Given $A \in \mathbb R^{m\times n}$, I know that:
$$\|A\|_2 \le \sqrt {m} \|A\|_\infty$$ $$\|A\|_\infty \le \sqrt {n} \|A\|_2$$
I am supposed to provide an example of a matrix such that the equality holds. I tried diagonal first, to no avail, and can't seem to realize how to proceed as the $2$-norm would be tough to calculate.
For $$\|A\|_2 \le \sqrt {m} \|A\|_\infty$$ consider $$A=\begin{bmatrix}1&0&0&0\\1&0&0&0\\1&0&0&0\\1&0&0&0\end{bmatrix}$$and for $$\|A\|_\infty \le \sqrt {n} \|A\|_2$$consider$$A=\begin{bmatrix}1&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}$$
Hint
For more general cases, consider $A=uv^T$ where $u$ and $v$ are $m\times 1$ and $n\times 1$ vectors respectively.