a compact operator on $l^2$ defined by an infinite matrix

Question

Let $A$ be an infinite matrix such that $\displaystyle \sum_{i,j}|a_{i,j}|^2<\infty$. Then $A$ defined a compact operator on $l^2$.

Answer 1

A subset $M\subset \ell_2$ is precompact if $$ \lim\limits_{n\to\infty}\sup\limits_{x\in M}\Vert(0,\ldots,0,x_n, x_{n+1},\ldots)\Vert_2=0 $$ The proof can be found here. It it enough to show that $M:=T(\operatorname{Ball}_{\ell_2}(0,1))$ is precompact for the operator $$ T:\ell_2\to\ell_2:(x_1,x_2,\ldots)\mapsto\left(\sum\limits_{j=1}^\infty a_{1j} x_j,\sum\limits_{j=1}^\infty a_{2j} x_j, \ldots\right) $$ For any $y\in M$ we have $x\in\operatorname{Ball}_{\ell_2}(0,1)$ such that $y=T(x)$, then $$ \begin{align} \Vert (0,\ldots,0,y_n,y_{n+1},\ldots)\Vert_2 &=\left(\sum\limits_{i=n}^\infty\left|\sum\limits_{j=1}^\infty a_{ij} x_j\right|^2\right)^{1/2}\\ &\leq\left(\sum\limits_{i=n}^\infty\sum\limits_{j=1}^\infty |a_{ij}|^2 \sum\limits_{j=1}^\infty|x_j|^2\right)^{1/2}\\ &=\left(\sum\limits_{i=n}^\infty\sum\limits_{j=1}^\infty |a_{ij}|^2\right)^{1/2} \left(\sum\limits_{j=1}^\infty|x_j|^2\right)^{1/2}\\ &=\left(\sum\limits_{i=n}^\infty\sum\limits_{j=1}^\infty |a_{ij}|^2\right)^{1/2} \Vert x\Vert_2\\ &\leq\left(\sum\limits_{i=n}^\infty\sum\limits_{j=1}^\infty |a_{ij}|^2\right)^{1/2}\\ \end{align} $$ Since the last expression does not depend on $y\in M$ we get that $$ 0\leq \sup\limits_{y\in M}\Vert (0,\ldots,0,y_n,y_{n+1},\ldots)\Vert_2\leq \left(\sum\limits_{i=n}^\infty\sum\limits_{j=1}^\infty |a_{ij}|^2\right)^{1/2} $$ And now we take the limit $n\to\infty$ $$ 0 \leq\lim\limits_{n\to\infty}\sup\limits_{y\in M}\Vert (0,\ldots,0,y_n,y_{n+1},\ldots)\Vert_2 \leq\lim\limits_{n\to\infty}\left(\sum\limits_{i=n}^\infty\sum\limits_{j=1}^\infty |a_{ij}|^2\right)^{1/2} =0\tag{1} $$ Note: the last limit equals to zero because the series $\sum_{i,j=1}^\infty|a_{ij}|^2$ converges. From $(1)$ we conclude that $M:=T(\operatorname{Ball}_{\ell_s}(0,1))$ is precompact, so $T$ is compact.