ ≤ 3 and one eigenvalue in 4 ≤ λ ≤ 6. All the above results (A), (B), (C), (D), and (E) give us a location of the eigenvalues inside some discs and if the radii of these discs are small then the centers of these circles give us a good approximations of the eigenvalues. However if these discs are of large radius then we have to improve these approximations substantially. We shall now discuss this aspect of computing the eigenvalues more accurately. We shall first discuss the problem of computing the eigenvalues of a real symmetric matrix.



COMPUTATION OF THE EIGENVALUES OF A REAL SYMMETRIC MATRIX

We shall first discuss the method of reducing the given matrix to a similar tridiagonal matrix and then computing the eigenvalues of a real symmetric tridiagonal matrix. Thus the process of determining the eigenvalues of A = (aij), a real symmetric method involves two steps:

STEP 1:

Find a real symmetric tridiagonal matrix T which is similar to A.

STEP 2:

Find the eigenvalues of T. (The eigenvalues of A will be same as those of T since A and T are similar).

We shall first discuss step 2.

Numerical analysis/ Computations of eigenvaues Lecture notes

VittalRao/IISc Bangaolre M4/L3/V1/May 2004/1

DETERMINATION OF THE EIGENVALUES OF A REAL SYMMETRIC TRIDIAGONAL MATRIX

Let

⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

=

−

−−−

nn

nnn

abbab

babbab

ba

T

1

112

332

221

11

0..........000..........0

...................................0.....000.....000....000

be a real symmetric tridiagonal matrix.

Let us find Pn (λ) = det [T - λI]

λ

λ

λλ

−−

−−−

−

−−

=

nn

nnn

abbab

babba

1

112

221

11

0..........00.....0

..............................0.....00..........0

The eigenvalues of T are precisely the roots of Pn (λ) = 0

(Without loss of generality we assume bi ≠ 0 for all i. For if bi = 0 for some i then the above determinant reduces to two diagonal blocks of the same type and thus the problem reduces to that of the same type involving smaller sized matrices).

We define Pi (λ) to be the ith principal minor of the above determinant. We have

( )( )( ) ( ) ( ) ( )λλλλ

λλλ

212

1

11

0 1

−−− −−=

−==

iiiii PbPaP

aPP

…….. (I)

What we are interested in finding the zeros of ( )λnP . To do this we analyse the polynomials ( ) ( ) ( )λλλ nPPP ,.....,, 10 .

Let C be any real number. Compute ( ) ( ) ( )CPCPCP n,.....,, 10 (which can be calculated recursively by (I)). Let N (C) denote the agreements in sign between two consecutive in


VittalRao/IISc Bangaolre

the above sequence of values, ( ) ( ) ( )CPCPCP n,.....,, 10 . [If for some i, , we take its sign to be the same as that of

( ) 0=CPi

( )CPi 1− ]. Then we have

(F) There are exactly N (C) eigenvalues of T that are ≥ C.

Example:

If for an example we have an 8 x 8 matrix T (real symmetric tridiagonal)

giving use to,

( )( )( )( )( )( )( )( )( ) 21

4101

1161

2131

2111

8

7

6

5

4

3

2

1

0

−===−=

=−=−=

==

PPPPPPPPP

Here

( ) ( )( ) ( )( ) ( )1,1

1,11,1

65

32

10

PPPPPP

(Because since P6 (1) = 0 we h

Thus three pairs of sign agreeeigenvalues of T greater than o

It is this idea of result (F) that repeated applications of (F) thmeans of an example.

agree in sign

agree in sign

M4/L3/V1/May 2004/2

ave to take its sign as the same as that of P5 (1).

ments are achieved. Thus N (C) = 3; and there will be 3 r equal to 1; and the remaining 5 eigen values are < 1.

will be combined with (A), (B), (C), (D) and (E) and clever at locate the eigenvalues of T. We now explain this by



Example 7:

Let

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

−=

31001240

04120021

T

Here we have

Absolute Row sum 1 = 3




and therefore,

7==∞

MARST

(Note since T is symmetric we have MARS = MACS and therefore TTT ==∞1

). Thus by our result (C) we have that the eigenvalues are all in the interval –7 ≤ λ ≤ 7

[ ]

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Now the Gerschgorin (discs) intervals are as follows:

G1 : Centre 1 radius : 2 ∴ G1 : [-1, 3]

G2 : Centre -1 radius : 6 ∴ G2 : [-7, 5]

G3 : Centre 2 radius : 5 ∴ G3 : [-3, 7]

G4 : Centre 3 radius : 1 ∴ G4 : [-2, 4]



G3

G1

[ ]

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

G4 G2

We see that G1, G2, G3 and G4 all intersect to form one single connected region [-7, 7]. Thus by (E) there will be 4 eigenvalues in [-7, 7]. This gives therefore the same information as we obtained above using (C). Thus so far we know all eigenvalues are in [-7, 7]. Now we shall see how we use (F) to locate the eigenvalues.

