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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

You have two positive requirements and one negative requirement. The positive ones are: being normal (which means having some orthonormal basis of eigenvectors) and being an isometry (which means having some orthonormal basis of eigenvectors with eigenvalues on the unit circle), which implies the first condition. The negative requirement is not being an involution, which given the positive requirements means that at least one eigenvalue is not$~\pm1$, and hence is a non-real complex number on the unit circle.

So take any operator whose matrix on some orthonormal basis is diagonal with diagonal entries on the unit circle but not all $\pm1$. In other words take any unitary transformation with at least one non-real eigenvalue. For example one can take the operator of multiplication by$~\mathbf i$, on any nonzero complex inner product space.

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