Page 99 - Matrices theory and applications

5. Nonnegative Matrices
82
2. For every x in C, Ax = 0. Then let us define on C a continuous map
f by
1
It is clear that f(x) ≥ 0and that f(x) 1 = 1. Finally,
1
1 f(x)= Ax 1 Ax.
Af(x)= AAx ≥ Aρ(A)x = ρ(A)f(x),
Ax 1 Ax 1
so that f(C) ⊂ C. Then Brouwer’s theorem (see [3], p. 217) asserts
N
that a continuous function from a compact convex subset of IR
into itself has a fixed point. Thus let y be a fixed point of f.Itis
a nonnegative eigenvector, associated to the eigenvalue r = Ay 1 .
Since y ∈ C,we have ry = Ay ≥ ρ(A)y and thus r ≥ ρ(A), which
implies r = ρ(A).
That proof can be adapted to the case where a real number r and a
nonzero vector y are given satisfying y ≥ 0and Ay ≥ ry. Just take for C

the set of vectors x such that x i =1, x ≥ 0, and Ax ≥ rx.We then
i
conclude that ρ(A) ≥ r.

5.3 The Perron–Frobenius Theorem: Strong Form

Theorem 5.3.1 Let A ∈ M n (IR) be a nonnegative irreducible matrix.
Then ρ(A) is a simple eigenvalue of A, associated to a positive eigenvector.
Moreover, ρ(A) > 0.

5.3.1 Remarks
1. Though the Perron–Frobenius theorem says that ρ(A)is a simple
eigenvalue, it does not tell anything about the other eigenvalues
of maximal modulus. The following example shows that such other
eigenvalues may exist:

0 1
.
1 0
The existence of several eigenvalues of maximal modulus will be
studied in Section 5.4.

2. One obtains another proof of the weak form of the Perron–Frobenius
theorem by applying the strong form to A + αJ,where J> 0and
α> 0, then letting α tend to zero.