Page 193 - Matrix Analysis & Applied Linear Algebra

188 Chapter 4 Vector Spaces

Basic Facts of Independence
For a nonempty set of vectors S = {u 1 , u 2 ,..., u n } in a space V, the
following statements are true.
• If S contains a linearly dependent subset, then S itself (4.3.13)
must be linearly dependent.
• If S is linearly independent, then every subset of S is (4.3.14)
also linearly independent.
• If S is linearly independent and if v ∈V, then the ex- (4.3.15)
tension set S ext = S∪{v} is linearly independent if and
only if v /∈ span (S) .
m
• If S⊆ and if n>m, then S must be linearly (4.3.16)
dependent.

Proof of (4.3.13). Suppose that S contains a linearly dependent subset, and,
for the sake of convenience, suppose that the vectors in S have been permuted
so that this dependent subset is S dep = {u 1 , u 2 ,..., u k } . According to the
definition of dependence, there must be scalars α 1 ,α 2 ,...,α k , not all of which
are zero, such that α 1 u 1 +α 2 u 2 +···+α k u k = 0. This means that we can write
α 1 u 1 + α 2 u 2 + ··· + α k u k +0u k+1 + ··· +0u n = 0,
where not all of the scalars are zero, and hence S is linearly dependent.
Proof of (4.3.14). This is an immediate consequence of (4.3.13).
Proof of (4.3.15). If S ext is linearly independent, then v /∈ span (S) , for
otherwise v would be a combination of vectors from S thus forcing S ext to
be a dependent set. Conversely, suppose v /∈ span (S) . To prove that S ext is
linearly independent, consider a linear combination
α 1 u 1 + α 2 u 2 + ··· + α n u n + α n+1 v = 0. (4.3.17)
It must be the case that α n+1 =0, for otherwise v would be a combination of
vectors from S. Consequently,
α 1 u 1 + α 2 u 2 + ··· + α n u n = 0.
But this implies that
α 1 = α 2 = ··· = α n =0
because S is linearly independent. Therefore, the only solution for the α ’s in
(4.3.17) is the trivial set, and hence S ext must be linearly independent.
Proof of (4.3.16). This follows from (4.3.3) because if the u i ’s are placed as
columns in a matrix A m×n , then rank (A) ≤ m<n.