First of al let us see how many eigenvalues will be ≥ 0. Let C = 0. Find N (0) and we will get the number of eigenvalues ≥ 0 to be N (0).

Now

λλ

λλ

λ

−−−−

−−−

=−

31001240

04120021

IT

( ) 10 =λP ( ) λλ −= 11P

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )λλλλ

λλλλλλλλ

234

123

012

316241

PPPPPPPPP

−−=−−=−+−=

Now, we have,

( )( )( )( )( ) 730

26050

1010

4

3

2

1

0

−=−=−=

==

PPPPP



We have

as three consecutive pairs having sign agreements.

( ) ( )( ) ( )( ) ( )0,0

0,00,0

43

32

10

PPPPPP

( ) 30 =∴ N

∴ Three are 3 eigenvalues ≥ 0 and ∴ one eigenvalue < 0.

i.e. there are eigenvalues in [0, 7]and there is 1 eigenvalue in [-7, 0]

One eigenvalue 3 eigenvalues

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Fig.1

Let us take C = -1 and calculate N (C). We have

( )( )( )( )( ) 1881

48141

2111

4

3

2

1

0

−=−−=−−=−

=−=−

PPPPP

Again we have N (-1) = 3. ∴ There are 3 eigenvalues ≥ -1 compare this with figure1. We get

One eigenvalue 3 eigenvalues



-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

(Fig.2)

Let us take the mid point of [-7, -1] in, which the negative eigenvalue lies.

So let C = -4.

P0 (-4) = 1 Again there are three pairs of sign agreements.

P1 (-4) = 5 ∴ N (-4) = 3. ∴ There are 3 eigenvalues ≥ -4.

P2 (-4) = 11 Comparing with fig. 2 we get

P3 (-4) = -14

P4 (-4) = -109

that the negative eigenvalue is in [-7, -4] ………..(*)

Let us try mid pt. C = -5.5

We have

P0 (-5.5) = 1

P1 (-5.5) = + 6.5 ∴ N (-5.5) = 4. ∴ 4 eigenvalues ≥ -5.5.

P2 (-5.5) = 25.25 Combining this with (*) and fig. 2 we get

P3 (-5.5) = 85.375 that negative eigenvalue is in [-5.5 – 4].

P4 (-5.5) = 683.4375

We again take the mid pt. C and calculate N (C) and locate in which half of this interval does this negative eigenvalue lie and continue this bisection process until we trap this negative eigenvalue in as small an interval as necessary.

Now let us look at the eigenvalues ≥ 0. We have from fig. 2 three eigenvalues in [0, 7]. Now let us take C = 1

P0 (1) = 1

P1 (1) = 0 ∴ N (1) = 3



P2 (1) = - 4 ∴ all the eigenvalues are ≥ 1 ………….. (**)

P3 (1) = - 4

P4 (1) = - 4

C = 2

P0 (2) = 1

P1 (2) = -1 ∴ N (2) = 2∴ There are two eigenvalues

P2 (2) = - 1 ≥ 2. Combining this with (**) we get one

P3 (2) = 16 eigenvalue in [1, 2) and two in [2, 7].

P4 (2) = 17

C = 3

P0 (3) = 1 ∴ N (3) = 1 ∴ one eigenvalue ≥ 3

P1 (3) = -2 Combining with above observation we get

P2 (3) = 4 one eigenvalue in [1, 2)

P3 (3) = 28 one eigenvalue in [2, 3)

P4 (3) = - 4 one eigenvalue in [3, 7)

Let us locate the eigenvalue in [3, 7] a little better. Take C = mid point = 5

P0 (5) = 1

P1 (5) = - 4 ∴ N (5) = 1

P2 (5) = 20 ∴ this eigenvalue is ≥ 5

P3 (5) = 4

P4 (5) = -28

∴ This eigenvalue is in [5, 7]

Let us take mid point C = 6

P0 (6) = 1

P1 (6) = - 5 ∴ N (6) = 0



P2 (6) = 31 ∴ No eigenvalue ≥ 6

P3 (6) = - 44 ∴ the eigenvalue is in [5, 6)

P4 (6) = 101

Thus combining all, we have,

one eigenvalue in [-5.5, -4)

one eigenvalue in [1, 2)



Each one of these locations can be further narrowed down by the bisection applied to each of these intervals.

