An operator is not compact

Question

let $T$ be the operator on $ \ell_{2}$ (the complex Hilbert space of square summable sequences), defined by

$$ T(x_{1},x_{2},x_{3},\dots)=(0,x_{2},x_{3},\dots).$$

Show that $T$ is not compact.

Answer 1

Let $x_n, n>1$ be the sequence such that its $n$ term is $1$ the other terms are zero. $\|x_n\|=1$ $T(x_n)=x_n$ and $\|x_n-x_m\|=\sqrt2$ so you cannot extract a converging sequence from $(x_n)$ and the image of the closed ball of radius $1$ is not relatively compact.