A First Course in Linear Algebra, 2017a

204 R n ⎡
⎢
⎣
1 2 1
1 3 0
1 3 −1
1 2 0
⎤
⎡
⎥
⎦ → ⎢
⎣
1 0 0
0 1 0
0 0 1
0 0 0
Therefore, S can be extended to the following basis of U:
⎧⎡
⎤ ⎡ ⎤ ⎡ ⎤⎫
1 2 1
⎪⎨
⎢ 1
⎥
⎣ 1 ⎦
⎪⎩
, ⎢ 3
⎥
⎣ 3 ⎦ , ⎢ 0
⎪⎬
⎥
⎣ −1 ⎦ ,
⎪⎭
1 2 0
Next we consider the case of removing vectors from a spanning set to result in a basis.
⎤
⎥
⎦
♠
Theorem 4.90: Finding a Basis from a Span
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
and not all of the c i = 0, then you could pick c j ≠ 0, divide by it and solve for ⃗u j in terms of the others,
(
⃗w j = ∑ − c )
i
⃗w i
i≠ j
c j
Then you could delete ⃗w j from the list and have the same span. Any linear combination involving ⃗w j
would equal one in which ⃗w j is replaced with the above sum, showing that it could have been obtained as
a linear combination of ⃗w i for i ≠ j. Thus k − 1 ∈ S contrary to the choice of k . Hence each c i = 0andso
{⃗u 1 ,···,⃗u k } is a basis for W consisting of vectors of {⃗w 1 ,···,⃗w m }.
♠
The following example illustrates how to carry out this shrinking process which will obtain a subset of
a span of vectors which is linearly independent.
Example 4.91: Subset of a Span
Let W be the subspace
⎧⎡
⎪⎨
span ⎢
⎣
⎪⎩
1
2
−1
1
⎤ ⎡
⎥
⎦ , ⎢
⎣
1
3
−1
1
⎤ ⎡
⎥
⎦ , ⎢
⎣
8
19
−8
8
⎤ ⎡
⎥
⎦ , ⎢
⎣
−6
−15
6
−6
⎤ ⎡
⎥
⎦ , ⎢
⎣
Find a basis for W which consists of a subset of the given vectors.
1
3
0
1
⎤ ⎡
⎥
⎦ , ⎢
⎣
1
5
0
1
⎤⎫
⎪⎬
⎥
⎦
⎪⎭