3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Martin Sleziak Jan 15 '17 at 20:26
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.