I need to show:
$$ \|A\|_1 \leq \sqrt{n} \|A\|_2 \leq n\|A\|_1 $$
and
$$ \|A\|_\infty \leq \sqrt{n} \|A\|_2 \leq n\|A\|_{\infty} $$
I have shown the first half of each using the Cauchy-Schwarz inequality, and suspect the Holder inequality is somehow relevant too. Can these be used to show the second half of each of the above inequalities as well or does another approach need to be taken?
For each of the first halves,
$$\begin{align} |\langle Ae_j,\vec{1}\rangle| & \leq \|Ae_j\|_2\|\vec 1\|_2 \\ \|A\|_1 & \leq \sqrt{n}\sqrt{\Sigma_{i=1}^n |(Ae_j)_i|^2}\\ & \leq \sqrt{n}\sqrt{\Sigma_{i=1}^n |(Ae_j)_i|^2} \leq \sqrt{n}\left( max_{\|x\|_2=1} \sqrt{\Sigma_{i=1}^n |(Ax)_i|^2}\right)\\ & \leq \sqrt{n} \|A\|_2 \\ \|A\|_1 & \leq \sqrt n \|A\|_2\\ &(QED, \text{part 1 of the first inequality})\\ & \\ \|A^T\|_1 & \leq \sqrt n\|A^T\|_2 \\ & \|A^T\|_1 = \|A\|_\infty \text{ and }\|A^T\|_2 = \|A\|_2 \\ \|A\|_\infty & \leq \sqrt n \|A\|_2 \\ &(QED, \text{part 1 of the second inequality})\\ \end{align}$$
For part 2 of the first inequality, we first assume $A\in\mathbb{R}^{m\times n}$ and let $x\in\mathbb{R}^{n}$. We have \begin{equation*} \Vert A\Vert_{2}=\max_{x\neq 0}\frac{\Vert Ax\Vert_{2}}{\Vert x\Vert_{2}}\leq\max_{x\neq 0}\frac{\Vert Ax\Vert_{1}}{(1/\sqrt{n})\Vert x\Vert_{1}}=\sqrt{n}\Vert A\Vert_{1}. \end{equation*}
For part 2 of the second inequality, we have \begin{equation*} \Vert A\Vert_{2}=\max_{x\neq 0}\frac{\Vert Ax\Vert_{2}}{\Vert x\Vert_{2}}\leq\max_{x\neq 0}\frac{\sqrt{m}\Vert Ax\Vert_{\infty}}{\Vert x\Vert_{\infty}}=\sqrt{m}\Vert A\Vert_{\infty}. \end{equation*}