We shall now discuss the method of obtaining a real symmetric tridiagonal T similar to a given real symmetric matrix A.

Numerical analysis/Computations of eigenvaues Lecture notes


TRIDIAGONALIZATION OF A REAL SYMMETRIC MATRIX

Let A = (aij) be a real symmetric nxn matrix. Our aim is to get a real symmetric tridiagonal matrix T such that T is similar to A. The process of obtaining this T is called the Givens – Householder scheme. The idea is to first find a reduction process which annihilates the off – tridiagonal matrices in the first row and first column of A and repeatedly use this idea. We shall first see some preliminaries.

Let be a real nx1 vector. Then (U ≠ θ

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nU

UU

UM

2

1

n)

H = UUt is an nxn real symmetric matrix. Let α be a real number (which we shall suitably choose) and consider

( )IUUIHIP t ...........αα −=−=

We shall choose α such that P is its own inverse. (Note that Pt = P). So we need

P2 = I

i.e.

(I - α H) (I - α H) = I

i.e.

(I - α UUt) (I - α UUt) = I

I – 2 α UUt + α2 UUt UUt = I

So we choose α such that

α2 UUt UUt = 2 α UUt

Obviously, we choose α ≠ 0. Because otherwise we get P = I; and we don’t get any new transformation.

∴ We need



α UUt UUt = 2. UUt

But UtU = U21 + U2

2 + ….. + U2n =

2U is a real number ≠ 0 and thus we have

α (Ut U). UUt = 2 UUt

and hence

)..(..........2 IIUU t=α

Thus if we U is an nx1 vector and different from θn and α is as in (II) then P defined as

)....(.......... IIIUUIP tα−=

is such that

)....(..........1 IVPPP t −==

Now we go back to our problem of tridiagonalization of A. Our first aim is to find a P of the form (IV) such that PAPAPPt = has off tridiagonal entries in 1st row and 1st column as zero. We can choose the P as follows:

Let ( )Vaaas n ..................... 12

312

2122 +++=

(the sum of the squares of the entries below the 1st diagonal entry in A)

Let s = nonnegative square root of s2.

Let

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛+

=

1

31

2121 .sgn0

na

aasa

UM

………. (VI)

Thus U is the same as the 1st column of A except that the 1st component is taken as 0 and second component is a variation of the second component in the 1st column of A. All others are same as 1st column of A.

Then



1

2

−

⎥⎦

⎤⎢⎣

⎡=

UU t

α

( )[ ] 11

241

231

222121 2/......sgn

−+++++= naaaasa

[ ] 11

231

221

221

2 2/.....2 −+++++= naasasa

( )[ ]{ } 121

21

231

221

2 2/2..... −+++++= assaaa n

21

2

1ass +

=

212

1ass +

=∴α (VII)

Thus if α is as in (VII) and U is as in (VI) where s is as in (V) then

P = I - α UUt

is s.t. P = Pt = P-1, and it can be shown that

A2 = PA1P = PAP (i.e let A1 = A)

is similar to A and has off tridiagonal entries in 1st row and 1st column as 0.

Now we apply this procedure to the matrix obtained by ignoring 1st column and 1st row of A2.

Thus we now choose

22

422

3222 ..... naaas +++=

(where now aij denote entries of A2)

(i.e. s2 is sum of squares of the entries below second diagonal entry of A2)

s = Positive square root of s2



⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

+=

2

42

3232 ).(00

na

asasigna

U

M

322

1ass +

=α

P = I - α UUt

Then

A3 = PA2P

has off tridiagonal entries 1n 1st, 2nd rows and columns as zero. We proceed similarly and annihilate all off tridiagonal entries and get T, real symmetric tridiagonal and similar to A.

Note: For an nxn matrix we get tridiagonalization in n – 2 steps.

Example:

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

4211241111541145

A

A is a real symmetric matrix and is 4 x 4. Thus we get tridiagonalization after (4 – 2) i.e. 2 steps.

