296 CHAPTER 11. LIMITS OF VECTORS AND MATRICES

Also let x0 denote the vector such that Ax0 = λ 0x0 with x0 >> 0. First note that xT0 v > 0

because both these vectors have all entries positive. Therefore, v may be scaled such that

vTx0 = xT0 v = 1. (11.6)

DefineP≡ x0v

T .

Thanks to 11.6,

Aλ 0

P = x0vT = P, P

(Aλ 0

)= x0v

T(

Aλ 0

)= x0v

T = P, (11.7)

andP2 = x0v

Tx0vT = vTx0 = P. (11.8)

Therefore, (Aλ 0−P)2

=

(Aλ 0

)2

−2(

Aλ 0

)P+P2

=

(Aλ 0

)2

−P.

Continuing this way, using 11.7 repeatedly, it follows((Aλ 0

)−P)m

=

(Aλ 0

)m

−P. (11.9)

The eigenvalues of(

Aλ 0

)−P are of interest because it is powers of this matrix which

determine the convergence of(

Aλ 0

)mto P. Therefore, let µ be a nonzero eigenvalue of this

matrix. Thus ((Aλ 0

)−P)x= µx (11.10)

for x ̸= 0, and µ ̸= 0. Applying P to both sides and using the second formula of 11.7 yields

0= (P−P) x=

(P(

Aλ 0

)−P2

)x= µPx.

But since Px= 0, it follows from 11.10 that

Ax= λ 0µx

which implies λ 0µ is an eigenvalue of A. Therefore, by Corollary 11.4.5 it follows thateither λ 0µ = λ 0 in which case µ = 1, or λ 0 |µ|< λ 0 which implies |µ|< 1. But if µ = 1,then x is a multiple of x0 and 11.10 would yield((

Aλ 0

)−P)x0 = x0

296 CHAPTER 11. LIMITS OF VECTORS AND MATRICESAlso let 29 denote the vector such that Axo = Apxo with xp >> O. First note that x) v>0because both these vectors have all entries positive. Therefore, v may be scaled such thatv' aj = avi. (11.6)DefineP= xov!Thanks to 11.6,A A AqoP = tow! =P, r(x) = av" (+) =aqv' =P, (11.7)andP=aov' av! =v' ap =P. (11.8)Therefore,lIa>| >NEnN|~Continuing this way, using 11.7 repeatedly, it follows(a)--(aY-rThe eigenvalues of (4) — P are of interest because it is powers of this matrix whichmdetermine the convergence of (4) to P. Therefore, let be a nonzero eigenvalue of thisA((q,) <P) e=ne (11.10)for « # 0, and u £0. Applying P to both sides and using the second formula of 11.7 yieldsA 20=(P—-P)x=(P z. —P’ )x=wuP2.0matrix. ThusBut since Px = 0, it follows from 11.10 thatAx =Aouxwhich implies Apu is an eigenvalue of A. Therefore, by Corollary 11.4.5 it follows thateither Agu = Ao in which case p = 1, or Ag || < Ao which implies |u| < 1. Butif u = 1,then x is a multiple of xq and 11.10 would yield(8)-)a-