definition of Vandermonde determinant
Topics
About: matrices space
The table of contents of this article
Starting Context
- The reader knows a definition of field.
- The reader knows a definition of determinant of square matrix over ring.
- The reader admits the proposition that for the polynomials ring over any field and any nonconstant polynomial, if and only if the evaluation of the polynomial at a field element is 0, the polynomial can be factorized with x - the element.
- The reader admits the Laplace expansion of the determinant of any square matrix over any commutative ring holds and its corollary.
Target Context
- The reader will have a definition of Vandermonde determinant.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( F\): \(\in \{\text{ the fields }\}\)
\( \{r_1, ..., r_n\}\): \(\subseteq F\)
\(*D_n (r_1, ..., r_n)\): \(= det \begin{pmatrix} {r_1}^{n - 1} & {r_1}^{n - 2} & ... & r_1 & 1 \\ ... \\ {r_n}^{n - 1} & {r_n}^{n - 2} & ... & r_n & 1 \end{pmatrix}\)
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Conditions:
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2: Note
In fact, \(D_n (r_1, ..., r_n) = (r_1 - r_2) ... (r_1 - r_n) (r_2 - r_3) ... (r_2 - r_n) ... (r_{n - 1} - r_n)\) (when \(n = 1\), it is \(1\)).
Let us prove that claim inductively.
Let us suppose that \(2 \le n\).
When \(\{r_1, ..., r_n\}\) is not distinct, \(D_n (r_1, ..., r_n) = 0\), by a property of determinant (some rows are the same), and \((r_1 - r_2) ... (r_1 - r_n) (r_2 - r_3) ... (r_2 - r_n) ... (r_{n - 1} - r_n) = 0\), so, \(D_n (r_1, ..., r_n) = (r_1 - r_2) ... (r_1 - r_n) (r_2 - r_3) ... (r_2 - r_n) ... (r_{n - 1} - r_n)\).
Let us suppose that \(\{r_1, ..., r_n\}\) is distinct hereafter.
\(D_n (x, r_2, ..., r_n)\) is a '\(n - 1\)'-degree polynomial of \(x\).
For each \(j \in \{2, ..., n\}\), \(D_n (r_j, r_2, ..., r_n) = 0\), by a property of determinant: the \(1\)-th row and the \(j\)-th row are the same.
So, \(D_n (x, r_2, ..., r_n) = r (x - r_2) ... (x - r_n)\), by the proposition that for the polynomials ring over any field and any nonconstant polynomial, if and only if the evaluation of the polynomial at a field element is 0, the polynomial can be factorized with x - the element.
\(r\) is the coefficient of \(x^{n - 1}\), but by the Laplace expansion of the determinant of any square matrix over any commutative ring holds and its corollary, \(r = D_{n - 1} (r_2, ..., r_n)\).
So, when \(2 \le n\), \(D_n (x, r_2, ..., r_n) = D_{n - 1} (r_2, ..., r_n) (x - r_2) ... (x - r_n)\), and so, \(D_n (r_1, r_2, ..., r_n) = (r_1 - r_2) ... (r_1 - r_n) D_{n - 1} (r_2, ..., r_n)\).
When \(n = 1\), \(D_1 (r_1) = 1\).
When \(n = 2\), \(D_2 (r_1, r_2) = (r_1 - r_2)\).
Let us suppose that for any \(n \in \mathbb{N} \setminus \{0\}\), \(D_n (r_1, ..., r_n) = (r_1 - r_2) ... (r_1 - r_n) (r_2 - r_3) ... (r_2 - r_n) ... (r_{n - 1} - r_n)\).
\(D_{n + 1} (r_1, r_2, ..., r_{n + 1}) = (r_1 - r_2) ... (r_1 - r_{n + 1}) D_n (r_2, ..., r_{n + 1}) = (r_1 - r_2) ... (r_1 - r_{n + 1}) (r_2 - r_3) ... (r_2 - r_{n + 1}) (r_3 - r_4) ... (r_3 - r_{n + 1}) ... (r_n - r_{n + 1})\).
So, the claim has been proved.
\(\{r_1, ..., r_n\}\) is distinct if and only if \(D_n \neq 0\).
