Weak convergence as convergence of matrix elements Let $H$ be a Hilbert space with orthonormal basis $(e_h)_{h \in \mathbb{N}}$ and let $(A_n)_{n \in \mathbb{N}}$ and $A$ be bounded linear operators. We say that $A_n$ converges weakly to $A$ if 
$$\forall \xi, \eta \in H, \quad (\eta, A_n\xi)\to (\eta, A\xi).$$

Question Is it true that $A_n$ converges weakly to $A$ if and only if 
$$\forall h, k \in \mathbb{N},\quad (e_h, A_n e_k) \to (e_h, Ae_k)?$$

Indeed, I'm wondering if it is correct to think at weak convergence as convergence of the matrix entries associated to the operators $A_n$ and $A$. 
Thank you.

If it is possible to write $\zeta = \sum_{i=0}^{\infty}a_{k}e_{k}, \eta=\sum_{i=0}^{\infty}b_{k}e_{k}$ with the $a_{k}, b_{k} = const. \forall k \in\mathbb{N}$, the the weak convergence as defined above follows directly from the fact that the direct product on the Hilbert space is linear, that the operator we're dealing with is linear, and that we can write the $\zeta, \eta$ can be written as linear combinations of the o.n.b..
The "if part" doesn't follow, however, since you only regard the diagonal entries of the operator. This goes in the same direction as the person answering the question before me already said. You should easily find plenty of examples that disprove the if part.
