A First Course in Linear Algebra, 2017a

4.10. Spanning, Linear Independence and Basis in R n 199
Theorem 4.78: Subspaces are Spans
Let V be a nonempty collection of vectors in R n . Then V is a subspace of R n if and only if there
exist vectors {⃗u 1 ,···,⃗u k } in V such that
V = span{⃗u 1 ,···,⃗u k }
Furthermore, let W be another subspace of R n and suppose {⃗u 1 ,···,⃗u k }∈W. Then it follows that
V is a subset of W.
Note that since W is arbitrary, the statement that V ⊆ W means that any other subspace of R n that
contains these vectors will also contain V .
Proof. We first show that if V is a subspace, then it can be written as V = span{⃗u 1 ,···,⃗u k }. Pick a vector
⃗u 1 in V .IfV = span{⃗u 1 }, then you have found your list of vectors and are done. If V ≠ span{⃗u 1 },then
there exists⃗u 2 a vector of V which is not in span{⃗u 1 }. Consider span{⃗u 1 ,⃗u 2 }.IfV = span{⃗u 1 ,⃗u 2 },weare
done. Otherwise, pick ⃗u 3 not in span{⃗u 1 ,⃗u 2 }. Continue this way. Note that since V is a subspace, these
spans are each contained in V . The process must stop with ⃗u k for some k ≤ n by Corollary 4.70, and thus
V = span{⃗u 1 ,···,⃗u k }.
Now suppose V = span{⃗u 1 ,···,⃗u k }, we must show this is a subspace. So let ∑ k i=1 c i⃗u i and ∑ k i=1 d i⃗u i be
two vectors in V, andleta and b be two scalars. Then
a
k
∑
i=1
c i ⃗u i + b
k
∑
i=1
d i ⃗u i =
k
∑
i=1
(ac i + bd i )⃗u i
which is one of the vectors in span{⃗u 1 ,···,⃗u k } and is therefore contained in V . This shows that span{⃗u 1 ,···,⃗u k }
has the properties of a subspace.
To prove that V ⊆ W , we prove that if ⃗u i ∈ V, then⃗u i ∈ W.
Suppose ⃗u ∈ V. Then⃗u = a 1 ⃗u 1 + a 2 ⃗u 2 + ···+ a k ⃗u k for some a i ∈ R, 1≤ i ≤ k. SinceW contain each
⃗u i and W is a vector space, it follows that a 1 ⃗u 1 + a 2 ⃗u 2 + ···+ a k ⃗u k ∈ W .
♠
Since the vectors ⃗u i we constructed in the proof above are not in the span of the previous vectors (by
definition), they must be linearly independent and thus we obtain the following corollary.
Corollary 4.79: Subspaces are Spans of Independent Vectors
If V is a subspace of R n , then there exist linearly independent vectors {⃗u 1 ,···,⃗u k } in V such that
V = span{⃗u 1 ,···,⃗u k }.
In summary, subspaces of R n consist of spans of finite, linearly independent collections of vectors of
R n . Such a collection of vectors is called a basis.