I have to check that the following operator $T$ is compact:
Define $T:l^{2} \to {l^2}$ by $Tx=y=(\eta_{j})$, where $x=(\zeta_{j})$ and
$(\eta_{j})=\sum_{k=1}^\infty \alpha_{jk} \zeta_{j}$ and $\sum_{j=1}^\infty \sum_{k=1}^\infty |\alpha_{jk}|^{2} \lt \infty$
Show that $T$ is compact.
My attempt:
I constructed a sequence of operators,$(T_{n})$,
$T_{n}:l^{2} \to {l^2}$ defined by,
$T_{n}x=(\sum_{k=1}^\infty \alpha_{1k} \zeta_{1},\sum_{k=1}^\infty \alpha_{2k} \zeta_{2},.......,\sum_{k=1}^\infty \alpha_{nk} \zeta_{n},0,0,....) $ and then i proved that each $T_{n}$ is bounded and the range of $T_{n}$ is finite and therefore each $T_{n}$ is a compact operator.
Then I considered,
\begin{align} ||(T_{n}-T)x||^{2}& = ||(\sum_{k=1}^\infty \alpha_{(n+1)k} \zeta_{n+1},\sum_{k=1}^\infty \alpha_{(n+2)k} \zeta_{n+2},.....)||^{2}\\ & = \sum_{j=n+1}^\infty \sum_{k=1}^\infty |\alpha_{jk} \zeta_{j}|^{2}\\ &= \sum_{j=n+1}^\infty \sum_{k=1}^\infty |\alpha_{jk}|^{2} |\zeta_{j}|^{2}\\ \end{align}
Now how do i use the condition,$\sum_{j=1}^\infty \sum_{k=1}^\infty |\alpha_{jk}|^{2} \lt \infty$ to prove that $T$ is compact? I know that once I get that $(T_{n})$ is uniformly operator convergent then compactness will follow, but how do i get that $(T_{n})$ converges to $T$ from here?
Also in a subpart of this question I have to illustrate using an example that that the condition given in the above problem is sufficient for compactness but not necessary, I don't quite understand what am i required to do here ?