On the inverse matrix of Vandermonde (Vandermonde) matrices and the principle of idft in fast Fourier transform (FFT)

On the relationship between inverse matrix of Vandermonde (Vandermonde) matrices and Lagrange (Lagrange) interpolation and the principle of idft in fast Fourier transform (FFT)

Tags: linear algebraic FFT Lagrangian interpolation of determinant matrices

As long as you see a little bit of linear algebra, you should all know Vandermonde determinant.
\[v (X_0,x_1,\cdots, x_{n-1}) =\begin{bmatrix}{1}&{1}&{\cdots}&{1}\{x_{0}}&{x_{1}}&{\cdots} &{x_{n-1}}\{x_{0}^2}&{x_{1}^2}&{\cdots}&{x_{n-1}^2}\{\vdots}&{\vdots}&{}&{\vdots}\ {X_{0}^{n-1}}&{x_{1}^{n-1}}&{\cdots}&{x_{n-1}^{n-1}}\\end{bmatrix} \]
The Vandermonde determinant, because of its own particularity, has the general formula:
\[v (X_0,x_1,\cdots, x_{n-1}) =\prod _{n > I > J \geq 0} (x _{i}-x _{j}) \]

We can also write the entries in the determinant into the matrix, that is, the Vandermonde phalanx
\[v=\begin{pmatrix}{1}&{1}&{\cdots}&{ 1}\{x_{0}}&{x_{1}}&{\cdots}&{x_{n-1}}\{x_{0}^2}&{x_{1}^2}&{\cdots}&{x_{n-1}^2}\{\vdots }&{\vdots}&{}&{\vdots}\{x_{0}^{n-1}}&{x_{1}^{n-1}}&{\cdots}&{x_{n-1}^{n-1}}\\end{ Pmatrix}\]

Considering the inverse matrix of Vandermonde matrices, we can calculate with the adjoint matrix.
For \ (v\) adjoint matrix \ (v^*\)
\[(v^*) _{ij}=c_{ij}\]
where \ (c_{ij}\) is \ (v\) the algebraic cofactor type
There \ (v^{-1}={v* \over det (V)}\)
So for each item, there \ ((V^{-1}) _{ij}={c_{ij} \over det (V)}\)
All we need to know is that each algebraic cofactor and determinant of the quotient can be.
But this method is more complex, especially for the missing row of the Vandermonde determinant is difficult to calculate, and the focus of this article is not in this, if you want to find detailed proof can go to see this blog Vandermonde matrix inverse matrix formula
Finally, you can get
\[(V^{-1}) _{ij}= ( -1) ^{j+1}{\sum\limits_{0 \leq p_1<\cdots < P_{n-j} < n;\ p_1,p_2,\cdots p_{n-j} \ne i} x_{p_1} x_{p_2} \cdots X_{p_{n-j}} \over \prod\limits_{0 \leq k < n;\ K\ne i} (x_k-x_i)} \]

The above method is too complex, then we consider the practical significance of Vandermonde square matrix to think.
Re-examine the square, found that the last Vandermonde square is equivalent to bring into the\ (n\)Points for evaluation, i.e.
\[\begin{pmatrix}{a_0}\{a_1}\{a_2}\{\vdots}\{a_{n-1}}\\end{pmatrix}\begin{pmatrix}{1}&{1}&{\cdots} &{1}\{x_{0}}&{x_{1}}&{\cdots}&{x_{n-1}}\{x_{0}^2}&{x_{1}^2}&{\cdots}&{x_{n-1}^2}\{ \vdots}&{\vdots}&{}&{\vdots}\{x_{0}^{n-1}}&{x_{1}^{n-1}}&{\cdots}&{x_{n-1}^{n-1}}\\end {pmatrix}=\begin{pmatrix}{y_1}\{y_2}\{y_3}\{\vdots}\{y_{n}}\end{pmatrix}\]
The equivalent of a polynomial\ (f (x) =\sum_{i=0}^{n-1} a_ix^i\), where\ (y_i=f (x_i) \)
The Vandermonde Square is the equivalent of bringing in\ (n\)Point, in turn, multiply its inverse matrix\ (n\)Interpolation of points.
That
\[\begin{pmatrix}{a_0}\{a_1}\{a_2}\{\vdots}\{a_{n-1}}\\end{pmatrix}=\begin{pmatrix}{y_1}\{y_2}\{y_3}\{\ Vdots}\{y_{n}}\end{pmatrix}\begin{pmatrix}{1}&{1}&{\cdots}&{1}\{x_{0}}&{x_{1}}&{\cdots} &{x_{n-1}}\{x_{0}^2}&{x_{1}^2}&{\cdots}&{x_{n-1}^2}\{\vdots}&{\vdots}&{}&{\vdots}\ {x_{0}^{n-1}}&{x_{1}^{n-1}}&{\cdots}&{x_{n-1}^{n-1}}\\end{pmatrix}^{-1} \]

