0©

² ¬ ¬ q0 u i

¬ i ¢

iph h

i

i h i h

The argument can be continued showing each time that

p±° ¯ i ¯®

¢s ¢c

² q0 s

¬

0© (s ¢

E

¬

and this is valid for . In the end, it will be proved that belongs to the boundary ©

W (

(

y

of the -th disc for an arbitrary .

An immediate corollary of the Gershgorin theorem and the above theorem follows.

˜¢{ ¦

¦ ¡¢ ¤B9

£A

0

' ## '

§

If a matrix is strictly diagonally dominant or irreducibly diago-

nally dominant, then it is nonsingular.

| ¢ Cv¥ ¡At ¢

| §| ¡¡C

CC

¡¨

£ 6 A

£

If a matrix is strictly diagonally dominant, then the union of the Gershgorin disks

¦©

¬

excludes the origin, so cannot be an eigenvalue. Assume now that it is only irre- Y

ducibly diagonal dominant. Then if it is singular, the zero eigenvalue lies on the boundary

of the union of the Gershgorin disks. In this situation, according to the previous theorem,

this eigenvalue should lie on the boundary of all the disks. This would mean that

« q0 u u c 0 f

0 uR c0 g q

¬ v

¬ r¤

for ‘ (V(

(

y

Is

ih

hs

which contradicts the assumption of irreducible diagonal dominance.

The following theorem can now be stated.

´ 9 Av ¡˜¤

¥¦ ¥

A ' P

1

§

If is a strictly diagonally dominant or an irreducibly diagonally dom-

©©

inant matrix, then the associated Jacobi and Gauss-Seidel iterations converge for any .

£ 6 A T

£

We ¬rst prove the results for strictly diagonally dominant matrices. Let be ©

X

T

¬

the dominant eigenvalue of the iteration matrix for Jacobi and w

% W“ # '

X

# ¬

² ©

for Gauss-Seidel. As in the proof of Gershgorin™s theorem, let

`

% 'W “ 0 R 0F

( ¥

‚0

¬ ¬

be an eigenvector associated with , with , and , for . Start from © 0F E

¢

y y y

equation (4.38) in the proof of Gershgorin™s theorem which states that for ,

« q™g fg «

u ¢ c0 u ¢ c0 gf

¢ 0© 0 u 0F 0 0

rg

q

0

¢ ¢ c0 0 ¢ ¢ c 0 h

0 y

h

i i

h h

T

This proves the result for Jacobi™s method.

© X &## © ¬ '

©

%²

For the Gauss-Seidel iteration, write the -th row of the equation

!

in the form

f f

w ¢F ¢ ¢c ‚ uFu ¢c ( „ u F u ¢ c

¬

©

¢ ¡u ¢ ¡u

¢

which yields the inequality

£ £

0 u F F0 u ¢ c 0 0 u ¢ c0

¢ ¡u ¢ ¡u

0

0 0© u c

£ £

0 ¢ ¢ c0 0 u F 50 u ¢ c 0 ¢ ¡u 0 ¢ ¢ cT 0

² a²X 0 ¢ 0 ¢ ¢¡u

¢

0

£ 7£ £ £

² £ W£

£ ( £

The last term in the above equation has the form with all nonnegative

(

W

£ ² W £ z£

² Y£

and . Therefore,

T £ £

0 0©

X

£ ² £ £ ²z£

£ w£ y

W

T

In the case when the matrix is only irreducibly diagonally dominant, the above proofs