Let $A$ be a $n$ by $n$ real matrix. Let $\|A\|$ denote the operator norm with respect to the usual Euclidean metric on $\mathbb{R}^n$. I want to prove there exist constants $c$ and $C$ depending only on $n$ such that $$ c \| A \|\leq \max_{i,j} |A_{i,j}| \leq C \| A \| $$
How can I prove this? Thank you!
ps if this question is too basic, I would also appreciate a reference
Since $\|A \|=\max_\limits{\|x\| = 1}\|Ax\|$, we have that: $$\|A\| \geq \|Ae_j\| \ \forall \ j=1,\dots, n \ ,$$ where $e_j$ is the $j$th vector of the canonical basis (i.e. all zeros and a $1$ in position $j$).
But, letting $A^{(j)}$ be the $j$th column of $A$, we have: $$\|Ae_j\| = \| A^{(j)} \| \geq \max_i |A_{i j}| \ .$$
Since it holds for each $j$, it means that: $$\| A\| \geq \max_{i,j=1,\dots, n} |A_{i j}| \ .$$
For the other side of the inequality, we have that: $$\| A\| = \max_{\|x\|=1}\|Ax\| = \max_{\|x\|=1}\left\|A\sum_{i=1}^n x_ie_i \right\| \ ,$$ where we re-write $x$ as a linear combination of the basis vectors.
Now, distributing the sum over the product, we have: $$\max_{\|x\|=1}\left\|A\sum_{i=1}^n x_ie_i \right\| = \max_{\|x\|=1}\left\| \sum_{i=1}^n x_iAe_i \right\| \ \leq \max_{\|x\|=1}\left\| \sum_{i=1}^n 1 \cdot A^{(i)} \right\| = \left\| \sum_{i=1}^n A^{(i)} \right\| \ ,$$ where we used the fact that $\|x\|=1 \implies |x_i| \leq 1 \ \forall \ i$.
Now we can conclude because: $$\left\| \sum_{i=1}^n A^{(i)} \right\| \leq \sum_{i=1}^n \|A^{(i)} \| \leq n\max_i \|A^{(i)} \| \leq n \cdot n \cdot \max_{i,j = 1,\dots,n} |A_{i j}| = n^2 \max_{i,j =1,\dots,n} |A_{i j}| \ .$$
