Linear Algebra, 2020a

Section II. Linear Independence 109
repeated data. We would have been trying to solve a two-unknowns problem
with essentially only one piece of information.
We take ⃗v to be a “repeat” of the vectors in a set S if ⃗v ∈ [S] so that it depends
on, that is, is expressible in terms of, elements of the set ⃗v = c 1 ⃗s 1 + ···+ c n ⃗s n .
1.2 Lemma Where V is a vector space, S is a subset of that space, and ⃗v is an
element of that space, [S ∪ {⃗v}] =[S] if and only if ⃗v ∈ [S].
Proof Half of the if and only if is immediate: if ⃗v /∈ [S] then the sets are not
equal because ⃗v ∈ [S ∪ {⃗v}].
For the other half assume that ⃗v ∈ [S] so that ⃗v = c 1 ⃗s 1 + ···+ c n ⃗s n for some
scalars c i and vectors ⃗s i ∈ S. We will use mutual containment to show that the
sets [S ∪ {⃗v}] and [S] are equal. The containment [S ∪ {⃗v}] ⊇ [S] is clear.
To show containment in the other direction let ⃗w be an element of [S ∪ {⃗v}].
Then ⃗w is a linear combination of elements of S ∪ {⃗v}, which we can write as
⃗w = c n+1 ⃗s n+1 + ···+ c n+k ⃗s n+k + c n+k+1 ⃗v. (Possibly some of the ⃗s i ’s from
⃗w’s equation are the same as some of those from ⃗v’s equation but that does not
matter.) Expand ⃗v.
⃗w = c n+1 ⃗s n+1 + ···+ c n+k ⃗s n+k + c n+k+1 · (c 1 ⃗s 1 + ···+ c n ⃗s n )
Recognize the right hand side as a linear combination of linear combinations of
vectors from S. Thus⃗w ∈ [S].
QED
The discussion at the section’s opening involved removing vectors instead of
adding them.
1.3 Corollary For ⃗v ∈ S, omitting that vector does not shrink the span [S] =
[S − {⃗v}] if and only if it is dependent on other vectors in the set ⃗v ∈ [S].
The corollary says that to know whether removing a vector will decrease the
span, we need to know whether the vector is a linear combination of others in
the set.
1.4 Definition In any vector space, a set of vectors is linearly independent if none
of its elements is a linear combination of the others from the set. ∗ Otherwise
the set is linearly dependent.
Thus the set {⃗s 0 ,...,⃗s n } is independent if there is no equality ⃗s i = c 0 ⃗s 0 +
...+ c i−1 ⃗s i−1 + c i+1 ⃗s i+1 + ...+ c n ⃗s n . The definition’s use of the word ‘others’
means that writing ⃗s i as a linear combination via ⃗s i = 1 · ⃗s i does not count.
∗ See also Remark 1.13.