A linear operator $T$ on a (complex) separable Hilbert space $H$ is said to be a weighted shift operator if there is some orthogonal basis $\{e_n\}_n$ and weight sequence $\{w_n\}_n$ such that $$Te_n=w_n e_{n+1}, \forall n$$ $T$ is unilateral if $n$ runs over $\mathbb{N}$ and bilateral if $\mathbb{Z}$. The adjoint is given by
$$T^* e_n=\overline{w}_{n-1}e_{n-1} \text{ for all } n $$ if $T$ is bilateral and
\begin{align*}
T^* e_n&= \overline{w}_{n-1} e_{n-1} \text{ for all } n\geq 1\\
T^* e_0&=0
\end{align*} if $T$
is unilateral
I saw in a paper that the unilateral shift is never invertible with the reason that $T^*$ is not invertible but the bilateral shift can be invertible given some conditions .
Do we have that an operator is invertible iff its adjoint is invertible? I can't see the reason behind the conclusion. Please I need hints. Thanks
Best Answer
$T$ is invertible iff $T^{*}$ is invertible and ${(T^{*})}^{-1}={(T^{-1})}^{*}$. You can easily verify this using the fact $(UV)^{*}=V^{*}U^{*}$.
Actually, the way you have defined a unilateral shift its range is orthogonal to $e_1$. Since the operator is not surjective it is not invertible.