Step 1:

24264.418

18114 2222

==

=++=

s

s

( )( ) 97056.341

424264.41811

212 =

+=

+=

assα = 0.02860



⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛+

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛+

=

1124264.440

.sgn0

41

31

2121

aa

asaU

With this α, U, we get

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−−−−−−

=−=

97140.002860.023570.0002860.097140.023570.0023570.023570.094281.000001

tUUIP α

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

−−−−

==

5.35.1105.15.31011624264.4

0024264.45

2 PAPA

Step 2

( ) ( )41421.12

211 222

==

=+−=

s

s

( )( ) 29289.041421.3

1141421.12

11

322 ==

+=

+=

assα

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−−

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

+=

141421.200

141421.11

00

.sgn00

42

3232

aasa

U



P = I - α UUt

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−−

=

70711.070711.00070711.070711.00000100001

A3 = PA2P

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛−

−

=

20000541421.10041421.1624264.40024264.45

which is tridiagonal.

Thus the Givens – Householder scheme for finding the eigenvalues involves two steps, namely,

STEP 1: Find a tridiagonal T (real symmetric) similar to T (by the method

described above)

STEP 2: Find the eigenvalues of T (by the method of sturm sequences and bisection described earlier)

However, it must be mentioned that this method is used mostly to calculate the eigenvalue of the largest modulus or to sharpen the calculations done by some other method. If one wants to calculate all the eigenvalues at the same time then one uses the Jacobi iteration which we now describe.



JACOBI ITERATION FOR FINDING EIGENVALUES OF A REAL SYMMETRIC MATRIXSome Preliminaries:

Let be a real symmetric matrix. ⎟⎟⎠

⎞⎜⎜⎝

⎛=

2212

1211

aaaa

A

Let ; (where we choose ⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

θθθθ

CosCos

Psin

sin4

πθ ≤ for

purposes of convergence of the scheme) Note

and ⎟⎟⎠

⎞⎜⎜⎝

⎛−

=θθθθ

CosCos

P t

sinsin

IPPPP tt ==

Thus P is an orthogonal matrix. Now

⎟⎟⎠

⎞⎜⎜⎝

⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

==θθθθ

θθθθ

cossinsincos

cossinsincos

2212

12111

aaaa

APPA t

⎟⎟⎠

⎞⎜⎜⎝

⎛+−++−+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=θθθθθθθθ

θθθθ

cossinsincoscossinsincos

cossinsincos

22122212

12111211

aaaaaaaa

( ) ( )( ) ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛

+−−++−−++−++

=θθθθθθθθθθθθθθθθ

22212

211

22122211

22122211

22212

211

coscossin2sinsincoscossinsincoscossinsincossin2cos

aaaaaaaaaaaa

Thus if we choose θ such that,

( ) ( ) 0sincoscossin 22122211 =−++− θθθθ aaa . . . (I)

We get the entries in (1,2) position and (2,1) position of A1 as zero.

(I) gives

( ) 02cos2sin2 12

2211 =+⎟⎠⎞

⎜⎝⎛ +− θθ aaa

θθ 2sin2

2cos 221112

aaa −=⇒



( )( )

2211

221112

2211

12 sgn222tanaa

aaaaa

a−

−=

−=⇒ θ

βα

= . . . . . (II)

Where )sgn(2 221112 aaa −=α . . . . . (III)

2211 aa −=β . . . . . (IV)

θθ 2tan12sec 22 +=∴

2

2

1βα

+= from (II)

2

22

ββα +

=

22

22 2cos

βαβθ+

=∴

22

2

221cos22cos

βα

βθβα

βθ+

=−⇒+

=∴

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++=⇒

221

21cos

βα

βθ . . . . . . . (V)

and

22

22 12cos12sincossin2

βαβθθθθ+

−=−==

2222

2

βα

αβα

α

+=

+=

22cos2

sinβαθ

αθ+

=∴ . . . . . .(VI)



(V) and (VI) give sinθ, cosθ and if we choose

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

θθθθ

cossinsincos

P with these values of cosθ, sinθ, then

PtAP = A1 has (2,1) and (1,2) entries as zero.

We now generalize this idea.

Let A = (aij) be an nxn real symmetric matrix.

