Exercise 2.3.17. Suppose that and are subspaces of , each with dimension 3. Show that and must have at least one vector in common other than the zero vector.
Answer: Since and each have dimension 3 their respective bases each contain three vectors. Let , , and be a basis for and , , and be a basis for .
Now consider the combined set of six vectors. Since we have six vectors in a vector space of dimension 5 the combined set of vectors is linearly dependent, with at least one vector expressible as a linear combination of the other five vectors. Without loss of generality assume that is dependent on the other five vectors, so that we have
for some set of weights through . We can rearrange the above equation as follows:
Now consider the vector . Since is a linear combination of the basis vectors , , and it is in the subspace . But from the above equation we also have so that is a linear combination of the basis vectors , , and and thus is also in the subspace .
So is a member of both and . Now suppose . We then have or so that is a linear combination of and and the set of vectors , , and is linearly dependent. But this contradicts the assumption that , , and form a basis for and are thus linearly independent. Since the assumption leads to a contradiction we conclude that .
We have thus shown that there must exist a nonzero vector that is a member of both and .