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Let $(X,\langle\cdot,\cdot\rangle)$ be a Hilbert Space over $K$ with orthonormal basis $(x_n)$, and let $(\lambda_n)\in K$ be a bounded sequence. The mapping $T:X\to X$ is defined by

$Tx:=\sum\limits_{n=1}^\infty\lambda_n\langle x,x_n\rangle x_n$, $x\in X$.

Find the adjoint operator $T^*$ of $T$.

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    $\begingroup$ Remember the defining characteristic of the adjoint operator. Write it out. Look at it. $\endgroup$ – Daniel Fischer Jan 30 '14 at 9:04
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    $\begingroup$ along with the comment of @DanielFischer, you may try for the finite dimensional case first. $\endgroup$ – RSG Jan 30 '14 at 9:14
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Clearly the adjoint of $T$ is defined as $$ T^*x:=\sum\limits_{n=1}^\infty\overline{\lambda}_n\langle x,x_n\rangle x_n. $$ To see that, note that $\langle Tx,x_n\rangle=\lambda_n\langle Tx,x_n\rangle$ and $$ \langle x,y\rangle=\sum_{n=1}^\infty \langle x,x_n\rangle\overline{\langle y,x_n\rangle}, $$ and hence $$ \langle Tx,y\rangle=\sum_{n=1}^\infty \langle Tx,x_n\rangle\overline{\langle y,x_n\rangle} =\sum_{n=1}^\infty \lambda_n\langle x,x_n\rangle\overline{\langle y,x_n\rangle}, $$ while $$ \langle x,T^*y\rangle=\sum_{n=1}^\infty \langle x,x_n\rangle\overline{\langle T^*y,x_n\rangle} =\sum_{n=1}^\infty \langle x,x_n\rangle\overline{\bar\lambda_n\langle y,x_n\rangle}, =\sum_{n=1}^\infty \lambda_n\langle x,x_n\rangle\overline{\langle y,x_n\rangle}. $$

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