$T:\ell_2\to \ell_2, T(x_1,x_2,\dotsc)=(x_1,x_2/2,x_3/3,\dotsc)$

Question

$T:\ell_2\to \ell_2, T(x_1,x_2,\dotsc)=(x_1,x_2/2,x_3/3,\dotsc)$ I need to know whether it is self adjoint and unitary operator given that $x_i\in\mathbb C$

I am not able to do it please tell me how to proceed.

In my earlier post one said eigenvalues of a self adjoint operator must be real, here I tried like say $\lambda$ be an eigenvalue for $T$ then $T(x)=\lambda x$ but then $(x_1,x_2/2,x_3/3,\dotsc)=\lambda(x_1,x_2,x_3,\dotsc)$, I am lost here,

Answer 1

First you should make sure that this is a well-defined operator, i.e. it maps $\ell^{2}$ to $\ell^{2}$. This is easy.

This is clearly not unitary as you can find by considering $||(0,1,0,0...)|| \neq ||(0,1/2,0,0...)||$.

An operator is self-adjoint iff $\langle Tx,y \rangle = \langle x,Ty\rangle$ for all $x,y \in \ell^{2}$.

Now let $x,y \in \ell^{2}$. Then

$$ \langle Tx,y \rangle = \sum\limits_{n=1}^{\infty}\frac{x_{n}}{n}\overline{y_{n}} = \sum\limits_{n=1}^{\infty}x_{n}\overline{\frac{y_{n}}{n}} = \langle x,Ty\rangle$$

which means that $T$ is self-adjoint.

Note that eigenvalues being real is a necessary condition for an operator to be self-adjoint but not sufficient. So even if you show that all the eigenvalues are real, you cannot claim that the operator is self-adjoint.

Answer 2

Hint: If $$\langle Tx, y \rangle = x_1 \overline{y_1} + \dfrac{x_2}{2} \overline{y_2} + \dfrac{x_3}{3} \overline{y_3} + \ldots = x_1 \overline{z_1} + x_2 \overline{z_2} + x_3 \overline{z_3} + \ldots = \langle x, z \rangle$$ for all $(x_1, x_2, x_3, \ldots) \in l_2$ then $z_1 = y_1$, $z_2 = \dfrac{y_2}{2}$, $z_3 = \dfrac{y_3}{3}$, $\ldots$.