Page 101 - Matrices theory and applications

5. Nonnegative Matrices
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5.3.2 A Few Lemmas
Lemma 5.3.1 Let r ≥ 0 and x ≥ 0 such that Ax ≥ rx and Ax = rx.

Then there exists r >r such that C r is nonempty.
Proof
n−1
Let y := (I n + A)
x.Since A is irreducible and x ≥ 0isnonzero, one
has y> 0. Similarly, Ay − ry =(I n + A) n−1 (Ax − rx) > 0. Let us define

r := min j (Ay) j /y j , which is strictly larger than r.Wethen have Ay ≥ r y,
so that C r contains the vector y/ y 1.
Lemma 5.3.2 The nonnegative eigenvectors of A are positive.
Proof
+
Givensucha vector x with Ax = λx,we observe that λ ∈ IR .Then
1 n−1
x = (I n + A) x,
(1 + λ) n−1
and the right-hand side is strictly positive, from Proposition 5.1.2.
Finally, we can state the following result.
Lemma 5.3.3 Let A, B ∈ M n (CC) be matrices, with A irreducible and
|B|≤ A.Then ρ(B) ≤ ρ(A).
In case of equality (ρ(B)= ρ(A)), the following hold:
•|B| = A;
• for every eigenvector x of B associated to an eigenvalue of modulus
ρ(A), |x| is an eigenvector of A associated to ρ(A).

Proof
In order to establish the inequality, we proceed as above. If λ is an
eigenvalue of B, of modulus ρ(B), and if x is a normalized eigenvector,
then ρ(B)|x|≤ |B|· |x|≤ A|x|,so that C ρ(B) is nonempty. Hence ρ(B) ≤
R = ρ(A).
Let us investigate the case of equality. If ρ(B)= ρ(A), then |x|∈ C ρ(A) ,
and therefore |x| is an eigenvector: A|x| = ρ(A)|x| = ρ(B)|x|≤ |B|· |x|.
Hence, (A−|B|)|x|≤ 0. Since |x| > 0 (from Lemma 5.3.2) and A−|B|≥ 0,
this gives |B| = A.

5.3.3 The Eigenvalue ρ(A) Is Simple

Let P A (X) be the characteristic polynomial of A. It is given as the compo-
sition of an n-linear form (the determinant) with polynomial vector-valued
functions (the columns of XI n −A). If φ is p-linear and if V 1 (X),... ,V p (X)