Finding matrix norm equivalence constants I've been given the following:
"Find the best positive constants $\alpha$ and $\beta$ such that $\alpha\left\|A\right\|_2\leq\left\|A\right\|_1\leq\beta\left\|A\right\|_2$ for all $A\in\mathbb{R}^{m\times n}$.
This is part b of a question along with "Prove that $\left\|x\right\|_2\leq\left\|x\right\|_1\leq\sqrt{n}\left\|x\right\|_2$ for all $x\in\mathbb{R}^n$," which I was able to get after awhile by expanding out the first half and using Cauchy-Schwarz on the second, which would lead me to believe that at least one of those would be related to solving this.
My book provides solutions for $\frac{1}{\sqrt{n}}\left\|A\right\|_\infty\leq\left\|A\right\|_2\leq\sqrt{m}\left\|A\right\|_\infty$ and $\left\|A\right\|_2\leq\left\|A\right\|_F\leq\sqrt{n}\left\|A\right\|_2$ but they use $\left\|x\right\|_2\leq\sqrt{n}\left\|x\right\|_\infty$ which I can see being true but isn't explicitly stated prior to this, and I don't think I can use without proving first. Would I simply be able to use the corresponding ratios from the first part to show this? I'm kinda lost to be honest.
(Also, I'm taking "best" to mean closest-bounding, unless there's only one valid pair of them, though it remains somewhat ambiguous.)
 A: You can use simply the fact that $\|x\|_2\leq\|x\|_1\leq\sqrt{n}\|x\|_2$ to get (upper bound in the numerator, lower bound in the denominator):
$$
\|A\|_1=\max_{x\neq 0}\frac{\|Ax\|_1}{\|x\|_1} \leq
\max_{x\neq 0}\frac{\sqrt{m}\|Ax\|_2}{\|x\|_2} = \sqrt{m}\|A\|_2.
$$
Similarly for the other:
$$
\|A\|_2=\max_{x\neq 0}\frac{\|Ax\|_2}{\|x\|_2} \leq 
\max_{x\neq 0}\frac{\|Ax\|_1}{(\sqrt{n})^{-1}\|x\|_1} = \sqrt{n}\|A\|_1.
$$
Hence
$$
\frac{1}{\sqrt{n}}\|A\|_2\leq\|A\|_1\leq\sqrt{m}\|A\|_2.
$$
The constants $\alpha=\frac{1}{\sqrt{n}}$ and $\beta=\sqrt{m}$ are best in the sense that they are the tightest constants such that the bounds hold for all matrices, that is, there exist matrices $A_{\alpha}$ and $A_{\beta}$ for which one of the bounds is attained: $\frac{1}{\sqrt{n}}\|A_{\alpha}\|_2=\|A_{\alpha}\|_1$ and $\|A_{\beta}\|_1=\sqrt{m}\|A_{\beta}\|_2$. These matrices can be, e.g.:
$$
A_{\alpha}=\begin{bmatrix}
1 & 1 & \cdots & 1 \\
0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots \\
0 & 0 & \cdots & 0
\end{bmatrix}, \qquad
A_{\beta}=\begin{bmatrix}
1 & 0 & \cdots & 0 \\
1 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots \\
1 & 0 & \cdots & 0 
\end{bmatrix}.
$$
