On the characteristic vectors of a matrix

LlNEAR

ALGEBRA

AND

ITS

On the Characteristic

APPLICATIONS

6,

189-196 (1972)

189

Vectors of a Matrix

G. N. DE OLIVEIRA

Institute

de ililatewuitica

Coinzbra,

Portugal

Communicated by Leon Mirsky

1. INTRODUCTION Given a square matrix determine

regions

of A must lie.

A of order n, there are several theorems

of the complex

Considering

plane where

the characteristic

each of them can be represented

vectors

As for the characteristic

of S where the points lie.

This amounts

the coordinates value,

these

characteristic

roots,

representing

roots

of A with norm 1,

to determining

of the unitary

which relations

can sometimes

n-dimensional

we might try to determine

the characteristic

of the characteristic

relations

which

by one point of the sphere S of radius

1 and center in the origin of the coordinates space.

the characteristic

vectors.

vectors must

In addition

regions

of A must

be satisfied

by

to their intrinsic

be used to obtain

bounds

for the

roots.

Let A =

[aif] be a positive n x n matrix with its dominant characteristic root w and (x1,. . . , x,) the corresponding characteristic vector, which we can assume

to be positive.

Let R, =

2

j=l

atj,

,

m = mm aii,

R = max Ri, i m = min ai3,

z

The

i+j

following

inequalities

r=minR.

1’

i

x7n = mm xi, 1

were proved

xu = max xi.

by Ostrowski

t

[5].

9%

R--r+m Copyright 0

(1.1)

1972 by American Elsevier Publishing Company, Inc.

G. N. DE OLIVEIRA

190 The first of them was later improved -

element

matrix,

the following

Let p be a least,

[6], is known. of A.

[l], who showed that

R + y + {(R - T)~ + 4m2}1/2< x, \ 2m xM .

If A is an n x n nonnegative Schneider

by Brauer

nonvanishing,

(1.2) result,

due to

nondiagonal

Then (I.31

In

[4],

bounds

for

are given. The objective

of this paper is to find some further

results of this kind.

Our results are based on an idea which, though very simple, improves, least

in some cases,

the inequalities

at

above.

2. BASIC RESULTS [aij] be an n x n complex matrix and sa@$ose

2.1. Let A =

THEOREM

that it has a characteristic root 1 # 0 to which corresponds a characteristic vector (x1,. . . , x,) which is known to be real and nonnegative. i and k, aij and akj, j = 1,. a, j

>,

If, for fixed

, ?a, are real and satisfy j =

ak/p

l,...,n,

then if

xi = Xk = 0,

Proof. fxi

must

Similarly which

As ilxi =

1 is not real,

xi >

xk,

if

i, > 0,

xi <

xk,

if

3,<0.

Cy=i

aijxj

and aij and xi, j = 1,.. . , n, are real,

be real.

If 31 is not real,

xk = 0.

Suppose

it is obvious

now that

il is real.

implies ,$

aiixj

(2.1)

3

gakirj.

that

xi must

We have

aijxj

be zero. > akjxj

ON THE

CHARACTERISTIC

Here the left-hand of the theorem

VECTORS

side is &,

191

OF A MATRIX

and the right-hand

side is lx,.

The last part

follows.

It can be easily seen that, if in aii 2 akj, j = 1,. . . , n, the sign > holds at least

once and the vector

in (2.1), 3 The

(<)

can be replaced

theorem

matrices.

we have

by

proved

For this class of matrices

as we show with the following

(xi,.

. . , xn), is positive

(xi,.

>

then, I is real and,

(0.

is at once applicable further

to nonnegative

results can be easily deduced,

theorem.

THEOREM 2.2. Let A = [aii] be an n x n nonnegative matrix and , x,) the nonnegative vector which corresponds to its dominant char-

acteristic root w.

Let w > 0.

If, for fixed i and k, i # k, aij f

0, j =

1,. . , n, and

8,, = max ‘e, i z, then

If eik = 0, we have akj = 0, j = 1,.

Proof.

and the theorem matrix

obtained

by Oi,.

We can obtain

2.2 is readily

further

results

2.2, i.e., constructing

Theorem

2.1 to this matrix

dominant

vector

by the method

if possible.

(x1,.

vector

of

. , n. Using

established.

a matrix

characteristic

characteristic

roots and

characteristic

similar

used in the proof

As an example

we have: matrix

root w to which corresponds

. , x,).

