393

whenever m is large enough. It follows

limm→∞

∣∣∣∣∣∣∣∣( Aλ0)m

− P

∣∣∣∣∣∣∣∣ = 0

as claimed.What about the case when A > 0 but maybe it is not the case that A >> 0? As before,

K ≡ {x ≥ 0 such that ||x||1 = 1} .

Now defineS1 ≡ {λ : Ax ≥ λx for some x ∈ K}

andλ0 ≡ sup (S1) (2.7)

Theorem B.0.7 Let A > 0 and let λ0 be defined in 2.7. Then there exists x0 > 0 suchthat Ax0 = λ0x0.

Proof: Let E consist of the matrix which has a one in every entry. Then from TheoremB.0.4 it follows there exists xδ >> 0 , ||xδ||1 = 1, such that (A+ δE)xδ = λ0δxδ where

λ0δ ≡ sup {λ : (A+ δE)x ≥ λx for some x ∈ K} .

Now if α < δ{λ : (A+ αE)x ≥ λx for some x ∈ K} ⊆

{λ : (A+ δE)x ≥ λx for some x ∈ K}

and so λ0δ ≥ λ0α because λ0δ is the sup of the second set and λ0α is the sup of the first. Itfollows the limit, λ1 ≡ limδ→0+ λ0δ exists. Taking a subsequence and using the compactnessof K, there exists a subsequence, still denoted by δ such that as δ → 0, xδ → x ∈ K.Therefore,

Ax = λ1x

and so, in particular, Ax ≥ λ1x and so λ1 ≤ λ0. But also, if λ ≤ λ0,

λx ≤ Ax < (A+ δE)x

showing that λ0δ ≥ λ for all such λ. But then λ0δ ≥ λ0 also. Hence λ1 ≥ λ0, showing thesetwo numbers are the same. Hence Ax = λ0x. ■

If Am >> 0 for some m and A > 0, it follows that the dimension of the eigenspace forλ0 is one and that the absolute value of every other eigenvalue of A is less than λ0. If it isonly assumed that A > 0, not necessarily >> 0, this is no longer true. However, there issomething which is very interesting which can be said. First here is an interesting lemma.

Lemma B.0.8 Let M be a matrix of the form

M =

(A 0

B C

)or

M =

(A B

0 C

)where A is an r × r matrix and C is an (n− r) × (n− r) matrix. Then det (M) =det (A) det (B) and σ (M) = σ (A) ∪ σ (C) .