Page 98 - Matrices theory and applications

5.2. The Perron–Frobenius Theorem: Weak Form
81
Let us assume that Ax ≥ 0 (respectively > 0) for every x ≥ 0 (respec-
tively ≥ 0and = 0). Then the ith column A
positive), since it is the image of the ith vector of the canonical basis. Hence
A ≥ 0 (respectively > 0).
Conversely, A ≥ 0and x ≥ 0 imply trivially Ax ≥ 0. If A> 0, x ≥ 0,
and x = 0, there exists an index l such that x l > 0. Then
(i) is nonnegative (respectively
(Ax) i = a ij x j ≥ a il x l > 0,
j
and hence Ax > 0.
An important point is the following:
Proposition 5.1.2 If A ∈ M n (IR) is nonnegative and irreducible, then
(I + A) n−1 > 0.
Proof
m
Let x be a nonnegative, nonzero vector and define x m =(I + A) x,
which is nonnegative. Let us denote by P m the set of indices of the nonzero
m
m
components of x : P 0 is nonempty. Since x m+1 ≥ x ,one has P m ⊂
i
i
P m+1 . Let us assume that the cardinality |P m | of P m is strictly less than
n. There are thus one or more zero components, whose indices form a
nonempty subset I, complement of P m .Since A is irreducible, there exists
some nonzero entry a ij ,with i ∈ I and j ∈ P m .Then x m+1 ≥ a ij x m > 0,
j
i
which shows that P m+1 is not equal to P m , and thus |P m+1 | > |P m |.By
induction, we deduce that |P m |≥ min{m +1,n}. Hence |P n−1 | = n.
5.2 The Perron–Frobenius Theorem: Weak Form

Theorem 5.2.1 Let A ∈ M n (IR) be a nonnegative matrix. Then ρ(A) is
an eigenvalue of A associated to a nonnegative eigenvector.

Proof
Let λ be an eigenvalue of maximal modulus and v an eigenvector,
normalized by v 1 =1. Then
ρ(A)|v| = |λv| = |Av|≤ A|v|.
n
Let us denote by C the subset of IR (actually a subset of the unit simplex

K n) defined by the (in)equalities x i =1, x ≥ 0, and Ax ≥ ρ(A)x.This
i
is a closed convex set, nonempty, since it contains |v|. Finally, it is bounded,
because x ∈ C implies 0 ≤ x j ≤ 1 for every j;thusit iscompact.Let us
distinguish two cases:
1. There exists x ∈ C such that Ax =0. Then ρ(A)x ≤ 0 furnishes
ρ(A) = 0. The theorem is thus proved in this case.