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It can be shown that the spectrum of a unitary operator from a Hilbert space to itself is a subset of the unit circle, as in $|\lambda|=1$

It can also be shown that the spectrum of the shift operator (more specifically the right shift operator from $\ell^2$ to $\ell^2$) contains points inside the unit circle

The first answer in this post also shows that the right shift operator is unitary (this is for $\ell^2(\mathbb{Z})$ but the right shift operator in $\ell^2(\mathbb{N})$ is also unitary, proof below)

What am I misunderstanding that is causing me to see this as contradiction? Is my definition of a unitary operator incorrect?

Proof that the right shift operator is unitary: an operator is unitary if $||Tx||=||x||$ and $\langle Tx,Ty\rangle=\langle x,y\rangle$

First property: $$||Tx||=\sum_{n=1}^\infty|(Tx)_n|^2=\sum_{n=2}^\infty|x_{n-1}|^2+0=\sum_{n=1}^\infty|x_n|^2=||x||$$

Second property: $$\langle Tx,Ty\rangle=\sum_{n=1}^\infty (Tx)_n(Ty)_n=\sum_{n=2}^\infty x_{n-1}y_{n-1}+0=\sum_{n=1}^\infty x_ny_n=\langle x,y\rangle$$

John Doe
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  • $\ell^2(\mathbb N) \ne\ell^2(\mathbb Z)$ – JonathanZ Jan 31 '24 at 21:09
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    That's true but the right shift operator over $\mathbb{N}$ is still unitary, I'll add a proof real quick. – John Doe Jan 31 '24 at 21:11
  • Both $UU$ and $UU$ need to equal $I$. I haven't gone into the details, but my guess is that only one of them is. – JonathanZ Jan 31 '24 at 21:13
  • (I should probably add to the first comment that one can show that the two space are actually isomorphic pretty easily, but these are two fairly different operators, i.e. they don't correspond under any isomorphism if the underlying spaces.) – JonathanZ Jan 31 '24 at 21:16
  • @JonathanZ I think you're right – John Doe Jan 31 '24 at 21:16
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    @JonathanZ yep you're right, my definition of a unitary operator was incomplete. Since the adjoint operator is the left shift operator, it's clear to see that $\text{Right shift}\circ\text{Left shift}\neq I$ since it turns the 1st element in the series to 0. – John Doe Jan 31 '24 at 21:17
  • I'll write it up as an answer and we can close the question. – JonathanZ Jan 31 '24 at 21:20
  • @JohnathanZ Sounds good – John Doe Jan 31 '24 at 21:21

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The part that probably slipped by you is that $\ell^2(\mathbb N) \ne\ell^2(\mathbb Z)$.

But to be complete we should note that while it is possible to give an easy isomorphism between the two spaces ($\mathbb Z$ and $\mathbb N$ are both countable infinite sets, and any bijection between them induces a Hilbert space isomorphism), the two right shift operators do not correspond to each other under such an isomorphism.

And what closes the book on this is noting that while the operator on $\ell^2(\mathbb Z)$ is actually unitary, $U = $ right shift on $\ell^2(\mathbb N)$ has $U^*$ = left shift, so $UU^* = $ "zero out the first component", so that operator is not unitary.

JonathanZ
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