If $N$ is normal, show that $\begin{Vmatrix} Nx \end{Vmatrix}$=$\begin{Vmatrix} N^{H}x \end{Vmatrix}$ for every vector $x$ If $ N $ is normal, show that $\begin{Vmatrix} Nx \end{Vmatrix}$=$\begin{Vmatrix} N^{H}x \end{Vmatrix}$ for every vector x.
Deduce that the ith row of N has the same legth as the ith column.
In here, I have to consider that N has complex elements? It affects the result?

Hi couldn't you also express this as this:
$$\|Nx\|^2= \langle Nx,Nx\rangle= \overline{\left(Nx\right)}^T\cdot\left(Nx\right)= \overline{x}^T\cdot\overline{N}^T\cdot N\cdot x =
\overline{x}^T\cdot N\cdot \overline{N}^T\cdot x = \overline{\left(N^Hx\right)}^T\cdot\left(N^Hx\right)=
\langle N^Hx,N^Hx\rangle = \|N^Hx\|^2$$
Since $N^H=\overline{N}^T$ and $N^H\cdot N=N\cdot N^H$. I know it's maybe more writing and so on, but I understand this process a bit better than the one in previous answer.
