Linear Algebra and Its Applications, Exercise 2.4.11

Exercise 2.4.11. Suppose that for a matrix A and any b the system Ax = b always has at least one solution. Show that in this case the system A^Ty = 0 has no solution other than y = 0.

Answer: Suppose A is an m by n matrix and b is an m-element vector. If x = \begin{bmatrix} x_1&x_2&\cdots&x_n \end{bmatrix}^T is a solution for Ax = b then b is a linear combination of the n columns of A with the coefficients being x_1, x_2, \dotsc, x_m.

If we can find a solution x for any b in \mathbf R^m then the columns of A span \mathbf R^m since any vector in \mathbf R^m can be expressed as a linear combination of the columns of A. The rank of A is therefore r = m. (The rank cannot be less than m since then there would not be enough linearly independent columns of A to span \mathbf R^m, and cannot be greater than m because it is impossible to have more than m linearly independent vectors in \mathbf R^m.)

Now consider the left nullspace \mathcal N(A^T) consisting of all vectors y such that A^Ty = 0. The dimension of the left nullspace is m - r where r is the rank of A. But from above we have r = m so that the dimension of \mathcal N(A^T) is m - r = m - m = 0. The only element of \mathcal N(A^T) is therefore the zero vector, and there are no  solutions to A^Ty = 0 other than y = 0.

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.

 Buy me a snack to sponsor more posts like this!

This entry was posted in linear algebra. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s