Review exercise 2.31. Consider the rank-one matrix . Under what conditions would ?

Answer: In order for to exist must be a square matrix; otherwise we could not multiply by since the number of columns of the first matrix would not match the number of rows of the second.

Since is an by matrix the vectors and both have entries, with and .

We then have

Since is 1 by and is by 1 their product is a 1 by 1 matrix, or in other words a scalar value

We thus have

One way for to be zero is to have . Another way is if . Note that this would be trivially true if but it is also possible for to be zero even if is nonzero.

For example, consider the case where and so that

We then have

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.