Prove that for a bounded operator $A$, then $\sum_{n=1}^\infty |a_{mn}|^2 \leq \|A\|^2$. Suppose $A$ is some bounded operator on the Hilbert space of $\ell^2(\mathbb{N})$, with a representing matrix $(a_{mn})_{m,n=1}^\infty$. How can I prove that 
\begin{equation}
\sum_{n=1}^\infty |a_{mn}|^2 \leq \|A\|^2
\end{equation}
I have started by noting that since $A$ is a bounded operator, then $\|A\| = \|A^\ast\|$ (the adjoint operator of $A$). The representing matrix of such an adjoint would have entries $((\overline{a_{mn}})_{m,n=1}^\infty)^t$. Could I perhaps use these two ideas in order to prove the result. Any help is appreciated!

Let $(e_n)_n$ the orthogonal basis of $l^2(\mathbb{N})$ we have by definition:
$$
A(e_n)=\sum_{j\geq 1} a_{nj}e_j\qquad \forall j\in \mathbb{N}
$$
So 
$$
\|A(e_n)\|^2=\left\|\sum_{j\geq 1} a_{nj}e_j \right\|^2=\sum_{j\geq 1} \left\|a_{nj}e_j \right\|^2=\sum_{j\geq 1} |a_{nj}|^2
$$
$$
\|A(e_n)\|^2\leq \|A\|^2 \|e_n\|^2=\|A\|^2 
$$
and
