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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\|. $

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    Colum$n$ and row were changed, sorry. As for second to third, it is simpler than that: note that it is an assertion about vectors, not matrices, so it is just saying that the lenght of a vector is the same whether you write the vector as a column or as a row.2012-11-25
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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.