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 .
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.