Is there a way to prove that the 2-norm of a matrix is greater than the largest element?

Question

I would like to prove

$$\max_{ij} |A_{ij}| \leq \| A \|_2$$

Any help would be greatly appreciated!

Answer 1

Hint: since the two-norm $\|A\|_2$ is defined as the maximum of some function over all vectors, all you need to do is find some vector for which that function is at least any given matrix entry; more concretely, you need some vector $x$ for which $$\frac{|Ax|}{|x|}\geq A_{ij},$$ where $|\cdot|$ denotes the standard (Euclidean) norm on vectors. For scaling reasons, you might as well pick $x$ for which $|x|=1$; can you find some unit vector for which $A_{ij}$ becomes an entry of $Ax$?

Answer 2

The notation $\|A\|_2$ usually refers to the Frobenius norm, also called the Hilbert-Schmidt norm, defined to be $$ \|A\|_2 = \sqrt{\sum_{i,j}|a_{i,j}|^2}. $$ If this is what the OP has in mind, then the question is trivial since it is obvious that $$ |a_{k,l}|^2 \leq \sum_{i,j}|a_{i,j}|^2, $$ for every $k$ and $l$.

Most other matrix norms share this property, so it would be nice to know exactly which matrix norm is of interest to the OP.