# Prove $\frac{1}{\sqrt{n}}\|A\|_{\infty} \leq\|A\|_{2} \leq \sqrt{m}\|A\|_{\infty}$

equivalent-metricsinequalitylinear algebramatrix-normsnormed-spaces

I was trying to prove the following inequality, where $$A \in \mathbb{C}^{m\times n}$$.
$$\frac{1}{\sqrt{n}}\|A\|_{\infty} \leq\|A\|_{2} \leq \sqrt{m}\|A\|_{\infty}$$
But I was really having difficulty. I thought relating the singular values with individual matrix entries would be helpful, for instance Inequality matrix norm. However, I can't get anywhere with this. Does anyone know how to prove this?

Edit: I managed to solve the first part of the problem, using that $$\|A\|_2 = \sigma_1(A)$$, the largest singular value. Now I need to solve $$\|A\|_{2} \leq \sqrt{m}\|A\|_{\infty}$$.

If $$\|x\|_2 = 1$$, then
$$\|Ax\|_2^2 = \sum_{i = 1}^m\left\lvert \sum_{j = 1}^n A_{ij}x_j\right\rvert^2 \le \sum_{i = 1}^m\left(\sum_{j = 1}^n |A_{ij}|^2\right) \le m\max_{1\le i \le m} \left(\sum_{j = 1}^n |A_{ij}|\right)^2 \le m \|A\|_\infty^2$$ Taking square roots, then taking the supremum over all $$x$$ with $$\|x\|_2 = 1$$, we obtain your right bound.