Page 39 - Linear Algebra Done Right

Linear Dependence Lemma: If (v 1 ,...,v m ) is linearly depen-
2.4
dent in V and v 1 = 0, then there exists j ∈{2,...,m} such that the
following hold: Span and Linear Independence 25
(a) v j ∈ span(v 1 ,...,v j−1 );
(b) if the j th term is removed from (v 1 ,...,v m ), the span of the
remaining list equals span(v 1 ,...,v m ).

Proof: Suppose (v 1 ,...,v m ) is linearly dependent in V and v 1 = 0.
Then there exist a 1 ,...,a m ∈ F, not all 0, such that
a 1 v 1 +· · ·+ a m v m = 0.

Not all of a 2 ,a 3 ,...,a m can be 0 (because v 1 = 0). Let j be the largest
element of {2,...,m} such that a j = 0. Then

a 1 a j−1
2.5 v j =− v 1 − ··· − v j−1 ,
a j a j
proving (a).
To prove (b), suppose that u ∈ span(v 1 ,...,v m ). Then there exist
c 1 ,...,c m ∈ F such that

u = c 1 v 1 +· · ·+ c m v m .

In the equation above, we can replace v j with the right side of 2.5,
which shows that u is in the span of the list obtained by removing the
j th term from (v 1 ,...,v m ). Thus (b) holds.

Now we come to a key result. It says that linearly independent lists
are never longer than spanning lists.

2.6 Theorem: In a finite-dimensional vector space, the length of Suppose that for each
every linearly independent list of vectors is less than or equal to the positive integer m,
length of every spanning list of vectors. there exists a linearly
independent list of m
Proof: Suppose that (u 1 ,...,u m ) is linearly independent in V and vectors in V. Then this
that (w 1 ,...,w n ) spans V. We need to prove that m ≤ n. Wedoso theorem implies that V
through the multistep process described below; note that in each step is infinite dimensional.
we add one of the u’s and remove one of the w’s.