Example of normal linear maps

Question

Let $V$ be a real inner product space and $g:V\rightarrow V$ a normal linear map.
Can you give me an example of:
1- A map $g$ that is not self-adjoint?
2- A map $g$ that is not an isometry?
3- A map $g$ that is not orthodiagonalizable?

Answer 1

Consider the matrix

$$ A = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \in M_2(\mathbb{R}). $$

We have

$$ AA^T = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \begin{pmatrix} a & b \\ -b & a \end{pmatrix} = \begin{pmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{pmatrix}, \\ A^TA = \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \begin{pmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{pmatrix} $$

so $A^TA = AA^T$ and $A$ is a normal matrix. If $b \neq 0$, then $A$ is not self-adjoint and not orthogonally diagonalizable (as this is equivalent). If $a^2 + b^2 \neq 1$, then $A$ is not orthogonal.

Thus, if $b \neq 0$ and $a^2 + b^2 \neq 1$, the corresponding map $T_A \colon \mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $T_A(x) = Ax$ (considering $\mathbb{R}^2$ with the standard inner product) is not self-adjoint, not an isometry and not orthogonally diagonalizable. In fact, it is not diagonalizable (as the roots of the characteristic polynomial $a \pm ib$ are complex).