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Give an example of a normal operator $T$ on a complex inner product space, which is an isometry but $T^2≠I_V$.

(This question did not give what the inner product is, so how should I do? If under dot product, does T=(0 1; -1 0) satisfy?

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I think the question suggests you are allowed to choose a complex inner product space of you liking, and define an operator on that. The answer you suggested (if that is a $2\times2$ matrix you intended to write down) supposes the space $\Bbb C^2$. However there are also answers that can be defined with the same formualtion on any nonzero complex inner product space. –  Marc van Leeuwen Dec 2 '12 at 6:14
Marc van Leeuwen: Can you give an example for on any nonzero complex inner product space? –  i_a_n Dec 2 '12 at 6:16
Multiplication by the scalar $i$ –  Marc van Leeuwen Dec 2 '12 at 6:17
May I suggest that instead of flooding this website with vast numbers of related questions, that you STOP and wait and digest the answers you get and see whether you can apply them to the rest of your questions? –  Gerry Myerson Dec 2 '12 at 6:20
This seems like a strange question. The condition $T^2 = I$ is strong - most isometries should not obey it. Just curious - did you forget a part of the question? –  Neal Dec 2 '12 at 6:22

1 Answer 1

Take any operator whose matrix on some basis is diagonal with diagonal entries on the unit circle but not all real (i.e., $\pm1$). In other words (since normal matrices are all diagonalisable) take any unitary transformation with at least one non-real eigenvalue.

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