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Let $n \in \mathbb{N^{*}}$ and define $T_{n}:\ell_{2} \rightarrow \ell_{2}$ by $$T_{n}(x) = (\frac{x_{n}}{n},\frac{x_{n + 1}}{n + 1},... ).$$

where $x = (x_{1},x_{2},.....).$

Then,$$T_{n}^{*}(x) = (0,...,0,\frac{x_{1}}{n},\frac{x_{2}}{n + 1},... ).$$ Where the zeros in the adjoint are $n-1$ times.

Is ker $T_{n} = \{0\}$? what about Im $T_{n}^{*}$? Could anyone help me in finding it?

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$$\mathrm{ker}(T_n)=\{x=(x_1,x_2,\cdots)\mid x_k=0~\forall~k\ge n\}$$ $$\mathrm{Im}(T_n^{\star})=(\mathrm{ker}(T_n))^{\perp}=\{x=(x_1,x_2,\cdots)\mid x_k=0~\forall~1\le k<n\}$$

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  • $\begingroup$ what about the image? $\endgroup$
    – user426277
    Jan 4, 2018 at 4:19
  • $\begingroup$ Use the fact that the image of the adjoint operator is othocomplement of the kernel of the operator. $\endgroup$
    – QED
    Jan 4, 2018 at 4:23
  • $\begingroup$ you mean orthogonal complement .....right ? I know this fact but do not know the details of using it. $\endgroup$
    – user426277
    Jan 4, 2018 at 4:44