390 APPENDIX B. POSITIVE MATRICES

Theorem B.0.4 Let A >> 0 be an n× n matrix and let λ0 be given in 2.1. Then

1. λ0 > 0 and there exists x0>> 0 such that Ax0 = λ0x0 so λ0 is an eigenvalue for A.

2. If Ax = µx where x ̸= 0, and µ ̸= λ0. Then |µ| < λ0.

3. The eigenspace for λ0 has dimension 1.

Proof: To see λ0 > 0, consider the vector, e ≡ (1, · · · , 1)T . Then

(Ae)i =∑j

Aij > 0

and so λ0 is at least as large as

mini

∑j

Aij .

Let {λk} be an increasing sequence of numbers from S1 converging to λ0. Letting xk bethe vector from K which occurs in the definition of S1, these vectors are in a compact set.Therefore, there exists a subsequence, still denoted by xk such that xk → x0 ∈ K andλk → λ0. Then passing to the limit,

Ax0 ≥ λ0x0, x0 > 0.

If Ax0 > λ0x0, then letting y ≡ Ax0, it follows from Lemma B.0.2 that Ay >> λ0y andy >> 0. But this contradicts the definition of λ0 as the supremum of the elements of Sbecause since Ay >> λ0y, it follows Ay >> (λ0 + ε)y for ε a small positive number.Therefore, Ax0 = λ0x0. It remains to verify that x0 >> 0. But this follows immediatelyfrom

0 <∑j

Aijx0j = (Ax0)i = λ0x0i.

This proves 1.Next suppose Ax = µx and x ̸= 0 and µ ̸= λ0. Then |Ax| = |µ| |x| . But this implies

A |x| ≥ |µ| |x| . (See the above abominable definition of |x|.)Case 1: |x| ≠ x and |x| ≠ −x.In this case, A |x| > |Ax| = |µ| |x| and letting y = A |x| , it follows y >> 0 and

Ay >> |µ|y which shows Ay >> (|µ|+ ε)y for sufficiently small positive ε and verifies|µ| < λ0.

Case 2: |x| = x or |x| = −xIn this case, the entries of x are all real and have the same sign. Therefore, A |x| =

|Ax| = |µ| |x| . Now let y ≡ |x| / ||x||1 . Then Ay = |µ|y and so |µ| ∈ S1 showing that|µ| ≤ λ0. But also, the fact the entries of x all have the same sign shows µ = |µ| and soµ ∈ S1. Since µ ̸= λ0, it must be that µ = |µ| < λ0. This proves 2.

It remains to verify 3. Suppose then that Ay = λ0y and for all scalars α, αx0 ̸= y. Then

ARey = λ0 Rey, A Imy = λ0 Imy.

If Rey = α1x0 and Imy = α2x0 for real numbers, αi,then y = (α1 + iα2)x0 and it isassumed this does not happen. Therefore, either

tRey ̸= x0 for all t ∈ R

ort Imy ̸= x0 for all t ∈ R.