3.2. SUBSPACES 57

Proof: Suppose W is a subspace. Why is it a vector space? To be a vector space, theoperations 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 isnot obvious is whether 0 is in W and whether −u ∈W whenever u is. But these followright away from Proposition 3.0.2 because if u ∈W,(−1)u = −u ∈W by the fact that Wis closed with respect to linear combinations, in particular multiplication by the scalar −1.Similarly, take u∈W. Then 0 = 0u∈W. As to + being an operation on W, this also followsbecause for u,v ∈W,u+ v ∈W . Thus if it is a subspace, it is indeed a vector space.

Conversely, suppose it is a vector space. Then by definition, it is closed with respect tolinear 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 Vwith field of scalars F. Then W is also a finite dimensional vector space.

Proof: Suppose span(v1, · · · ,vn) = V . Using the same construction of Proposition3.1.8, the same process must stop after k ≤ n steps since otherwise one could obtain alinearly independent set of vectors with more vectors in it than a spanning set. Thus it hasa basis with no more than n vectors. ■

Example 3.2.4 Show that W ={(x,y,z) ∈ R3 : x−2y− z = 0

}is a subspace of R3. Find

a basis for it.

You have from the equation that x = 2y+ z and so any vector in this set is of the form 2y+ zyz

 : y,z ∈ R

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

y

 210

+ z

 101

where y,z are scalars. Hence it equals the span of the two vectors in R3 in the above. Arethe two vectors linearly independent? If so, they will be a basis. Suppose then that

y

 210

+ z

 101

=

 000

Then from the second position, y = 0. It follows then that z = 0 also and so the two vectorsform a linearly independent set. Hence a basis for W is

 210

 ,

 101



The dimension of this subspace is also 2.