Page 42 - Linear Algebra Done Right

Chapter 2. Finite-Dimensional Vector Spaces
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Subtracting the last equation from 2.9, we get
0 = (a 1 − b 1 )v 1 + ··· + (a n − b n )v n .
This implies that each a j −b j = 0 (because (v 1 ,...,v n ) is linearly inde-
pendent) and hence a 1 = b 1 ,...,a n = b n . We have the desired unique-
ness, completing the proof in one direction.
For the other direction, suppose that every v ∈ V can be written
uniquely in the form given by 2.9. Clearly this implies that (v 1 ,...,v n )
spans V. To show that (v 1 ,...,v n ) is linearly independent, suppose
that a 1 ,...,a n ∈ F are such that

0 = a 1 v 1 +· · ·+ a n v n .

The uniqueness of the representation 2.9 (with v = 0) implies that
a 1 =· · ·= a n = 0. Thus (v 1 ,...,v n ) is linearly independent and
hence is a basis of V.

A spanning list in a vector space may not be a basis because it is not
linearly independent. Our next result says that given any spanning list,
some of the vectors in it can be discarded so that the remaining list is
linearly independent and still spans the vector space.

2.10 Theorem: Every spanning list in a vector space can be reduced
to a basis of the vector space.

Proof: Suppose (v 1 ,...,v n ) spans V. We want to remove some
of the vectors from (v 1 ,...,v n ) so that the remaining vectors form a
basis of V. We do this through the multistep process described below.
Start with B = (v 1 ,...,v n ).

Step 1
If v 1 = 0, delete v 1 from B.If v 1 = 0, leave B unchanged.

Step j
If v j is in span(v 1 ,...,v j−1 ), delete v j from B.If v j is not in
span(v 1 ,...,v j−1 ), leave B unchanged.

Stop the process after step n, getting a list B. This list B spans V
because our original list spanned B and we have discarded only vectors
that were already in the span of the previous vectors. The process