# 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?

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By definition, $$\|Nx\|^2=\langle Nx,Nx\rangle=\langle N^HNx,x\rangle=\langle NN^Hx,x\rangle=\langle N^Hx,N^Hx\rangle=\|N^Hx\|^2.$$ The $i^{\rm th}$ column of $N$ is $Ne_i$, where $e_1,\ldots,e_n$ are the elements of the canonical basis. The $i^{\rm th}$ row of $N$ is given by $e_i^TN=(N^He_i)^H$. So $$\|e_i^TN\|=\|(N^He_i)^H\|=\|N^He_i\|=\|Ne_i\|.$$
Thanks. But I wonder maybe the ith row is $e_i^{T} N$? – email Nov 25 '12 at 10:59
And from the second to third, it is because the length is the same even though it is transposed? I mean, $\|A^H\|$=$\|A\|$? – email Nov 25 '12 at 12:04