I've found what I think might be a typo in the solution manual for Poole's Linear Algebra, 4th Edition. Exercise 59 in Section 7.4 reads as follows:
Prove that for any square matrix $A$, $\Vert A\Vert_2^2 \le \Vert A\Vert_1\Vert A\Vert_\infty$. [Hint: $\Vert A\Vert_2^2$ is the square of the largest singular value of $A$ and hence is equal to the largest eigenvalue of $A^TA$. Now use Exercise 34 in Section 7.2.]
The alleged solution is as follows:
Let $\sigma$ be the largest singular value of $A$. Then $\Vert A\Vert_2^2=\sigma ^2=\vert\lambda\vert$, where $\vert\lambda\vert$ is the eigenvalue of $A^TA$ of largest magnitude. Next, since $\Vert A^T\Vert_1=\Vert A\Vert_\infty$, we have $\Vert A^TA\Vert_1=\Vert A^T\Vert_1\Vert A\Vert_1=\Vert A\Vert_1\Vert A\Vert_\infty$. But by Exercise 34 in Section 7.2, $\Vert A^TA\Vert_1 \ge \vert\lambda\vert$. Thus $$\Vert A\Vert_2^2 =\vert\lambda\vert \le \Vert A^TA\Vert_1=\Vert A\Vert_1\Vert A\Vert_\infty$$
Here's where I believe there may be an error in the solution: since the operator 1-norm is a matrix norm, it (by definition) satisfies the condition that $\Vert AB\Vert_1 \le \Vert A\Vert_1\Vert B\Vert_1$ for all square matrices $A$ and $B$, so in this case $\Vert A^TA\Vert_1 \le \Vert A^T\Vert_1\Vert A\Vert_1$. I'm not sure why the solution asserts that equality holds for this relationship. One counterexample that I can think of is the matrix $$A=\begin{bmatrix} 1&-3&2\\4&-1&-2\\-5&1&3\end{bmatrix}$$ so that $$A^TA=\begin{bmatrix} 42&-12&21\\-12&11&-1\\-21&-1&17\end{bmatrix}$$ Then $\Vert A^TA\Vert_1=75 \neq 90=(9)(10)= \Vert A^T\Vert_1\Vert A\Vert_1$, using the fact that the operator 1-norm of a matrix is the largest absolute column sum of the matrix. It can still be shown that $\Vert A\Vert_2^2 \le \Vert A\Vert_1\Vert A\Vert_\infty$, except that now the final line of the solution is $$\Vert A\Vert_2^2 =\vert\lambda\vert \le \Vert A^TA\Vert_1 \le \Vert A^T\Vert_1\Vert A\Vert_1=\Vert A\Vert_1\Vert A\Vert_\infty$$ Is my correction justified? Also, is there a special class of matrices for which the equality $\Vert A^TA\Vert_1=\Vert A^T\Vert_1\Vert A\Vert_1$ would hold? Can this be generalized to other matrix norms, so that $\Vert A^TA\Vert=\Vert A^T\Vert\Vert A\Vert$? I have no reason to believe so, but I'm curious if anyone knows.