Showing two matrix norms are equivalent Let $A$ be an $n \times m $ matrix. We have 
$$ ||A||_F = \sqrt{ \sum_{i=1}^n \sum_{j=1}^m |a_{ij}|^2 } , ||A||_2 = \sup_{||x||_2 = 1 } || A x ||$$
I am trying to show that these two norms are equivalents. My prof said it is enough to show 
$$ || A ||_2 \leq ||A||_F $$
It doesnt make sense to me. Why dont we need to show the other inequality? Can someone explain to me why the above inequality is enough? thanks
 A: This isn't a super answerable question because it depends on what we can assume. In general the property you are asking about can be stated in terms of norms on spaces of arbitrary dimension, and requires the open mapping theorem. But in finite dimensional space we have that all norms are equivalent, which is a  famous proof and subsumes your question. 
Here's my elementary proof. Assume we know that the closed unit ball $B$ induced by the $|| ||_2$ norm is compact (this should be something we can assume because that's the standard euclidean $m \times n$ dimensional space, so the set is closed and bounded, thus compact. Now, we have a closed set $B_F$ which is the unit ball induced by $|| ||_F$.
if $||x||_2 > 1$, $||x||_F >1$, so $B_F \subset B$, and $B_F$ is the preimage of $[0,1]$ under the norm $|| ||_F$ (which is easily verified to be continuous) so it is a closed subset of a compact set so is compact. Thus, it achieves its nonzero minimum with repsect to the $|| ||_2$ norm, say at the point $x_0$, and say that $||x_0||_2 = c$. Now, we consider the set $\frac{1}{c}B_F$. This set contains $B$, so given any vector $x$,  $||x||_2 \leq \frac{1}{c}||x||_F$. 
A: Let $e_1,\ldots,e_m$ be the canonical base of $\mathbb{R}^m$. We have:
$$\|A\|_F^2 = \|Ae_1\|_2^2+\ldots+\|Ae_m\|_2^2 \leq m\cdot\|A\|_2^2$$
but for any $v\in\mathbb{R}^m$ such that $\|v\|_2=1$ we have:
$$ \|Av\|_2^2 = \langle v,r_1\rangle^2+\ldots+\langle v,r_n\rangle^2\leq \|v\|_2^2\left(\|r_1\|_2^2+\ldots+\|r_n\|_2^2\right)=\|A\|_F^2 $$
by the Cauchy-Schwarz inequality, where $r_1,\ldots,r_n$ are the rows of $A$. It follows that:
$$ \frac{1}{\sqrt{m}}\|A\|_F\leq \|A\|_2 \leq \|A\|_F \leq \sqrt{m}\|A\|_2 $$
hence the two norms are equivalent.