If,

of

to A and then applying

2.3. Let A = [aii] be an n x 9~ nonnegative

THEOREM

[bij] be the

B =

A and B have the same characteristic

Theorem

negative

Let

the ith row by 8,, and dividing

to ZJ. It is easy to see that b,, > bkj, j = 1,.

2.1, Theorem

a positive

Bik > 0.

eikxi, x~+~,. ., x,) is a nonnegative

B corresponding Theorem

Now suppose

from A by multiplying

the ith column (x1,. . ., xi-l.

holds.

, n, which implies xk = 0,

for fixed

with

the non-

h, k, i,

where

ht, i # k,

j=

1,. ..,z-1,i+1,...,

ahi

-

ahk

+

a&

-

ski

6

aii

-

akk,

92, (2.2)

Proof.

Let us add the ith row of A to its kth row and subtract

kth column is similar

from the ith column.

to A.

characteristic that

To the characteristic

Obviously,

Remark. x,/xi 3

, x,).

the

[ciJ ] which the

It is easily seen

. , n, and so xi + lclc3 xh.

is positive,

then

From Theorem

(%.2), the sign < holds

X, + _Y,;> _YiL. Similarly,

(2.2) the sign < is replaced by 3,

type of (i) and

C =

root w of C there corresponds

if in at least one of the inequalities

(x,, . . . , x,)

inequalities

a matrix

vector (x1,. . , x+~, xI; + xi, xl<, 1,.

ckj 3 chj, j = 1,.

and

We obtain

if in the

we shall have xi + xJ; < x,~.

2.2 we can easily derive inequalities

(ii) of Theorem

2.1 of 141.

In fact,

xi/xj 3

of the

l/Oi, and

l/Ojj imply

(2.3) The inequalities

(2.3) are, very often, better than (i) and (ii) of Theorem

2.1 of 141. Moreover the dominant 3.

IMPROVED

the bounds given by (2.3) can be computed

characteristic

root being

without

known.

BOUNDS

Let (x,,

, x,) be a real characteristic

the characteristic

root A. Suppose,

vector

for example,

of A corresponding

x(A) and alower bound O(A) f or ~,,/x~~~.If s is a positive integer, (x1,. will be a characteristic compute

bounds

vector

for x,/xJI

of As corresponding

in terms

to As. Thus

arises :

How

complete

answer

seems

better

do O(AB) and x(A”) to be difficult,

bounds by choosing s conveniently.

for a nonnegative

irreducible

matrix.

, x,) we can

of A”:

For 0 and x we can use some of the bounds given above. question

to

we have an upper bound

behave

but in many

A

cases we can get

This is particularly We consider

Then the

as s increases? interesting

two cases.

ON

THE

CHARACTERISTIC

Case 1.

A is primitive.

s 3 sO. Thus, matrices, not,

\‘ECTOKS

There

OF

A MATRIX

exists

an s,, such that

s 3 sa, we can use bounds

taking very

As > 0 for

valid only for positive

i.e., we can avoid the use of Schneider’s

in general,

193

bound

(1.3) which is

sharp.

Case 2. A is of the form 0

A,

0

.a.

0 -

0

A,

.a.

0

'*' 0 0 0 ......................

A, where the diagonal

h > 1,

0

0

... A,_,

0

0

a..

blocks

0 We have

are square.

Ah=

with vector

Bi = AiAi+l...A,A,... A,_l.We partition we are considering

of coordinates

as follows:

vector

which the coordinates

of the block

notice that each as in the first It

the

, yh), where

of yi is equal to the order of the block

yi is a characteristic

is known

is similar

(yi,.

of

characteristic the

number

Bi,i = 1,. . . , h.

Bi, and thus we can deduce relations yi must satisfy.

Bi is a primitive matrix

It is interesting

to

[3, p. 821 and so we can proceed

case. that,

under

conditions,

of

For instance,

a nonnegative

matrix

if

matrices.

Now let f(x) vector

matrix.

certain

A is irreducible and has a positive column (row), we can find a matrix S such that SAS-l > 0 [2]. If x is the Perron characteristic vector of A, Sx will be the Perron characteristic vector of SAS-l. We can now use for Sx any formula applicable to positive

to a positive

be a polynomial.

f(A)corresponding

(xi,.

