Linear Algebra and Its Applications, Review Exercise 1.22

Review exercise 1.22. Answer the following questions:

(a) If A has an inverse, does A^T also have an inverse? If so, what is it?

(b) If A is both invertible and symmetric, what is the transpose of A^{-1}?

(c) Illustrate (a) and (b) when A = \begin{bmatrix} 2&1 \\ 1&1 \end{bmatrix}.

Answer: (a) Assume A is invertible, so that A^{-1} exists. Then we have AA^{-1} = I and therefore (AA^{-1})^T = I^T = I. But (AA^{-1})^T = (A^{-1})^TA^T so we have (A^{-1})^TA^T = I and (A^{-1})^T is a left inverse for A^T.

Similarly we have A^{-1}A = I and therefore (A^{-1}A)^T = I^T = I. But (A^{-1}A)^T = A^T(A^{-1})^T so we have A^T(A^{-1})^T = I and (A^{-1})^T is a right inverse for A^T.

Since (A^{-1})^T is both a left inverse and a right inverse for A^T we see that A^T is invertible and (A^T)^{-1} = (A^{-1})^T.

(b) If A is symmetric then we have A = A^T. If A is also invertible then A^{-1} exists and from (a) above we know that (A^T)^{-1} = (A^{-1})^T. We then have (A^{-1})^T = (A^T)^{-1} = A^{-1}. Since (A^{-1})^T = A^{-1} we see that A^{-1} is also symmetric if A is.

(c) If we have

A = \begin{bmatrix} 2&1 \\ 1&1 \end{bmatrix} = A^T

then

A^{-1} = \frac{1}{2 \cdot 1 - 1 \cdot 1} \begin{bmatrix} 1&-1 \\ -1&2 \end{bmatrix} = \begin{bmatrix} 1&-1 \\ -1&2 \end{bmatrix} = (A^{-1})^T

Note that A^{-1} is symmetric just as A is.

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.

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

One Response to Linear Algebra and Its Applications, Review Exercise 1.22

  1. Pingback: Linear Algebra and Its Applications, Review … – My Math Homework

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