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I wasn't sure whether this question was appropriate for math.stackexchange or physics.stackexchange. I don't really have an understanding of bounded linear operators or Hilbert spaces so this question may be slightly informal. Nevertheless, let $U$ be a linear operator with adjoint $U^\dagger$. I've read that for only finite dimensional vector spaces, $U^\dagger U=id \Rightarrow UU^\dagger=id$. Could someone please explain why this is the case and provide an example of an infinite dimensional vector space for which the implication doesn't hold?

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Consider the right-shift operator on $l^2$. Its adjoint is the left-shift operator. We have that $LS \circ RS=id$, but...

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In fact, for two arbitrary $n\times n$ matrices, if $AB=I$ then $BA=I$.

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  • $\begingroup$ This was flagged (by somebody) as low quality, so I was summoned to the scene. I am wondering how do you think this helps the asker who specifically made the question about infinite dimensional vector spaces? $\endgroup$ – Jyrki Lahtonen Jan 3 '16 at 17:32
  • $\begingroup$ I was replying to "I've read that for only finite dimensional vector spaces, $U^\dagger U=id \Rightarrow UU^\dagger=id$. Could someone please explain why this is the case", having (perhaps mistakenly) interpreted this to ask, in part, why the implication at issue was true in finite dimensions. $\endgroup$ – John Dawkins Jan 3 '16 at 17:48
  • $\begingroup$ Thanks John, your answer was helpful. Your interpretation wasn't a mistake. $\endgroup$ – Muster Mark Jan 4 '16 at 9:29

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