Exercise 3.1.20. Suppose is a subspace of . Show that . What does this mean?
Answer: We first consider the case where ; in other words, contains only the zero vector. From exercise 3.1.18 we know that . The only vector that is orthogonal to all vectors in is the zero vector, so the zero vector is the only member of . We thus have when .
We next consider the case where ; in other words, contains at least one nonzero vector (and thus has dimension of at least 1). Then there must exist a linearly independent set of one or more basis vectors through that span . Consider the by matrix that has the basis vectors through as its rows. The row space of is then the set of all linear combinations of the basis vectors through and is thus equal to the itself.
Per 3D on page 138 (the Fundamental Theorem of Linear Algebra, part 2) the nullspace of is the orthogonal complement of the row space and the row space of is the orthogonal complement of the nullspace of . But the row space is , so the nullspace of is the orthogonal complement of . The orthogonal complement of the nullspace of is then and is equal to the row space .
So for all subspaces in we have .
To recap: For we have trivially. Any nonzero subspace is the row space of some matrix and is the nullspace of that matrix. The fact that then follows from the fact that the row space of a matrix is the orthogonal complement of the nullspace and vice versa. The basis sets of and are mutually orthogonal, and together they form a basis set for . Thus no (nonzero) vector can be orthogonal to 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.