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Let $T$: $l^2 \rightarrow l^2$ denote the operator \begin{align} T(x_1,x_2,\dots, x_n,\dots) = (x_2,x_4,\dots,x_{2n},\dots). \end{align} There are several questions regarding this operator that I need to answer, but the ones I have trouble with is:

1) Find all eigenvalues of $T$ (That is, all $\lambda$ such that $(T-\lambda I)^{-1}$ does not exist).

For this question I have found that $\| T\| = 1$ and I know that all eigenvalues are contained within the ball of radius $\|T \|$ and center 0 (for bounded operators between Hilbert spaces). So I know that all eigenvalues $\lambda$ satisfy $|\lambda| \leq \|T\| = 1$. My guess is, that all values within that unit-ball are eigenvalues, but I don't know how to show this.

2) Determine the operator $T^*$.

Usually, finding the adjoint of an operator doesn't cause me trouble, but this one is different. By definition of the adjoint operator I have to find $T^*$ such that \begin{align} \langle Tx,y\rangle = \langle x, T^*y\rangle, \hspace{10px} \forall x,y \in l^2, \end{align} but I don't know which manipulations I can use so that every element of $x$ is used in \begin{align} \langle Tx, y \rangle &= \sum\limits_{k=1}^{\infty}Tx_k\cdot \overline{y}_k = \sum\limits_{k=1}^{\infty}x_{2k}\cdot \overline{y}_k. \end{align} I want to end of with something like \begin{align} \langle Tx, y \rangle = \sum\limits_{k=1}^{\infty} x_k \cdot \overline{T^*y_k} = \langle x, T^*y \rangle, \end{align} for some operator $T^*$.

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    $\begingroup$ This can be treated very generally. Your operator is a surjective partial isometry, therefore has norm $1$. $T^*$ is an isometry (it embeds the underlying space to a "half"), its spectral properties are essentially that of the unilateral shift. $\endgroup$ – Michael Nov 28 '13 at 13:54
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    $\begingroup$ Hint: $(T^*(x))_k = x_n$ if $k = 2n$ and $0$ otherwise. ($T$ is called the "down sampling operator" and $T^*$ "up sampling" in signal processing.) $\endgroup$ – Michael Nov 28 '13 at 14:07
  • $\begingroup$ Thank you. What do you mean with "unilateral shift"? $\endgroup$ – Eff Nov 28 '13 at 14:17
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The adjoint is defined by $$T^{*}(x_1, x_2, \ldots, x_{2n}, \ldots ) = (0, x_1, \ldots, 0, x_n 0, x_{n+1} \ldots),$$ that is, $x_k \mapsto 0$ if $k$ is odd, $x_k \mapsto x_{\frac{k}{2}}$ if $k$ is even.

Then you can see that $\left<x, T^*y \right> = \sum_{n=1} x_n z_n$, where $z_n = \left\{ \begin{array}{ll}0 &\mbox{ for } n \mbox{ odd} \\ y_{\frac{n}{2}} &\mbox{ for } n \mbox{ even}\end{array} \right.$.

Thus $\sum_{n=1}^{\infty} x_n z_n=\sum_{n \in 2\mathbb{N}} x_ny_{\frac{n}{2}} $ substituting $n=2k$ we obtain that
$$\left<x, T^*y \right>= \sum_{n \in 2\mathbb{N}} x_ny_{\frac{n}{2}} = \sum_{k=1}^{\infty}x_{2k}y_{k}= \left<Tx, y\right>.$$

First part of the question: [EDIT]

Answer to your comment above

Unilateral shift

$S \colon \ell^2 \rightarrow \ell^2 $ defined by $S(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots)$. $S$ is an isometry. Its adjoint is called a backward shift.

Lemma. The spectrum of $S$ is a closed unit disc, the point spectrum of $S$ is empty.

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  • $\begingroup$ Thanks a lot. So what you are saying is that $x = (1, \lambda^2, \lambda^4, \dots) \in l^2 \iff |\lambda|<1$, therefore the eigenvalues are all $\lambda$ where $| \lambda | < 1$? $\endgroup$ – Eff Nov 28 '13 at 14:56

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