Let 1 ≤ g < p < n. (Instead of (1,2) position above choose (q, p) position)

Consider,

)sgn(2 ppqqqp aaa −=α . . . . . . . (A)

ppqq aa −=β . . . . . . . (B)

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++=

221

21cos

βα

βθ . . . . . . . (C)

22cos21sin

βα

αθ

θ+

= . . . . . . . (D)

q p


Vittal rao/IISc.Bangalore

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

−=

1

cossin

1sincos

11

O

O

O

θθ

θθP

then A1 = Pt AP has the entries in (q, p) position and (p, q) pos

In fact A1 differs from A only in qth row, pth row and qth columbe shown that these new entries are

θθ

θθ

cossin

sincos1

1

piqipi

piqiqi

aaa

aaa

+−=

+= i ≠ q, p (qth row

θθ

θθ

cossin

sincos1

1

ipiqip

ipiqiq

aaa

aaa

+−=

+= i ≠ q, p (qth column p

.0

ccossin2sin

scossin2cos

11

21

21

==

+−=

++=

pqqp

ppqpqqpp

ppqpqqqq

aa

aaaa

aaaa

θθθ

θθθ

Now the Jacobi iteration is as follows.

Let A = (aij) be nxn real symmetric.

q

p

M4/L5/V1/May 2004/4

ition as zero.

n and pth column and it can

pth row) . .(E)

th column) . .(F)

os

in2

2

θ

θ

. . . (G)



Find 1 ≤ g < p ≤ n such that qpa is largest among the absolute values of all the off diagonal entries in A.

For this q, p find P as above. Let A1 = Pt AP. A1 can be obtained as follows:

All rows of A1 are same as A except qth row,

Column

pth row, qth column, pth column which are obtained from (E), (F), (G).

Now A1 has 0 in (q, p), (p, q) position.

Replace A by A1 and repeat the process. The process converges to a diagonal matrix the diagonal entries of which give the eigenvalues of A.

Example:

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−−

=

3241242242931237

A

Entry with largest modulus is at (2, 4) position.

∴ q = 2, p = 4.

( ) ( ) 244422 .sgn2.sgn2 aaaaaa qpppqq −=−=α

( )( )( ) .8412 ==

639 =−=−= ppqq aaβ

10022 =+∴ βα ; 1022 =+ βα

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

++=∴

221

21cos

βα

βθ



89442.08.054

1061

21

===⎟⎠⎞

⎜⎝⎛ +=

108

)89442.0(21

cos21sin

22=

+=

βαα

θθ = 0.44721

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛−

=∴

89442.0044721.000100

44721.0089442.000001

P

A1 = PtAP will have a124 = a1

42 = 0.

Other entries that are different from that of A are a121, a1

22, a123 ; a1

41, a142, a1

43, a144 ; (of

course by symmetric corresponding reflected entries also change).

We have,

1305.3sincos 4121211 =+= θθ aaa

44721.0cossin 4121411 −=+−= θθ aaa

89443.0sincos 4323231 −=+= θθ aaa

68328.2cossin 4323431 =+−= θθ aaa

11sincossin2cos 24424

22222

1 =++= θθθθ aaaa

1coscossin2sin 24424

22244

1 =+−= θθθθ aaaa

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−−

−−

=∴

00000.168328.2044721.068328.2489443.020000.089443.0111305.344721.021305.37

1A

Now we repeat the process with this matrix.

The largest absolute value is at (1, 2) position.

∴ q = 1, p = 2.



441172211 =−=−=−=−= aaaa ppqqβ

( ) ( )( )11305.32sgn2 −=−= ppqqgp aaaα = - 6.2610.

42968.7

200121.5522

22

=+

=+

βα

βα

87704.0121cos

22=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++=

βα

βθ ;

48043.0cos21sin

22−=

+=

βα

αθ

θ

∴ The entries that change are

71484.12coscossin2sin

28516.5sincossin2cos

21485.0cossin

39222.0sincos

17641.0cossin

18378.2sincos

0

22212

21122

1

22212

21111

1

2414241

2414141

2313231

2313131

211

121

=+−=

=++=

−=+−=

−=+=

=+−=

=+=

==

θθθθ

θθθθ

θθ

θθ

θθ

θθ

aaaa

aaaa

aaa

aaa

aaa

aaa

aa

and the new matrix is



⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−−

−−

168328.221485.039222.068328.2417641.018378.221485.017641.071484.12039222.018378.2028516.5

Now we repeat with q = 3, p = 4 and so on.

And at the 12th step we get the diagonal matrix

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−09733.2000

060024.5000071986.12000078305.5

giving eigenvalues of A as 5.78305, 12.71986, -5.60024, 2.09733.

Note: At each stage when we choose (q, p) position and apply the above transformation to get new matrix A1 then sum of squares of off diagonal entries of A1 will be less than that of A by 2a2

qp.


