Normal transformations that are not self adjoint / unitary Is it possible to have a normal transformation $T$ with only real eigenvalues that is not self adjoint?
Similarly, could a normal transformation have only eigenvalues whose absolute value is $1$ and yet not be Unitary?
What are simple examples of such occurrences?
 A: The answer is negative to both questions (in the finite dimensional case, at least). A normal matrix is unitarily equivalent to a diagonal matrix with its eigenvalues on the main diagonal. It is obvious that a diagonal matrix with real entries is self-adjoint and that a diagonal matrix with complex entries of modulus $1$ is unitary.
A: It's not possible when working with complex inner product spaces. Let me show that if $T$ is normal and has real eigenvalues, then $T$ is self-adjoint. The situation with a unitary map is similar.
Since $T$ is normal, we can find an orthonormal basis $e_1, \ldots, e_n$ of eigenvectors of $T$. Write $Te_i = \lambda_i e_i$ with $\lambda_i \in \mathbb{R}$. Calculating, we have
$$ \left< Tv, w \right> = \left< T(\left( \sum_{i=1}^n \left<v, e_i \right> e_i \right), \sum_{j=1}^n \left< w, e_j \right> e_j \right> = \left< \sum_{i=1}^n \left< v, e_i \right> \lambda_i e_i, \sum_{j=1}^n \left< w, e_j \right>  e_j \right> = \sum_{i=1}^n \left< v, e_i \right> \lambda_i \overline{\left< w, e_i \right>} = \sum_{i=1}^n \lambda_i \left<v, e_i \right> \overline{\left<w, e_i \right>}. $$
Similarly,
$$ \left< v, Tw \right> = \left< \sum_{i=1}^n \left<v, e_i \right> e_i, T \left( \sum_{j=1}^n \left<w, e_j \right> e_j \right) \right> = \left< \sum_{i=1}^n \left< v, e_i \right> e_i, \sum_{j=1}^n \left< w, e_j \right> \lambda_j e_j \right> = \sum_{i=1}^n \left<v, e_i \right> \overline{ \left< w, e_i \right> \lambda_i} = \sum_{i=1}^n \lambda_i \left<v, e_i \right> \overline{\left<w, e_i \right>} $$
using the fact that $\lambda_i$ are real and so $\overline{\lambda_i} = \lambda_i$. Since $\left< Tv, w \right> = \left< v, Tw \right>$ for all $v,w \in V$, we have $T = T^{*}$ and $T$ is self-adjoint.
If you work with real inner product spaces, then it's possible that $T$ is normal with only real eigenvalues but $T$ is not self-adjoint. For example, take $V = \mathbb{R}^2$ with the standard inner product and $T$ to be the rotation by $\frac{\pi}{2}$ degrees counter-clock wise. Then $T$ doesn't have any eigenvalues (and so vacuously all the eigenvalues of $T$ are real) but $T$ is not self-adjoint. 
