Page 41 - Linear Algebra Done Right

Bases
done. If U = span(v 1 ,...,v j−1 ), then choose a vector v j ∈ U such
that
v j ∉ span(v 1 ,...,v j−1 ). 27
After each step, as long as the process continues, we have constructed
a list of vectors such that no vector in this list is in the span of the
previous vectors. Thus after each step we have constructed a linearly
independent list, by the linear dependence lemma (2.4). This linearly
independent list cannot be longer than any spanning list of V (by 2.6),
and thus the process must eventually terminate, which means that U
is finite dimensional.

Bases

A basis of V is a list of vectors in V that is linearly independent and
spans V. For example,

(1, 0,..., 0), (0, 1, 0,..., 0),...,(0,..., 0, 1)
n
n
is a basis of F , called the standard basis of F . In addition to the

n
standard basis, F has many other bases. For example, (1, 2), (3, 5)

2
is a basis of F . The list (1, 2) is linearly independent but is not a

2
2
basis of F because it does not span F . The list (1, 2), (3, 5), (4, 7)
2
spans F but is not a basis because it is not linearly independent. As
m
another example, (1,z,...,z ) is a basis of P m (F).
The next proposition helps explain why bases are useful.
2.8 Proposition: A list (v 1 ,...,v n ) of vectors in V is a basis of V
if and only if every v ∈ V can be written uniquely in the form
2.9 v = a 1 v 1 +· · ·+ a n v n ,
where a 1 ,...,a n ∈ F.
Proof: First suppose that (v 1 ,...,v n ) is a basis of V. Let v ∈ V. This proof is
Because (v 1 ,...,v n ) spans V, there exist a 1 ,...,a n ∈ F such that 2.9 essentially a repetition
holds. To show that the representation in 2.9 is unique, suppose that of the ideas that led us
b 1 ,...,b n are scalars so that we also have to the definition of
linear independence.
v = b 1 v 1 +· · ·+ b n v n .