Av ¡˜¤

¥¦ ¥ ¤9

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' P

1

§

(Gershgorin) Any eigenvalue of a matrix is located in one of the ©

IR c

R

closed discs of the complex plane centered at and having the radius

u

« ¬R gf

0 vR c 0 rg q

u

iph

sth

In other words,

T

« fu

¥¦( § ¤¢© a± a i ¯®

X

0 IR c v0 uvR c 0

0©

²

£¡ R

such that

E

u

R£u

W

£ 6 A

£

©

Let be an eigenvector associated with an eigenvalue , and let be the index © !

© © A0

¬ 0 R 0F

of the component of largest modulus in . Scale so that , and , for 0F ¢

¥ y y

¬E ©

. Since is an eigenvector, then

!

T

« rqg fg ² ¬

X

¢ ¢c (uFu ¢c

²

© ¢F

iph h

which gives

« rqg fg « ™qg fg a± a i ¯®

0 0 u F 500

¢ ¢c u ¢ c0 ¢ c0

0©

² ¬0 u ¢

h

i iph

h h

This completes the proof.

µ£ „ ¢

|5¥ j qz A C¡y 5

„ 5| | 5£§|

¢ ¡

¡¡C

C @¡

¨§ £ "!

§

Since the result also holds for the transpose of , a version of the theorem can also be

formulated based on column sums instead of row sums.

¤

The discs de¬ned in the theorem are called Gershgorin discs. The theorem states that

¤ §

the union of these discs contains the spectrum of . It can also be shown that if there are

Gershgorin discs whose union is disjoint from all other discs, then contains exactly

! ¦ ¦

eigenvalues (counted with their multiplicities). For example, when one disc is disjoint

!

from the others, then it must contain exactly one eigenvalue.

An additional re¬nement which has important consequences concerns the particular

§

case when is irreducible.

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¥¦ ¥¡ vB9

9A

' 1

§ §

Let be an irreducible matrix, and assume that an eigenvalue of ©

¤

lies on the boundary of the union of the Gershgorin discs. Then lies on the boundary ©

of all Gershgorin discs.

£ A™¤ ¢¡6

£

©

As in the proof of Gershgorin™s theorem, let be an eigenvector associated with 0 R 0F ¥

c0

¬ ¬

, with , and , for . Start from equation (4.38) in the proof of

© 0F E

¢ !

y y

Gershgorin™s theorem which states that the point belongs to the -th disc. In addition, © ©

!

T

belongs to the boundary of the union of all the discs. As a result, it cannot be an interior X

¢ ¢c

² ¬„0

point to the disc . This implies that . Therefore, the inequalities

¢ ©

0©

# ( ¢

in (4.38) both become equalities:

« ™qg fg q0 c 0© « rqg fg q0 u F 500 u r± a i ¯®

¢ c0 0 u ¢ c0

¬² ¬ q

¬ ¢

¢¢

ph

i iph

h h

§ ¤ y

Let be any integer . Since is irreducible, its graph is connected and, therefore,

there exists a path from node to node in the adjacency graph. Let this path be !

¬

$( £ $(

#( ! !( t

! !

W H ¥

i ¢ ¢c ¬

By de¬nition of an edge in the adjacency graph, . Because of the equality in Y

u F

k0 u 0F

¬

(4.39), it is necessary that for any nonzero . Therefore, must be equal to 0F 0i

¢

y

one. Now repeating the argument with replaced by shows that the following equality ! !

W

holds:

« rqg f g q0 i « rqg f g q0 u F F00 u i r± ¯ i ¯®