Exercise: Show an operator in Hlbert space is compact

Question

Let $\;l^2\;=\{ (x_1,x_2,\dots) : \sum_{n=1}^{\infty} {\vert x_n \vert}^2 \lt +\infty\}\;$ be our Hilbert space and consider the operators $\;T_i: l^2 \rightarrow l^2 \;$ as

$T_1x=(a_1x_1,0,\dots)\;\\T_2x=(a_1x_1,a_2x_2,0,\dots)\\ .\;\\.\; \\T_nx=(a_1x_1,a_2x_2,\dots,a_nx_n,0,\dots)\;$

where $\;a_i \in \mathbb C\;$ $\forall 1\le i \le n\;$

Prove that each $\;T_i\;$ is compact.

My thought:

I know a finite range operator is compact. And it seems this is the statement I need in order to solve this exercise. However it's a bit unclear to me why $\;dimRan(T_i)\lt +\infty\;\;\forall i$ since $\;T_ix\;$ includes infinite number of zeros. Also each $\;T_i\;$ reminds me -in a kind of way- of the linear span of $\;x\;$ but it isn't the linear span...

I 'm having a really hard time getting my head around this so any help would be valuable.

Thanks in advance!

Answer 1

Note that $T_n(x) \in \operatorname{sp} \{ e_1,...,e_n\}$, hence $T_n$ has finite dimensional range.

Answer 2

All the operators you list are finite-rank, so compact.

More generally, if you define $$ Tx=(a_1x_1,a_2x_2,\ldots) $$ then $T$ is compact if and only if $\lim_{n\to\infty}a_n=0$.