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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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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
    
Column 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. –  Martin Argerami Nov 25 '12 at 14:47

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