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1.5. Vector subspaces.

A related notion to a vector space is that of a vector subspace. Suppose that V  is a vector space and let W ⊂ V  be a subset. This means that W  consists of some (but not necessarily all) of the vectors in V  . As we saw above with the example of the upper half-plane, not every subset W  will itself be a vector space. Since V  is a vector space, we know that we can add vectors in W  and multiply them by scalars, but will only make W  into a vector space in its own right if the results of these operations are back in W  . Formally, we have

Definition 1.11. A subset W ⊂ V  of a vector space is a vector subspace of V  if the following two axioms are satisfied

  • S1 If v and w are vectors in W  , then so is v + w ; and
  • S2 For any scalar λ ∈ ℝ  , if w is any vector in W  , then so is λ w .

One also often uses the phrases ‘ W  ’ is a subspace of V  ’ and ‘ W  ’ is a linear subspace of ‘ V  .’

Problem 1.12. Let V  be a vector space. Prove that a (non-empty) subset U  of V  is a subspace if and only if for all u1,u2 ∈ U and for all λ,μ ∈ ℝ,

λu1 + μu2 ∈ U .

Axiom S1 says that W  is closed under vector addition, whereas S2 says that W  is closed under scalar multiplication.

Let us make sure we understand what these two properties mean. For v and w in W  , v+ w belongs to V  because V  is a vector space. The question is whether v + w belongs to W  , and S1 says that it does. Similarly, if w ∈ W  is a vector in W  and λ ∈ ℝ  is any scalar, then λw belongs to V  because V  is a vector space. The question is whether λw also belongs to W  , and S2 says that it does.

You may ask whether we should not also require that the zero vector 0  also belongs to W  . In fact this is guaranteed by S2, because for any w ∈ W  , 0 = 0w (why?) which belongs to W  by S2. From this point of view, it is S2 that fails in the example of the upper half-plane (1), since scalar multiplication by a negative scalar λ < 0  takes vectors in the upper half-plane to vectors in the lower half-plane. Similarly, if W ⊂ V  is a subspace, then if w ∈ W  , - w ∈ W  .

Problem 1.13. Find a subset 2S ⊂ ℝ  which is closed under scalar multiplication but not under vector addition. Do the same for   3ℝ  .

Consider the set ℝ3  of ordered triples of real numbers:

 3ℝ  = {(v1,v2,v3) ∣ vi ∈ ℝ for i = 1,2,3},
 and consider the following subsets

  • W1 = {(v1,v2,0) ∣ vi ∈ ℝ for i = 1,2} ⊂ ℝ3  ,
  • W2 = {(v1,v2,v3) ∣ vi ∈ ℝ for i = 1,2,3 and v3 ≥ 0} ⊂ ℝ3  , and
  • W3 = {(v1,v2,1) ∣ vi ∈ ℝ for i = 1,2} ⊂ ℝ3  .

I will leave it to you as an exercise to show that W  1 obeys both S1 and S2 whence it is a vector subspace of ℝ3 , whereas W  2  does not obey S2, and W  3  does not obey either one. Can you think of a subset of ℝ3  which obeys S2 but not S1? (See Problem 1.13.)