The notion of a subspace is of great importance in applications. Here is what is meant by a subspace.
Definition 3.2.1 Let V be a vector space with field of scalars F. Then let W ⊆ V,W≠∅. That is, W is a nonempty subset of V . Then W is a subspace of V if whenever α,β are scalars and u,v are vectors in W, it follows that

In words, W is closed with respect to linear combinations.
The fundamental result about subspaces is that they are themselves vector spaces.
Theorem 3.2.2 Let W be a nonzero subset of V a vector space with field of scalars F. Then it is a subspace if and only if is itself a vector space with field of scalars F.
Proof: Suppose W is a subspace. Why is it a vector space? To be a vector space, the operations of addition and scalar multiplication must satisfy the axioms for a vector space. However, all of these are obvious because it is a subset of V . The only thing which is not obvious is whether 0 is in W and whether −u ∈ W whenever u is. But these follow right away from Proposition 3.0.2 because if u ∈ W,
Conversely, suppose it is a vector space. Then by definition, it is closed with respect to linear combinations and so it is a subspace. ■
This leads to the following simple result.
Proposition 3.2.3 Let W be a nonzero subspace of a finite dimensional vector space V with field of scalars F. Then W is also a finite dimensional vector space.
Proof: Suppose span
Example 3.2.4 Show that W =
You have from the equation that x = 2y + z and so any vector in this set is of the form

Conversely, any vector which is of the above form satisfies the condition to be in W. Therefore, W is of the form

where y,z are scalars. Hence it equals the span of the two vectors in ℝ^{3} in the above. Are the two vectors linearly independent? If so, they will be a basis. Suppose then that

Then from the second position, y = 0. It follows then that z = 0 also and so the two vectors form a linearly independent set. Hence a basis for W is

The dimension of this subspace is also 2.
Example 3.2.5 Show that

is a basis for ℝ^{3}.
There are two things to show, that the set of vectors is independent and that it spans ℝ^{3}. Thus we need to verify that there is exactly one solution to the system of equations

for any choice of the right side. Recall how to do this. You set up the augmented matrix and then row reduce it.

After some row operations, this yields

Thus there is a unique solution to the system of equations. This shows that the set of vectors is a basis because one solution when the right side of the system equals the zero vector is x = y = z = 0. Therefore, from what was just done, it is the only solution and so the vectors are linearly independent. As to the span of the vectors equalling ℝ^{3}, this was just shown also.
Example 3.2.6 Show that

is not a basis for ℝ^{3}.
You can do it the same way. It is really a question about whether there exists a unique solution to the system

for any choice of the right side. The augmented matrix is

After row reduction, this yields

Thus there is no solution to the equation unless b = a. It follows the span of the given vectors is not all of ℝ^{3} and so this cannot be a basis.
Example 3.2.7 Show that

is not a basis for ℝ^{3}.
If the span of these vectors were all of ℝ^{3}, this would contradict Theorem 3.1.5 because it would be a spanning set which is shorter than a linearly independent set
Example 3.2.8 Show that

is not a basis for ℝ^{3}.
If it were a basis, then it would need to be linearly independent but this cannot happen because it would contradict Theorem 3.1.5 by being an independent set of vectors which is longer than a spanning set.
Theorem 3.2.9 If V is an n dimensional vector space and if
Proof: Say
Now say V = span
Now suppose for k < n,