A First Course in Linear Algebra, 2017a

9.4. Subspaces and Basis 487
Then from Lemma 9.23, {⃗u 1 ,···,⃗u k ,⃗u k+1 } is also linearly independent. Continue adding vectors in this
way until n linearly independent vectors have been obtained. Then
span{⃗u 1 ,···,⃗u n } = V
because if it did not do so, there would exist⃗u n+1 as just described and {⃗u 1 ,···,⃗u n+1 } would be a linearly
independent set of vectors having n + 1 elements. This contradicts the fact that {⃗v 1 ,···,⃗v n } is a basis. In
turn this would contradict Theorem 9.35. Therefore, this list is a basis.
♠
Recall Example 9.24 in which we added a matrix to a linearly independent set to create a larger linearly
independent set. By Theorem 9.43 we can extend a linearly independent set to a basis.
Example 9.44: Adding to a Linearly Independent Set
Let S ⊆ M 22 be a linearly independent set given by
{[
1 0
S =
0 0
Enlarge S to a basis of M 22 .
]
,
[
0 1
0 0
]}
Solution. Recall from the solution of Example 9.24 that the set R ⊆ M 22 given by
{[ ] [ ] [ ]}
1 0 0 1 0 0
R = , ,
0 0 0 0 1 0
is also linearly [ independent. ] However this set is still not a basis for M 22 as it is not a spanning set. In
0 0
particular, is not in spanR. Therefore, this matrix can be added to the set by Lemma 9.23 to
0 1
obtain a new linearly independent set given by
{[ ] [ ] [ ] [ ]}
1 0 0 1 0 0 0 0
T = , , ,
0 0 0 0 1 0 0 1
This set is linearly independent and now spans M 22 . Hence T is a basis.
♠
Next we consider the case where you have a spanning set and you want a subset which is a basis. The
above discussion involved adding vectors to a set. The next theorem involves removing vectors.
Theorem 9.45: Basis from a Spanning Set
Let V be a vector space and let W be a subspace. Also suppose that W = span{⃗w 1 ,···,⃗w m }.Then
there exists a subset of {⃗w 1 ,···,⃗w m } which is a basis for W.
Proof. Let S denote the set of positive integers such that for k ∈ S, there exists a subset of {⃗w 1 ,···,⃗w m }
consisting of exactly k vectors which is a spanning set for W. Thus m ∈ S. Pick the smallest positive
integer in S. Callitk. Then there exists {⃗u 1 ,···,⃗u k }⊆{⃗w 1 ,···,⃗w m } such that span{⃗u 1 ,···,⃗u k } = W. If
k
∑
i=1
c i ⃗w i =⃗0