Let $H$ a separable Hilbert space, $(e_n) \subset H$ an orthonormal basis and $(\alpha_n) \subset \mathbb{C}$ a bounded sequence.
Define $T:H \to H$ by $T(x) = \sum_{n=1}^{\infty}\alpha_n \langle x, e_n \rangle e_n$
I am trying to find $T^*$
I have $\langle T^*x,y \rangle = \langle x,Ay \rangle = \langle x, \sum_{n=1}^{\infty} \alpha_n \langle y, e_n \rangle e_n \rangle = \sum_{n=1}^{\infty} \alpha_n \overline{\langle e_n,y \rangle} \langle x,e_n \rangle = \langle \alpha_n \langle x, e_n \rangle \overline{e_n}, \overline{y} \rangle$
The problem is I get $\overline{y}$ instead of $y$ in the second coordinate. How can we obtain the proper shape? Or is there a different approach for this?
Best Answer
You conjugated $\langle y,e_n\rangle$ when you extracted it from the inner product, but then you also flipped the entries: that is, you conjugated twice. You also forgot to conjugate $\alpha_n$. So your second to last term should have been $$ \sum_n \overline{\alpha_n}\,\langle e_n,y\rangle\,\langle x,e_n\rangle, $$ to get $$ \left\langle \sum_n\overline{\alpha_j}\langle x,e_n\rangle,y\right\rangle. $$ If will help you understand what's happening if you notice that what you are doing is basically $$ \begin{bmatrix} \alpha_1&0\\0&\alpha_2\end{bmatrix} ^*=\begin{bmatrix} \overline{\alpha_1}&0\\0&\overline{\alpha_2}\end{bmatrix} $$