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Is operator $T_n$ a compact operator?

$$T_n:l_2\rightarrow l_2$$

$$T_nx=(\underbrace{0,0,\ldots,0,}_{n\text{ zeros}}, x_1,x_2,x_3,\ldots)\text{ where }x=(x_1,x_2,x_3,\ldots)\in l_2,\ \sum^\infty_{k=1} |x_k|^2\lt\infty$$

could you please help.

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    $\begingroup$ Well, to start, what is the definition of compact that you have learned? $\endgroup$ – Christopher A. Wong May 28 '14 at 21:53
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This operator is not compact. Recall that any compact operator takes a bounded sequence to another sequences that has a convergent subsequence. If $\{ e_n \}_{n =1}^\infty$ is an orthonormal basis, then $T_n$ takes this basis to another orthonormal set, which clearly does not have a limit point. So $T_n$ is not compact.

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