## 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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