
Norm inequality
Hi,
I need to show that $\displaystyle \ A \_2 \leq \sqrt{n} \A\_2$ where $\displaystyle A$ is $\displaystyle m\times n$. THe $\displaystyle A$ notation means we take the absolute value of all elements of $\displaystyle A$.
I've tried a few things (with no luck), such that first trying to show that $\displaystyle \A\_{\infty}\geq \ A \_2$,
and then use the fact that $\displaystyle \sqrt{n}\A\_2\geq \A\_{\infty}$.
Would be great if someone could give me a hint or two.
Thanks.

Re: Norm inequality
I don't know what the norm notations refer, but did you try to use CauchySchwarz inequality?