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## 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

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

• 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 ‘ .’

Problem 1.12. Let be a vector space. Prove that a (non-empty) subset of is a subspace if and only if for all and for all ,

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

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 , .

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

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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