Consider the Hilbert space $$\ell^2=\{(x_1,x_2,...,x_n),x_i\in\mathbb C\text{ for all }i\text{ and }\sum_{i=1}^\infty |x_i|^2<\infty\}$$ with the inner product
$$\langle(x_1,x_2,\dots,x_n)(y_1,y_2,\dots,y_n)\rangle = \sum_{i=1}^\infty x_i \overline y_i.$$ Ddefine $T:\ell^2 \to \ell^2$ by $$T((x_1,x_2,...,x_n))=(x_1,\frac{x_2}2,\frac{x_3}3,...).$$ Then $T$ is
a)neither self-adjoint or unitary
b)both self-adjoint and unitary
c)unitary but not adjoint
d)self-adjoint but unitary

$$\langle Tx,y\rangle=\langle T((x_1,x_2,...,x_n)),(y_1,y_2,...,y_n)\rangle= \langle (x_1,x_2/2,x_3/3,...),(y_1,y_2,..,y_n)\rangle= \sum (x_i\overline y_i/i)= \sum x_i(\overline y_i/i)= \langle (x_1,x_2,...,x_n),(y_i,y_2/2,y_3/3,..y_n/n)\rangle=\langle x,Ty\rangle$$ so T is self-adjoint

  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ Jan 15, 2017 at 20:26
  • $\begingroup$ It is a diagonal operator with real coefficients tending to $0.$ Therefore it is self-adjoint and non-unitary. $\endgroup$ Feb 16, 2022 at 12:33

1 Answer 1


That´s right,because for $T:\ell^2(\mathbb{N})\rightarrow \ell^2(\mathbb{N})$ defined by $(Tx)_j=\frac{1}{j}x_j$ for $j=1,2,...$ one has $(Tx,y)=\sum_{j=1}^{\infty}\frac{1}{j}x_j\bar{y_j}=\sum_{j=1}^{\infty}x_j\overline{\frac{1}{j}y_j}=(x,Ty)$ for all $x,y\in\ell^2(\mathbb{N})$ it follows $T^*=T$ and thus $(T^*Tx)_j=(TT^*x)_j=(T^2x)_j=\frac{1}{j^2}x_j$ for all $x\in\ell^2(\mathbb{N})$ which means $T^*T\neq id_{\ell^2(\mathbb{N})}$, so $T$ is not unitary.


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