Generalization of the matrix in $l^2_{\mathbb{C}}(\mathbb{Z})$

Question

If we take the following operator:

$$A: l^2_{\mathbb{C}}\mathbb{Z})\rightarrow l^2_{\mathbb{C}}(\mathbb{Z}),$$ to each $x\in l^2$, we have for all $i\in \mathbb{Z}$ : $(Ax)_i=\sum_{j\in\mathbb{Z}}a_{ij}.x_j$.

And we suppose that: $\sum_j|a_{ij}|^2$ is finite for all $i\in \mathbb{Z}$, and also the operator $A$ is bounded.

Now let's prove that: $A$ is one-to-one if and only if $F=Vect\{a_i=(a_{ij})_j; i\in \mathbb{Z}\}$ is dense in $l^2$. for the inverse implication, I supposed that $A$ is not one to one, then there is $x$ in $l^2-\{0\}$ such that $Ax=0.$

and since $F$ is dense, then there is $x_n\in F$ such that $x_n\rightarrow x , \;(n\rightarrow \infty).$

and then: for all $n\in \mathbb{N}$: $x_n=\sum_k \lambda_{n,k}.a_k$.

so I triend to apply $A$, on $x_n$, and then tend $n$ to $\infty$ but then we'll just have $$\lim_n\sum_k\lambda_{n,k}\sum_ja_{ij}a_{ki}=0$$

Answer 1

If $Ax=0$, then $\sum_ja_{ij}x_j = 0$ for all $i$, so $\overline x := (\overline{x_j})_{j\in\mathbb Z}$ is perpendicular to each $a_i$. Since the span $S$ of the $a_i$ is dense in $\ell^2$, it follows that $\overline x\in S^\perp = \{0\}$ and thus $x=0$.