Let’s go back to the previous section for a second and examine Example 1 and Example 6.  In Example 1 we saw that  was a vector space with the standard addition and scalar multiplication for any positive integer n.  So, in particular  is a vector space with the standard addition and scalar multiplication.  In Example 6 we saw that the set of points on a line through the origin in  with the standard addition and vector space multiplication was also a vector space.

So, just what is so important about these two examples?  Well first notice that they both are using the same addition and scalar multiplication.  In and of itself that isn’t important, but it will be important for the end result of what we want to discus here.  Next, the set of points in the vector space of Example 6 are also in the set of points in the vector space of Example 1.  While it’s not important to the discussion here note that the opposite isn’t true, given a line we can find points in  that aren’t on the line. 

What we’ve seen here is that, at least for some vector spaces, it is possible to take certain subsets of the original vector space and as long as we retain the definition of addition and scalar multiplication we will get a new vector space.  Of course, it is possible for some subsets to not be a new vector space.  To see an example of this see Example 7 from the previous section.  In that example we’ve got a subset of  with the standard addition and scalar multiplication and yet it’s not a vector space.

We want to investigate this idea in more detail and we’ll start off with the following definition.

Now, technically if we wanted to show that a subset W of a vector space V was a subspace we’d need to show that all 10 of the axioms from the definition of a vector space are valid, however, in reality that doesn’t need to be done. 

Many of the axioms (c, d, g, h, i, and j) deal with how addition and scalar multiplication work, but W is inheriting the definition of addition and scalar multiplication from V.  Therefore, since elements of W are also elements of V the six axioms listed above are guaranteed to be valid on W.

The only ones that we really need to worry about are the remaining four, all of which require something to be in the subset W.  The first two (a, and b) are the closure axioms that require that the sum of any two elements from W is back in W and that the scalar multiple of any element from W will be back in W.   Note that the sum and scalar multiple will be in V we just don’t know if it will be in W.  We also need to verify that the zero vector (axiom e) is in W and that each element of W has a negative that is also in W (axiom f).

As the following theorem shows however, the only two axioms that we really need to worry about are the two closure axioms.  Once we have those two axioms valid, we will get the zero vector and negative vector for free.

Proof : To prove this theorem all we need to do is show that if we assume the two closure axioms are valid the other 8 axioms will be given to us for free.  As we discussed above the axioms c, d, g, h, i, and j are true simply based on the fact that W is a subset of V and it uses the same addition and scalar multiplication and so we get these for free.

We only need to verify that assuming the two closure condition we get axioms e and f as well.  From the second condition above we see that we are assuming that W is closed under scalar multiplication and so both 0u and  must be in W, but from Theorem 1 from the previous section we know that,

But this means that the zero vector and the negative of u must be in W and so we’re done.

Be careful with this proof.  On the surface it may look like we never used the first condition of closure under addition and we didn’t use that to show that axioms e and f were valid.  However, in order for W to be a vector space it must be closed under addition and so without that first condition we can’t know whether or not W is in fact a vector space.  Therefore, even though we didn’t explicitly use it in the proof it was required in order to guarantee that we have a vector space.

Next we should acknowledge the following fact.

Fact

Because V can be thought of as a subset of itself we can also think of it as a subspace of itself.  Also, the zero space which is the vector space consisting only of the zero vector,  is a subset of V and is a vector space in its own right and so will be a subspace of V.

At this point we should probably take a look at some examples.  In all of these examples we assume that the standard addition and scalar multiplication are being used in each case unless otherwise stated.