., x,) will be a characteristic

to f(A), and therefore

W(A))d

2 < XV(A)).

G. N. Ll!iOL,I\~lilKA

194 We can ask:

How do we choose f(x) to maximize

0(/(A)) or to minimize

X(f(A)) ? 4. BOUNDS

FOR

THEOREM

CHARACTERISTIC

4.1.

ROOTS

Let A be an n x ?a nonnegative

its dominant characteristic root, and (x,,

irreducible matrix,

If tij aye ntlmbers such that xi/x, > li,,

characteristic vector.

The proof is obvious.

Assuming

lij = 1/6Jij (Oij as defined side of (1.2)) ; tij

improves

Theorem

3 of

A > 0, we can take,

in Theorem

left-hand

then

for instance,

2.2) if l/Bij > y (y denotes

= y 0tl rerwise.

111.

wl

. , x,) the corresfionding positive

We get an inequality

To compute

the

Eij, note

that

the

which l/O,j =

mink(%lajJ. THEOREM

4.2.

Let A, W, (x,,....

xJ,

and 5, j have the same meaning

as in Theorem 4.1. Let ,u be a characteristic root of A different from W. Then

Proof.

In

141 it has been proved

2

ipl
j=,

We have min

‘2

L i The theorem 5. NUMERICAL

that

! i ai

min ~ ‘ x,

_u,.

xj > niin(ai,Ej,)

t

11

follows. EXAMPLE

Let

A=

5

1

1

11

4

7

6

3

-3

1.

ON THE

CHARACTERISTIC

Theorem

3.1 of [4] gives

Theorem

VECTORS

2.1 and inequality

195

OF A MATRIX

(3.6) of the same

paper give (5.2)

XT”‘ and I,/_L~ < w-

1.684,

(5.3)

respectively. To compute characteristic The

all these bounds it is necessary

inequalities

The second

(2.3) of the present

paper

1.833 < cz”=l? x.?

< 6.334.

2.2 provides

a better

inequalities

(5.4),

which

(5.6)

information

(5.5),

about

in (5.1).

the vector

in (5.2).

Obviously

The

Theorem

(x1,. . . , x,) than the

and (5.6).

4.2 gives

improves

We conclude

(5.3). with a result

THEOREM 5.1.

that

is valid for an arbitrary

Let A be an arbitrary n x n matrix.

its characteristic roots, and let (x1,. . . , x,) corresponding

give

of (5.5) and (5.6) are not contained

parts

first parts of (5.4) and (5.6) are not contained

Theorem

to know that the dominant

root of A is 12.

to 2.

matrix.

Let ii be one of

be a characteristic vector of A

If 1 5 aii, then

(5.7)

196

c;.

Proof.

which

N.

DE

OLlVEIRh

We have

implies lil -

aji( lXij <

k

Iajjl . rfzlXp!.

J_l,j#i As 1, is supposed to be different theorem

from aij, maxkzilXk( cannot

Other theorems do not pursue Applying

of the type of Theorem

the matter

Theorem

of the matrix

further

A above,

Thanks the

are due to E. Seneta

research

binatbria

characteristic

2.h)

that

root

and vector

w = 12):


for his suggestions

was supported

C. Gulbenkian, e Teoria

Brauer,

2 G. N. de

in the correspondence

exchanged

by

and

Estudos

Institute

Gerais

de Alta

ITnivcrsitArios

Cultura

(Projecto

de Mqambique, de Anjlise

Duke

Math.

Oliveira.

J.

“4(1957).

HED. FUG. Ci. The

Theory

265-274.

Univ.

Coimbra

of Matrices,

Vol.

41(1968), 2, Chelsea

l&Z_“l. Publ.

Co., New York,

1959. Lynn

and

W.

5 )I. 11. Ostrowski, 6 H.

Schneider,

Received

July,

Con-

de Matrizcs).

3 F. R. Gantmacher, 4 M.

We

author.

I;unda@o

I A.

5.1 can be easily derived.

we get (we recall

x :j -

This

The

here.

5.1 to the Perron

max(.z,,

with

vanish.

follows.

Timlake, J.

Proc. 1969

London

Linear Mat/z.

I
Algebra Sot.

Math.

and

27(1952),

Sot.

2nd

A+@.

?(1969),

144P152.

25%256. SW. 11(1959),

127-130.