At first, we might try set operations: union, intersection, and difference. The set difference we can rule out right away: since and must both contain the zero vector, cannot.
What about the union, ? Before trying to understand this in general, let’s try a concrete example: take , and let (the axis, essentially), and (the axis). Is their union a subspace?
Any subspace has to be closed under addition. If we add the vector (which lies along the axis) to the vector (which lies along the axis), we get the vector , which does not lie along either axis.
False.
Any subspace has to be closed under addition. If we add the vector (which lies along the axis) to the vector (which lies along the axis), we get the vector , which does not lie along either axis.
With a motivating example under our belts, we can try to tackle the general result. (Note that this result remains true even if is infinite-dimensional!)
For the other direction, it’s not clear how to get started with our hypothesis. When a direct proof seems difficult, remember that we can also try proving the contrapositive: If and , then is not a subspace.
Now we have more to work with: negation turns the “or” into an “and”, and proving that something is not a subspace is easier: we just have to show that one part of the subspace test fails. As our motivating example suggests, we should expect closure under addition to be the condition that fails.
An important point to keep in mind with this proof: closure under addition means that if a subspace contains and , then it must contain . But if a subspace contains , that does not mean it has to contain and . As an example, consider the subspace of . It contains the vector , but it does not contain or .
Now, suppose that , and . Since , there must be some element such that . Since , there must be some element such that . We know that , so we consider the sum, .
By a similar argument, if , we can conclude that , contradicting the assumption that . So does not belong to or , so it cannot belong to . Since is not closed under addition, it is not a subspace.
The key here is that the intersection contains only those vectors that belong to both subspaces. So any operation (addition, scalar multiplication) that we do in can be viewed as taking place in either or , and we know that these are subspaces. After this observation, the rest is the Subspace Test.
Let and be subspaces of . Since and , we have . Now, suppose . Then , and . Since and is a subspace, . Similarly, , so . If is any scalar, then is in both and , since both sets are subspaces, and therefore, . By the Subspace Test, is a subspace.
The intersection of two subspaces gives us a subspace, but it is a smaller subspace, contained in the two subspaces we’re intersecting. Given subspaces and , is there a way to construct a larger subspace that contains them? We know that doesn’t work, because it isn’t closed under addition. But what if we started with , and threw in all the missing sums? This leads to a definition:
Now, suppose is a subspace of such that and . Let . Then for some and . Since and ,. Similarly, . Since is a subspace, it is closed under addition, so . Therefore, .
This is a proof that would be difficult (if not impossible) without using a basis. Your first thought might be to choose bases for the subspaces and , but this runs into trouble: some of the basis vectors for might be in , and vice-versa.
Of course, those vectors will be in , but it gets hard to keep track: without more information (and we have none, since we want to be completely general), how do we tell which basis vectors are in the intersection, and how many?
Instead, we start with a basis for . This is useful, because is a subspace of both and . So any basis for can be extended to a basis of , and it can also be extended to a basis of .
The rest of the proof relies on making sure that neither of these extensions have any vectors in common, and that putting everything together gives a basis for . (This amounts to going back to the definition of a basis: we need to show that it’s linearly independent, and that it spans .)
Now, consider the set . We claim that is a basis for . We know that is linearly independent, since it’s a basis for , and that . It remains to show that none of the are in the span of ; if so, then is independent by Lemma 1.7.10.
Since , it suffices to show that none of the belong to . But we know that , so if , then . But if , then , which would imply that is linearly dependent, and since is a basis, this is impossible.
Notice how a vector can be written as a sum of a vector in and a vector , but not uniquely, in general: in the above proof, we can change the values of the coefficients and , as long as the sum remains unchanged. Note that these are the coefficients of the basis vectors for , so we can avoid this ambiguity if and have no nonzero vectors in common.
If the sum is direct, then we have simply . The other reason why direct sums are preferable, is that any can be written uniquely as where and , since we no longer have the ambiguity resulting from the basis vectors in .
Suppose that , and suppose that we have , for . Then , which implies that
.
Now, , since is a subspace, and similarly, . But we also have , which implies that . (Since is in , and this is the same vector as .) Therefore, , which implies that , so . But we must also then have , so .
Conversely, suppose that every can be written uniquely as , with and . Suppose that . Then and , so we also have , since is a subspace. But then , where and . On the other hand, , and belongs to both and . It follows that . Since was arbitrary, .
We end with one last application of the theory we’ve developed on the existence of a basis for a finite-dimensional vector space. As we continue on to later topics, we’ll find that it is often useful to be able to decompose a vector space into a direct sum of subspaces. Using bases, we can show that this is always possible.
Let be a basis of . Since , the set is a linearly independent subset of . Since any linearly independent set can be extended to a basis of , there exist vectors such that
We know , since contains the basis for we’ve constructed. To show the sum is direct, it suffices to show that . To that end, suppose that . Since , we have
for scalars . Since , we can write
for scalars . But then
Since is a basis for , it’s independent, and therefore, all of the must be zero, and therefore, .
The subspace constructed in the theorem above is called a complement of . It is not unique; indeed, it depends on the choice of basis vectors. For example, if is a one-dimensional subspace of ; that is, a line, then any other non-parallel line through the origin provides a complement of . Later we will see that an especially useful choice of complement is the orthogonal complement.
One way to solve this is to ask yourself, what vectors are not in the span of the basis you found above? You can do this by solving an appropriate system of equations.