Exercise 2.3.12.Suppose that the set of vectors , , , and , is a basis for and that is a subspace of . Provide a counterexample to the conjecture that some subset of , , , and is necessarily a basis for .

Answer: Suppose that , , , and are equal to the vectors , , , and and suppose that is the subspace consisting of all vectors whose first two elements are equal to each other and whose last two elements are equal to each other; i.e., vectors in are of the form . (It is fairly simple to verify that is in fact a subspace, so I omit that here.) In this case none of the vectors , , , and are in the subspace and thus cannot be part of a basis for .

Instead we could use, for example, the vectors and as a basis for the subspace , with any vector in the subspace expressible as

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.