Page 90 - Matrices theory and applications

4.6. Exercises
73
for which |x i | = x ∞ and let J be its complement. If i ∈ I,then

|a ij ||x j |.
|a ij |≤ |λ − a ii | x ∞ =
a ij x j ≤

x ∞

j =i
j =i
j =i

( x ∞ −|x j |)|a ij |≤ 0, where all the terms in the sum
It follows that
j =i
are nonnegative. Each term is thus zero, so that a ij =0 for j ∈ J.Since A
is irreducible, J is empty. One has thus |x j | = x ∞ for every j,and the
previous inequalities show that λ belongs to every circle.
Definition 4.5.1 A square matrix A ∈ M n (CC) is said to be
1. diagonally dominant if

|a ii |≥ |a ij |, 1 ≤ i ≤ n;
j =i
2. strongly diagonally dominant if in addition at least one of these n
inequalities is strict;
3. strictly diagonally dominant if the inequality is strict for every index
i.
Corollary 4.5.1 Let A be a square matrix. If A is strictly diagonally dom-
inant, or if A is irreducible and strongly diagonally dominant, then A is
invertible.
In fact, either zero does not belong to the Gershgorin domain, or it is
not interior to the disks. In the latter case, A is assumed to be irreducible,
and there exists a disk D j that does not contain zero.
4.6 Exercises
n
1. Under what conditions on the vectors a, b ∈ CC does the matrix M
defined by m ij = a i b j satisfy M p =1 for every p ∈ [1, ∞]?
2. Under what conditions on x, y,and p does the equality in (4.2) or
(4.1) hold?
3. Show that
lim x p = x ∞, ∀x ∈ E.
p→+∞
n
4. A norm on K is a strictly convex norm if x = y =1, x = y,and
0 <θ < 1imply θx +(1 − θ)y < 1.
(a) Show that · p is strictly convex for 1 <p< ∞, but is not so
for p =1, ∞.