Page 19 - Matrices theory and applications

1. Elementary Theory
2
be a map such that
(a + b)x = ax + bx,
One says that E is a vector space over K (one often speaks of a K-vector
space) if moreover,
a(bx)= (ab)x, a(x + y)= ax + ay.
1x = x,
hold for all a, b ∈ K and x ∈ E. The elements of E are called vectors.In a
vector space one always has 0x = 0 (more precisely, 0 K x =0 E ).
When P, Q ⊂ K and F, G ⊂ E, one denotes by PQ (respectively P +
Q, F +G, PF) the set of products pq as (p, q) ranges over P ×Q (respectively
p+q, f+g, pf as p, q, f, g range over P, Q, F, G). A subgroup (F, +) of (E, +)
that is stable under multiplication by scalars, i.e., such that KF ⊂ F,is
again a K-vector space. One says that it is a linear subspace of E,or justa
subspace. Observe that F, as a subgroup, is nonempty, since it contains 0 E .
The intersection of any family of linear subspaces is a linear subspace. The
sum F + G of two linear subspaces is again a linear subspace. The trivial
formula (F + G)+ H = F +(G + H) allows us to define unambiguously
F + G + H and, by induction, the sum of any finite family of subsets of E.
When these subsets are linear subspaces, their sum is also a linear subspace.
I
Let I be a set. One denotes by K the set of maps a =(a i ) i∈I : I → K
where only finitely many of the a i ’s are nonzero. This set is naturally
endowed with a K-vector space structure, by the addition and product
laws
(a + b) i := a i + b i , (λa) i := λa i .
Let E be a vector space and let i → f i be a map from I to E.A linear
combination of (f i ) i∈I is a sum

a i f i ,
i∈I
where the a i ’s are scalars, only finitely many of which are nonzero (in other
I
words, (a i ) i∈I ∈ K ). This sum involves only finitely many terms. It is a
vector of E. The family (f i ) i∈I is free if every linear combination but the
trivial one (when all coefficients are zero) is nonzero. It is a generating
family if every vector of E is a linear combination of its elements. In other
words, (f i ) i∈I is free (respectively generating) if the map
K I → E,

(a i ) i∈I → a i f i ,
i∈I
is injective (respectively onto). Last, one says that (f i ) i∈I is a basis of E if
it is free and generating. In that case, the above map is bijective, and it is
actually an isomorphism between vector spaces.