## 1.5. Vector subspaces.A related notion to a vector space is that of a vector subspace. Suppose that is a vector space and let be a subset. This means that consists of some (but not necessarily all) of the vectors in . As we saw above with the example of the upper half-plane, not every subset will itself be a vector space. Since is a vector space, we know that we can add vectors in and multiply them by scalars, but will only make into a vector space in its own right if the results of these operations are back in . Formally, we have - S1 If and are vectors in , then so is ; and
- S2 For any scalar , if is any vector in , then so is .
One also often uses the phrases ‘ ’ is a subspace of ’ and ‘ ’ is a linear subspace of ‘ .’ Axiom S1 says that Let us make sure we understand what these two properties mean. For and in , belongs to because is a vector space. The question is whether belongs to , and S1 says that it does. Similarly, if is a vector in and is any scalar, then belongs to because is a vector space. The question is whether also belongs to , and S2 says that it does. You may ask whether we should not also require that the zero vector also belongs to . In fact this is guaranteed by S2, because for any , (why?) which belongs to 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 takes vectors in the upper half-plane to vectors in the lower half-plane. Similarly, if is a subspace, then if , . Consider the set of ordered triples of real numbers: and consider the following subsets - • ,
- • , and
- • .
I will leave it to you as an exercise to show that obeys both S1 and S2 whence it is a vector subspace of , whereas does not obey S2, and does not obey either one. Can you think of a subset of which obeys S2 but not S1? (See Problem 1.13.)
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