A First Course in Linear Algebra, 2017a

490 Vector Spaces
Now the following is a fundamental result about subspaces.
Theorem 9.48: Basis of a Vector Space
Let V be a finite dimensional vector space and let W be a non-zero subspace. Then W has a basis.
That is, there exists a linearly independent set of vectors {⃗w 1 ,···,⃗w r } such that
span{⃗w 1 ,···,⃗w r } = W
Also if {⃗w 1 ,···,⃗w s } is a linearly independent set of vectors, then W has a basis of the form
{⃗w 1 ,···,⃗w s ,···,⃗w r } for r ≥ s.
Proof. Let the dimension of V be n. Pick ⃗w 1 ∈ W where ⃗w 1 ≠⃗0. If ⃗w 1 ,···,⃗w s have been chosen such
that {⃗w 1 ,···,⃗w s } is linearly independent, if span{⃗w 1 ,···,⃗w r } = W, stop. You have the desired basis.
Otherwise, there exists ⃗w s+1 /∈ span{⃗w 1 ,···,⃗w s } and {⃗w 1 ,···,⃗w s ,⃗w s+1 } is linearly independent. Continue
this way until the process stops. It must stop since otherwise, you could obtain a linearly independent set
of vectors having more than n vectors which is impossible.
The last claim is proved by following the above procedure starting with {⃗w 1 ,···,⃗w s } as above. ♠
This also proves the following corollary. Let V play the role of W in the above theorem and begin with
abasisforW , enlarging it to form a basis for V as discussed above.
Corollary 9.49: Basis Extension
Let W be any non-zero subspace of a vector space V. Then every basis of W can be extended to a
basis for V .
Consider the following example.
Example 9.50: Basis Extension
Let V = R 4 and let
Extend this basis of W to a basis of V.
⎧⎡
⎪⎨
W = span ⎢
⎣
⎪⎩
1
0
1
1
⎤ ⎡
⎥
⎦ , ⎢
⎣
0
1
0
1
⎤⎫
⎪⎬
⎥
⎦
⎪⎭
Solution. An easy way to do this is to take the reduced row-echelon form of the matrix
⎡
⎤
1 0 1 0 0 0
⎢ 0 1 0 1 0 0
⎥
⎣ 1 0 0 0 1 0 ⎦ (9.3)
1 1 0 0 0 1
Note how the given vectors were placed as the first two and then the matrix was extended in such a way
that it is clear that the span of the columns of this matrix yield all of R 4 . Now determine the pivot columns.