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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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Could you explain why you accepted an answer to this question which states this result, only to now turn around and ask for an explanation of why the result is true? What you quoted in that question seems like an explanation for this! This suggests you did not understand the answer there, or that they didn't really give you the necessary tools to understand it. Why put forth the same question again, but now without all the context? –  Arturo Magidin May 23 '12 at 2:10
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You might also note that the set-up in that question assumed that $n$ was a power of $2$ and that $w$ was a complex $n$th root of unity, conditions you omitted here. As stated, the formula does not make sense, since a Vandermonde matrix is not determined by a single parameter; so absent *that* text, this question does not even make sense. Even if we were to successfully guess that $M_n(w)$ is meant to be the Vandermonde matrix associated to $1$, $w$, $w^2,\ldots,w^{n-1}$, as stated the claim is false (cont) –  Arturo Magidin May 23 '12 at 2:24
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(cont) since, for example, the inverse of $M_3(3)$ is not $\frac{1}{3}M_3(1/3)$. –  Arturo Magidin May 23 '12 at 2:25
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That's my point. The text does answer the question you are asking. The text explains why the inversion formula is true; the fact that you did not understand the text is the reason why you are asking this follow-up question. At the very least, you should link this question to the last and explain that you were unable to understand the text leading up to the formula. (Though, in my opinion, you should have continued to try to get answers to the other question that led you to actually understand the text.) –  Arturo Magidin May 23 '12 at 2:30
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It is a waste of everyone's time perpetrated by you to ask a connected question without even referencing the previous question it connects to (and in the process making it unintelligible and incorrect). It is a waste of your responders' time, perpetrated by you to accept answers that do not, in fact, answer your questions. Do you really want to talk about wasting people's time? It is not an "invalid point"; it's just one you refuse to acknowledge. –  Arturo Magidin May 23 '12 at 2:32

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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.

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Thank You Generic! –  user26649 May 29 '12 at 17:46

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