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This question is related to, and a follow-up for, this question. The notation here follows that of the text quoted there.

Let $\matrix M_n$ be a Vandermonde matrix of size $n$ by $n$. The columns of $\matrix M$ are orthogonal to each other, therefore each column is mapped onto a coordinate axis. $w$ is a $n^{th}$ root of unity, and $n$ is a power of $2$. The inversion formula is supposedly:

$\matrix M_n(w)^{-1} =\frac{1}{n}\matrix M_n\left(w^{-1}\right)$

The book explains why multiplying one column by another results in 1. Therefore I understand why multiplying $M(w)$ by its complex conjugate results in $1$. What is $\frac{1}{n}$ though? Why is it part of the equation and what is it?

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    @$A$rturoMagidin You seem to be taking this personally. Apologies if my comments seem harsh or insulting. I just feel this is a valid question. I will however keep your suggestion to link to relevant questions and necessary text in mind, and ensure my questions provide all the necessary information required for a solution.2012-05-23

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Note: $w$ must be a primitive $n$-th root of unity, not just any $n$-th root (for example, $M_n(1)$ does not satisfy orthogonality of the columns).

Let $A=M_n(w)$, $B=M_n(w^{-1})$, $C=AB$.

Recall the definition of matrix multiplication: $c_{ij}=\sum_{k=1}^n a_{ik} b_{kj}$ On the diagonal $i=j$, $a_{ik}b_{ki}=w^{ik}w^{-ik}=1$ (is this what you meant when saying that multiplication results in 1?). But since we are summing over $n$ such terms, we have $c_{ii}=n$. Hence we need the $1/n$ factor to obtain an identity matrix: $AB = n I_n$ and $A^{-1}=\frac 1 n B$.

In other words, the columns of $M_n$ are orthogonal, but not orthonormal: the squared norm of each column is $n$. You could define $M'_n(w)=\frac{1}{\sqrt n} M_n(w)$ to make the columns orthonormal and avoid this factor.