So we consider Lagrange interpolation, there are
\[f (x) =\sum_{i}y_i\prod_{j\ne i} {x-x_j \over x_i-x_j} \]
Obviously,\ ((v^{-1}) _{ij}\) is the coefficient of \ (\prod\limits_{k\ne i} {x-x_k \over x_i-x_k}\) in the \ (x^{j-1}\) key.

The core idea of the fast Fourier transform is to transform the coefficient vectors into point-valued vectors quickly, and then quickly restore the point-value vectors to the coefficient vectors, in which the operations we call \ (idft\).
With \ ( 1\) \ (n\) secondary root \ (w\), where \ (w_i=w^i\)
When we do a fast Fourier transform, we multiply a matrix of \ (V (w_0,w_1,\cdots,w_{n-1}) \) .
While in \ (idft\) , we need to multiply the \ (V (w_0,w_1,\cdots,w_{n-1}) ^{-1}\), but in practice, we will directly multiply the $ {1 \over n}v (w_0,w_{-1}, \CDOTS,W_{-N+1}) $. Next I will prove that the two matrices are the same. (Of course we default n is 2 recurses)

\[\prod\limits_{j\ne i} {(x-w^j) \over (w^i-w^j)}={\prod\limits_{j\ne i} (x-w^j) \over \prod\limits_{j\ne i} (w^i-w^j)}\ ]

Make \[g (x) =\prod_{0 \leq J < n} (x-w^j) \]
and \ (w^{0},w^1,\cdots,w^{n-1}\) are 1 n times complex root, according to the basic theorem of algebra, obviously have \[g (x) =x^n-1\]
So consider the primitive denominator \[\prod\limits_{j\ne i} (w^i-w^j) = \lim _{x \to w^i}{g (x) \over {x-w^i}}\]
According to the indeterminate form rule, the value of this limit is equivalent to the quotient of the upper and lower derivative.
\[\lim _{x \to w^i}{g (x) \over {x-w^i}}=\lim _{x \to w^i} G ' (x) =n \times w^{i (n-1)}=n \times w^{-i}\]

Proto-molecule
\[{\prod\limits_{j\ne i} (x-w^j)}={g (x) \over {x-w^i}}={1-x^n \over {w^i-x}}\=w^{-i} \times \begin{pmatrix}{1 \over 1-x W ^{-i}}-{x^n \over 1-xw^{-i}}\end{pmatrix}\=w^{-i} \times \begin{pmatrix}{\sum_{j=0}^{\infty} w^{-ij}x^j-\sum_{j=n}^ {\infty} w^{-i (j-n)}x^j} \end{pmatrix}\=w^{-i} \times \sum_{j=0}^{n-1} W^{-ij} x^j\]

The numerator is divided by the denominator.
\[original ={w^{-i} \times \sum\limits_{j=0}^{n-1} w^{-ij} x^j \over n \times w^{-i}}\=\sum_{j=0}^{n-1} {W^{-ij} \over n}x^i\]
Comparing the coefficients, it is not difficult to draw the same two matrices, namely
\[\BEGIN{PMATRIX}{1}&{1}&{\CDOTS}&{1}\{W_{0}}&{W_{1}}&{\CDOTS}&{W_{N-1}}\{W_{0}^2} &{w_{1}^2}&{\cdots}&{w_{n-1}^2}\{\vdots}&{\vdots}&{}&{\vdots}\{w_{0}^{n-1}}&{w_{1} ^{n-1}}&{\cdots}&{w_{n-1}^{n-1}}\\end{pmatrix}^{-1}={1 \over N}\begin{pmatrix}{1}&{1}&{\cdots} &{1}\{w_{0}}&{w_{-1}}&{\cdots}&{w_{-n+1}}\{w_{0}^2}&{w_{-1}^2}&{\cdots}&{w_{-n+1}^ 2}\{\vdots}&{\vdots}&{}&{\vdots}\{w_{0}^{n-1}}&{w_{-1}^{n-1}}&{\cdots}&{w_{-n+1}^{n-1} }\\end{pmatrix}\]

On the inverse matrix of Vandermonde (Vandermonde) matrices and the principle of idft in fast Fourier transform (FFT)