## 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.

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