3

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?

Muster Mark
  • 495

2 Answers2

4

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

Aloizio Macedo
  • 34,292
1

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

John Dawkins
  • 25,733
  • 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? – Jyrki Lahtonen Jan 03 '16 at 17:32
  • 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. – John Dawkins Jan 03 '16 at 17:48
  • Thanks John, your answer was helpful. Your interpretation wasn't a mistake. – Muster Mark Jan 04 '16 at 09:29