Page 100 - Matrices theory and applications

5.3. The Perron–Frobenius Theorem: Strong Form
83
3. Without the irreducibility assumption, ρ(A) may be a multiple eigen-
value, and a nonnegative eigenvector may not be positive. This holds
for a matrix of size n =2m that reads blockwise

B
0 m
A =
.
B
I m
Here, ρ(A)= ρ(B), and every eigenvalue has an even algebraic mul-
tiplicity. Moreover, if ρ(B) is a simple eigenvalue of B, associated to
the eigenvector Z ≥ 0, then the kernel of A − ρ(A)I n is spanned by

0 m
X = ,
Z
which is not positive.
Proof
For r ≥ 0, we denote by C r the set of vectors of IR n defined by the
(in)equalities
x ≥ 0, x 1 =1, Ax ≥ rx.
Each C r is a convex compact set. We saw in the previous section that if λ
is an eigenvalue associated to an eigenvector x of unit norm x 1 =1, then
|x|∈ C |λ| .Inparticular, C ρ(A) is nonempty. Conversely, if C r is nonempty,
then for x ∈ C r ,
r = r x 1 ≤ Ax 1 ≤ A 1 x 1 = A 1 ,
and therefore r ≤ A 1 . Furthermore, the map r → C r is nonincreasing
with respect to inclusion, and is “left continuous” in the following sense. If
r> 0, one has
C r = ∩ s<r C s .
Let us then define
R =sup{r | C r = ∅},
so that R ∈ [ρ(A), A 1 ]. The monotonicity with respect to inclusion shows
that r< R implies C r = ∅.
If x> 0and x 1 =1, then Ax ≥ 0and Ax =0, since A is nonnegative
and irreducible. From Lemma 5.3.1 it follows that R> 0. The set C R ,being
the intersection of a totally ordered family of nonempty compacts sets, is
nonempty.
Let x ∈ C R . Lemma 5.3.1 below shows that x is an eigenvector of A
associated to the eigenvalue R. We observe that this eigenvalue is not less
than ρ(A) and infer that ρ(A)= R. Hence ρ(A) is an eigenvalue associated
to the eigenvector x,and ρ(A) > 0. Lemma 5.3.2 below ensures that x> 0.
The proof of the simplicity of the eigenvalue ρ(A) will be given in Section
5.3.3.