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Let $T$ be a self adjoint operator on a finite dimensional inner product space $V$. Then $ \| T(x)\pm ix \|^2=\| T(x) \|^2+\| x\|^2$ for all $x \in V$. Deduce that $T-iI$ is inverible.

Since $\| T(x)\pm ix \|=0$ iff $\| T(x)\|=0$ and $\| x \|=0$ and it means $N(T)=0$ and it is one-to-one. Since $V$ is finite dimensional, $T$ is onto. Thus $T$ is invertible. Hence, $T-iI$ is invertible.

Is this proof complete?

nonam
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    Yes, this is correct. Note that this is also true on an infinite-dimensional Hilbert space, as $T$ self-adjoint has a real spectrum, which therefore never contains $i$. – Julien Jun 01 '13 at 14:23
  • @julien In infinite-dimensional space $T$ is injective but not necessary invertible. Isn't it? –  Jun 01 '13 at 14:25
  • @julien Sorry but from your last comment "this is also true on an infinite...." can I understand that $T-iI$ is also invertible? I'm not sure for this result. –  Jun 01 '13 at 14:31
  • @SamiBenRomdhane On an infinite Hilbert space, $T+iI$ will be bounded below, so the image is closed. To show the operator is invertible, we only need to show the orthogonal complement of the image is trivial. If $\langle Tx+ix,z\rangle=0$ for all $x$, you get $\langle x,Tz-iz\rangle$ for all $x$, hence $Tz-iz=0$. By the same argument as before, conclude $z=0$. – Harald Hanche-Olsen Jun 01 '13 at 14:35
  • @SamiBenRomdhane I did not say the argument applied verbatim. On any Hilbert space, if $T$ is (bounded) self-adjoint, then $T-\lambda Id$ is invertible (i.e. $\lambda \not\in \sigma(T)$) for any $\lambda\in\mathbb{C}\setminus\mathbb{R}$. That's equivalent to the assertion: $\sigma(T)\subseteq \mathbb{R}$. The key point is below boundedness, as mentioned by Harald. – Julien Jun 01 '13 at 14:40
  • @HaraldHanche-Olsen Thanks very much. –  Jun 01 '13 at 14:40
  • @julien Ok now I see thanks. –  Jun 01 '13 at 14:44
  • @RudevanNistelrooy Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Sep 14 '13 at 13:53

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Yes, this is correct. Note that this is also true on an infinite-dimensional Hilbert space, as $T$ self-adjoint has a real spectrum, which therefore never contains $i$. -- 1015

In an infinite-dimensional space, being bounded from below does not imply invertibility in general. So, an additional argument is needed for this case. -- Sami Ben Romdhane

Since $T+iI$ is bounded below, its image is closed. To show the operator is invertible, we only need to show the orthogonal complement of the image is trivial. If $\langle Tx+ix,z\rangle=0$ for all $x$, you get $\langle x,Tz-iz\rangle$ for all $x$, hence $Tz-iz=0$. By the same argument as before, conclude $z=0$. -- Harald Hanche-Olsen

Comments edited to improve the flow.

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