Page 108 - Matrices theory and applications

5.6. Exercises
91
,with I

∪ l,m J

and J
and j ∈ J ,
:= I l ∩ I
:= J l ∩ J .If i ∈ I

pq
m
lm
m
lm
lm
lm
with (l, m) =(p, q), we have m ij = 0. From Corollary 5.5.2, applied to

| = |J

|. Eliminating the empty
M and to its transposition, we have |I
lm
lm
parts, we obtain therefore a decomposition of M that is finer than the first

two, in the sense of inclusion order: Each I l (or I ) is a union of some parts
l

of the form I .
p
Since the set of decompositions of M is finite, the previous argument
shows that there exists a finest one. We shall call it the canonical decom-
position of M. It is the only decomposition for which the blocks of indices
I l × J l are themselves of class S∆.
5.6 Exercises
1. We consider the following three properties for a matrix M ∈ M n (IR).
P1 M is nonnegative.
T
T
P2 M e = e,where e =(1,... , 1) .
P3 M 1 ≤ 1.
(a) Show that P2 and P3 imply P1.
(b) Show that P2 and P1 imply P3.
(c) Does P1 and P3 imply P2?
2. Here is another proof of the simplicity of ρ(A) in the Perron–
Frobenius theorem, which does not require Lemma 5.3.3.
(a) We assume that A is irreducible and nonnegative, and we denote
by x a positive eigenvector associated to the eigenvalue ρ(A). Let
K be the set of nonnegative eigenvectors y associated to ρ(A)
such that y 1 =1. Show that K is compact and convex.
(b) Show that the geometric multiplicity of ρ(A)equals 1 (Hint:
Otherwise, K would contain a vector with at least one zero
component.)
(c) Show that the algebraic multiplicity of ρ(A)equals 1 (Hint:
Otherwise, there would be a nonnegative vector y such that Ay−
ρ(A)y = x> 0.)
3. Let M ∈ M n (IR) be either a strictly diagonally dominant, or an
irreducible strongly diagonally dominant, matrix. Assume that m jj >
0 for every j =1,... ,n and m ij ≤ 0 otherwise. Show that M is
invertible and that the solution of Mx = b,when b ≥ 0, satisfies
x ≥ 0. Deduce that M −1 ≥ 0.
4. Here is another proof of Theorem 5.3.1, due to Perron himself. We
proceed by induction on the size n of the matrix. The statement is
obvious if n = 1. We therefore assume that it holds for matrices
of size n. We give ourselves an irreducible nonnegative matrix A ∈