Page 132 - A Course in Linear Algebra with Applications

116 Chapter Five: Basis and Dimension

possible to find arbitrarily large linearly independent subsets
of V.
We will prove the theorem by showing that the subset
X is a basis of V. Write X = {vi, V2, •.., v n }. Suppose that
u is a vector in V which does not belong to X. Then the
subset {vi, V2,..., v n , u} must be linearly dependent since it
properly contains X. Hence there is a linear relation of the
form
C1V1 + c 2 v 2 H h c n v n + du = 0

where not all of the scalars c±, C2,..., c n , d are zero. Now if
the scalar d were zero, it would follow that c\V\ + C2V2 +
• • • + c n v n = 0, which, in view of the linear independence of
vi, V2,..., v n , could only mean that c\ = c 2 = • • • = c n = 0.
But now all the scalars are zero, which is not true. Therefore
d 7^ 0. Consequently we can solve the above equation for u to
obtain

_1
1
_1
r
u = (-o? c 1 )vi + (-d~ c 2)\ 2 H 1- (-rf c n )v n .
Hence u belongs to < i , . . . , v n > . Prom this it follows that
v
the vectors v i , . . . , v n generate V; since these are also linearly
independent, they form a basis of V.

Corollary 5.1.5
Every non-zero finitely generated vector space V has a basis.

Indeed by hypothesis V contains a non-zero vector, say v.
Then {v} is linearly independent and by 5.1.4 it is contained
in a basis of V.

Usually a vector space will have many bases. For exam-
ple, the vector space R 2 has the basis