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?
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?
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.