Page 131 - A Course in Linear Algebra with Applications

5.1: Existence of a Basis 115

Theorem 5.1.3
If{ v l j v 2 > • • • j v n
} is a basis of a vector space V, then each
vector v in V has a unique expression of the form
v = civi + c 2 v 2 H h c n v n

/or certain scalars Ci.

Proof
If there are two such expressions for v, say civi + • • • + c n v n
and diVi + • • • + d nv n, then, by equating these, we arrive at
the equation

(ci - di)vi H h (c n - d n)v n — 0.

By linear independence of the Vi this can only mean that c^ =
di for all i, so the expression is unique as claimed.

Naturally the question arises: does every vector space
have a basis? The answer is negative in general. Since a zero
space has 0 as its only vector, it has no linearly independent
subsets at all; thus a zero space cannot have a basis. However,
apart from this uninteresting case, every finitely generated
vector space has a basis, a fundamental result that will now
be proved. Notice that such a basis must be finite by 5.1.2.

Theorem 5.1.4
Let V be a finitely generated vector space and suppose that XQ
is a linearly independent subset of V. Then XQ is contained in
some basis XofV.

Proof
Suppose that V is generated by m elements. Then by 5.1.2
no linearly independent subset of V can contain more than m
elements. From this it follows that there exists a subset X
of V containing X 0 which is as large as possible subject to
being linearly independent. For if this were false, it would be