The Q R decomposition: Let A be an nxn real nonsingular matrix. Then we can find an orthogonal matrix Q and an upper triangular matrix R (with rii >0) such that A=QR called the QR decomposition of A. The Q and R are found as follows: Let a(1) ; a(2) ; ……. , a(n) be the columns of q(1) ; q(2) ; ……… , q(n) be the columns of Q r(1) , r(2) , ………., r(n) be the columns of R. Note: Since Q is Hermitian we have ( ) ( ) ( ) ( )Aqqq n ...................1

22

2

2

1 ====

( ) ( )( ) 0, =ji qq if i ≠ j …………….. (B). and since R is upper triangular we have

( ) ( )Cr

rr

r ii

i

i

i ......................

0

0

2

1

⎟⎟⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

=

M

M

Also the ith column of QR is; Qr(i) and ∴ ith column of

( ) ( ) ( ) ( )DqrqrqrQR iiiii .........................2

21

1 +++= We want A = QR. Comparing 1st column on both sides we get a(1) = QR ’s first column = Qr(1)

= r11q(1) by (D)

( ) ( ) ( )2

1112

1112

1 qrqra ==∴

= r11 ∵ r11 > 0 and ( )Abyq 1

2

1 =

Vittal rao/IISc.Bangalore M4/L6/L1/May2004/1


( )

2

111 ar =∴ and ( ) ( ) ( )Ea

rq .................1 1

11

1 =

giving 1st columns of R and Q. Next comparing second columns on both sides we get a(2) = Qr(2) = r12 q(1) +r22 q(2) ……….. (*) Therefore from (*) we get

( ) ( )( ) ( ) ( )( ) ( ) ( )( )1222

1112

12 ,,, qqrqqrqa += ( ) ( )( ) ( ) ( )Abyqqqr 1, 2

211112 === Q

( ) ( )( ) ( )Bbyqqand 0, 12 =

( ) ( )( ) ( )Fqar ............, 1212 =∴

∴ (*) gives

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )[ ] ( )Hqrar

q

and

Gqrar

qraqrand

qraqr

......................1

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

112

2

22

2

2

112

222

2

112

2

2

222

112

2222

−=

−=∴

−=∴

−=

(F), (G), (H) give 2nd columns of Q and R. We can proceed having got the first i - 1 columns of Q and R we get ith columns of Q and R as follows:

( ) ( )( ) ( ) ( )( ) ( ) ( )( )11

212

11 ,...,,.........,;, −

− === iiiii

ii qarqarqar

( ) ( ) ( ) ( )

2

11

22

11 ....... −

−−−−= iiiii

iii qrarqrar

( ) ( ) ( ) ( ) ( )[ ]1

12

21

1 .......... −−−−−= i

iiii

ii

i qrqrqrariq



Example:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

110101121

A

1st column of Q and R( ) 211 221

11 =+== ar

( )

( )

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

==

02

12

1

1 1

11

1 ar

q

2nd column of Q and R:

( ) ( )( ) 22

2

02

12

1

,102

, 1212 ==

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛== qar

( ) ( ) 311

1

011

102

22

2

112

222 =

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=−= qrar

( ) ( ) ( )[ ]

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

−=−=

31

313

1

31 1

1222 qraq



3rd column of Q and R:

( ) ( )( ) 22

2

02

12

1

,111

, 1313 ==

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛== qar

( ) ( )( )

31, 23

23 == qar

( ) ( ) ( )2

231

133

33 qrqrar −−=

231

313

1

31

011

111

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

−−⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

32

32

94

91

91

323

13

1

2

==++=

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛−

=

and

( ) ( ) ( ) ( )[ ]223

113

3

33

3 1 qrqrar

q −−=



⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛−

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛ −

=

326

16

1

323131

23

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

−

−

=∴

32

310

61

31

21

61

31

21

Q ;

⎟⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜

⎝

⎛

=

3200

3130222

R

and

AQR =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

110101121

giving us QR decomposition of A.


Numerical analysis/Computations of eigenvaules Lecture notes


QR algorithmLet A be any nonsingular nxn matrix. Let A = A1 = Q1 R1 be its QR decomposition. Let A2 = R1 Q1. Then find the QR decomposition of A2 as A2 = Q2 R2 Define A3 = R2 Q2 ; find QR decomposition of A3 as A3 = Q3 R3. Keep repeating the process. Thus A1 = Q1 R1 A2 = R1 Q1and the ith step is Ai = Ri-1 Qi-1Ai = Qi Ri Then Ai ‘ converges’ to an upper triangular matrix exhibiting the eigenvalues of A along the diagonal.

numerical analysis - nptel · computational linear algebra syllabus numerical analysis linear...

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