286 CHAPTER 11. LIMITS OF VECTORS AND MATRICES

where I is r× r for r the multiplicity of the eigenvalue 1 and N is a nilpotent matrix forwhich Nr = 0. I will show that because of Condition 2, N = 0.

First of all,Jri (λ i) = λ iI +Ni

where Ni satisfies Nrii = 0 for some ri > 0. It is clear that Ni (λ iI) = (λ iI)N and so

(Jri (λ i))n =

n

∑k=0

(nk

)Nk

λn−ki =

ri

∑k=0

(nk

)Nk

λn−ki

which converges to 0 due to the assumption that |λ i| < 1. There are finitely many termsand a typical one is a matrix whose entries are no larger than an expression of the form

|λ i|n−k Ckn(n−1) · · ·(n− k+1)≤Ck |λ i|n−k nk

which converges to 0 because, by the root test, the series ∑∞n=1 |λ i|n−k nk converges. Thus

for each i = 2, . . . , p,limn→∞

(Jri (λ i))n = 0.

By Condition 2, if ani j denotes the i jth entry of An, then either

p

∑i=1

ani j = 1 or

p

∑j=1

ani j = 1, an

i j ≥ 0.

This follows from Lemma 11.1.2. It is obvious each ani j ≥ 0, and so the entries of An must

be bounded independent of n.It follows easily from

n times︷ ︸︸ ︷P−1APP−1APP−1AP · · ·P−1AP = P−1AnP

thatP−1AnP = Jn (11.1)

Hence Jn must also have bounded entries as n→ ∞. However, this requirement is incom-patible with an assumption that N ̸= 0.

If N ̸= 0, then Ns ̸= 0 but Ns+1 = 0 for some 1≤ s≤ r. Then

(I +N)n = I +s

∑k=1

(nk

)Nk

One of the entries of Ns is nonzero by the definition of s. Let this entry be nsi j. Then this

implies that one of the entries of (I +N)n is of the form(n

s

)ns

i j. This entry dominates thei jth entries of

(nk

)Nk for all k < s because

limn→∞

(ns

)/

(nk

)= ∞