Exercise 3.1.17. Suppose that and are subspaces of and are orthogonal complements. Is there a matrix such that the row space of is and the nullspace of is ? If so, show how to construct using the basis vectors for .

Answer: Suppose through are the basis vectors for . (Since is a subspace of we will have .) Let be an by matrix with row 1 equal to , row 2 equal to , and so on through row equal to .

The row space of is the set of all linear combinations of the rows, which is equivalent to the set of all linear combinations of through . But since through is a basis set for they span and the set of all linear combinations of through is itself. The row space of is therefore equal to .

Now consider the null space of consisting of all vectors such that . If this is the case then the inner product of with each row of must be zero or (put another way) must be orthogonal to each row of . Since the rows of are through this means that any vector in the nullspace of is orthogonal to all of through .

Since is orthogonal to all of through it is also orthogonal to all linear combinations of through , and since through form a basis set for this means that is orthogonal to any vector in . All vectors in the nullspace are thus orthogonal to . Moreover, any vector orthogonal to (i.e., orthogonal to all vectors in ) must be orthogonal to all of through and is therefore in the nullspace of .

The nullspace of thus contains all vectors orthogonal to . But the set of all vectors orthogonal to is , the orthogonal complement to . The nullspace of is therefore equal to .

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.