Let $(e_n)$ be a complete orthonormal sequence in a separable Hilbert space $H$ and $T: H \to H$ be a right shift operator i.e. $Te_n = e_{n+1}$ for $n = 1, 2,...$, then find Hilbert adjoint operator of $T$.

I tried like this: Let $x=\sum\alpha_ie_i, y= \sum \beta_ie_i$, $T^*$ be the Hilbert adjoint of $T$. Now, $$\langle x,T^*y\rangle =\langle Tx,y\rangle =\langle \sum \alpha_iTe_i,\sum \beta_ie_i\rangle =\langle \sum\alpha_ie_{i+1},\sum \beta_ie_i\rangle =\sum \alpha_i\bar \beta_{i+1}=\langle \sum \alpha_ie_i,\sum\beta_{i+1}e_i\rangle \implies T^*y=\sum \beta_{i+1}e_i \implies T^*e_n=e_{n-1}$$

This implies that $T^*$ is a left shift operator. But I want to know, what will be $T^*e_1$? Since $e_1$ is in the domain of $T^*$, so how to get $T^*e_1$?

[Definition of Hilbert adjoint operator. Let $T: H_1 \to H_2$ be a bounded linear operator, where $H_1$ and $H_2$ are Hilbert spaces. Then the Hilbert-adjoint operator $T^*:H_2 \to H_1$ of $T$ is the operator such that for all $x \in H_1$ and $y \in H_2$ ,$\langle Tx, y\rangle =\langle x, T^*y\rangle $. ]


1 Answer 1


For any $k$, $\langle e_k, T^* e_1\rangle = \langle Te_k, e_1\rangle=\langle e_{k+1}, e_1\rangle = 0$. As $\{e_k\}_{k=1,2,3,\dots}$ is a complete orthonormal set, $T^* e_1 = 0$.

  • $\begingroup$ I suppose if it is finite dimension, then $e_n$ will be sent to $e_1$ $\endgroup$
    – dEmigOd
    Oct 20, 2017 at 11:02
  • $\begingroup$ @dEmigOd Sorry, don't follow? Please elaborate. $\endgroup$ Oct 20, 2017 at 11:03
  • $\begingroup$ $\langle e_n,T^*e_1\rangle = \langle Te_n, e_1\rangle =\langle e_1, e_1 \rangle = 1$. I mean let $\{ e_k \}$ be finite. $\endgroup$
    – dEmigOd
    Oct 20, 2017 at 11:05
  • $\begingroup$ so a left shift operator make sense $\endgroup$
    – dEmigOd
    Oct 20, 2017 at 11:07
  • $\begingroup$ If $e_k$ form a complete orthonormal set then in particular, finite subsets are linearly independent and so $e_k$ cannot repeat, so $e_1 ≠ e_{1+n} = Te_n$. $\endgroup$ Oct 20, 2017 at 11:09

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