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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

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$T$ 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.

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  • $\begingroup$ Thank you very much. You used the fact that a subset of a Hilbert space is dense iff the complement is ${0}$. I define $\mathbb{N}$ to include $0$. In this case I think you should use $e_0$. $\endgroup$ – stackuser Jun 10 at 19:02
  • $\begingroup$ Can I also use the same approach to show that a bilateral shift is invertible iff $(\frac{1}{w_n})$ is bounded. Clearly $T$ would be injective since $ker(T)={0}$. It suffices to show surjectivity. Any hints sir? $\endgroup$ – stackuser Jun 10 at 19:06
  • $\begingroup$ @stackuser If $\frac 1 {w_n}$ is bounded then the shift with weights $(\frac 1 {w_n})$ is the inverse of the one with weights $(w_n)$. $\endgroup$ – Kabo Murphy Jun 10 at 23:16
  • $\begingroup$ I came up with this: Suppose $\{\frac{1}{w_n}\}_n$ is bounded. Then the operator $S$ defined by $Se_n=\frac{1}{w_{n-1}} e_{n-1}$ for all $n\in Z$ is well defined. We see that $TS=ST=I$. This implies $T^{-1}$ exists and equals $S$. Moreover, $S$ is a bilateral shift operator: $Sv_n=s_n v_{n+1},$ where $v_n=e_{-n}$ and $s_n=\frac{1}{w_{-(n+1)}}$. Thus $T^{-1}$ is a bilateral weighted shift operator. But I can't see the converse. Please can you help @kavi $\endgroup$ – stackuser Jun 11 at 12:37
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    $\begingroup$ @stackuser 'bilateral shift is invertible given some conditions' The conditions are that $\{w_n\}$ and $\frac 1 {w_n}$ are both bounded. $\endgroup$ – Kabo Murphy Jun 11 at 23:18

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