Help to finish my proof: inequality with norm and Schwarz ineq Let $A=\left [ a_{ij} \right ]$ be the matrix of a linear mapping $A\in L\left ( \mathbb{R}^{n},\mathbb{R}^{m} \right )$.
-Prove that: $\left \| A \right \|\leq \left ( \sum_{i=1}^{m}\sum_{j=1}^{n}a{_{ij}}^{2} \right )^{^{\frac{1}{2}}}$
Note that: $\left \| A \right \|=sup_{\left \| x \right \|=1} \left \| Ax \right \|$
My tentative: $A$ is an $m*n$ matrix. Let $A_{i}$ be the $i$th row of $A$. Then the $ith$ row of the vector $Ax$ is the dot product $\left \langle A_{i},x \right \rangle$. 
Then: $\left \| Ax \right \|=\sqrt{\left \langle  A_{1},x\right \rangle^{2}+\left \langle A_{2},x \right \rangle^{2}+...+\left \langle A_{m},x \right \rangle^{2}}$ and using Schwarz inequality:$\left | \left \langle A_{i},x \right \rangle  \right |< \left \| A_{i} \right \|.\left \| x \right \|$ we get that the left hand side is less than or equal to $\sqrt{\left ( \left \| A_{1} \right \|^{2}+...+\left \| A_{n} \right \|^{2} \right ).\left \| x \right \|^{2}}=\sqrt{\left ( \left \| A_{1} \right \|^{2}+...+\left \| A_{n} \right \|^{2} \right )}$ because $\left \| x \right \|=1$
Is there anything wrong with my proof? Please let me know if there are any details I should add to the proof to finish it off. Also, if you guys have any alternative proofs, please share. Thanks
 A: Just one thing: the inequality $\langle A_i,x\rangle \leq \lVert A_i\rVert\cdot \lVert x\rVert$ is not strict in general.  
We note that the inequality $\lVert A\rVert\leq \left(\sum_{i=1}^m\sum_{j=1}^na_{ij}^2\right)^{\frac 12}$ is not an equality in general, since for example with $m=n$ and $A=I_n$, the identity matrix, we have $\lVert A\rVert =1$ whereas $\left(\sum_{i=1}^n\sum_{j=1}^na_{ij}^2\right)^{\frac 12}=\sqrt n$.
A: I think your proof is correct. Just let me show you a different one, coming from my Operator Algebras background. 
It's done for $A$ with real entries, but it applies to the complex case with the only change of "transpose" by "adjoint". 
We have $\|A\|^2=\|A^tA\|$. The matrix $A^tA$ is positive-semidefinite and so $\|A^tA\|$ is the largest eigenvalue of $A^tA$. Since the trace of $A^tA$ is the sum of its (nonnegative) eigenvalues, we have
$$
\|A\|^2=\|A^tA\|\leq\text{Tr}(A^tA)= \sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}^{2} 
$$
(the last equality being a straightforward calculation).
