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Here is a proof, taken from physicsforums.com. If this does not violate copyright, perhaps someone should clean it up and add it to the Wikipedia page.
"To prove the formula given in the article, replace the ith row by
1 x x^2 ... x^(n-1)
Then, take the determinant. The determinant is a function of x. Lets call it V(x).
It is a polynomial in x of degree (n-1). Hence, it has (n-1) roots. Furthermore, V(a_i) is the value of the vandermonde determinant, and you know that letting x=a_j for any j between 1 and n(excluding i) makes the determinant 0. Hence, a_i is a root of the polynomial V(x). Thus, (a_i - a_j) is a factor in the expansion of V(a_i). Repeating this with each of the i rows tells us that that
det(V)= C product (a_j - a_i), where the product is taken over 1<=i<j<=n. C is a constant. The fact that C=1 follows from induction. To show that C=1, just consider the cofactor expansion along the last column and examine the coefficient of the highest power of a_n. This is again a vandermonde determinant. Hence, C is the same constant as the smaller Vandermonde determinant. Of course, you need to check that in the case n=2 that C=1." 68.51.75.84 ( talk) 03:13, 30 November 2009 (UTC)particle25
I have added a link to a proof (not the same as yours, I considered the one I found simpler). 84.226.97.134 ( talk) 15:39, 6 August 2010 (UTC)
I switched to the transpose, since the equation for polynomial interpolation was incorrect and the transposed matrix works better in companion matrix as well.
Something is wrong with the formula for the confluent Vandermonde matrix; the index surely cannot be 0? AxelBoldt 18:05, 11 Jul 2004 (UTC)
In the formula for the determinant, what is the set ? It is not defined or referenced. -- Grubber
COMMENT BY ANOTHER PERSON: In its current incarnation, "the polynomial interpolation problem is ill-posed" comes before the mention of polynomial interpolation as an application. This needs to be switched around. --anon
I came across some more properties. Do you think they're woth to be included?
a) the eigenvalues a1,a2,...an of a vandermonde-matrix of dimension n have the following property:
all eigenvalues are irrational (roots of order n)
a1 + a2 + a3 + ... + an = integer a1*a2 + a1*a3 + ... + a2*a3 + a2*a4 + ... +a(n-1) *an = integer a1*a2*a3 + ... + a(n-2)*a(n-1)*an = integer ...
a1*a2*a3...*an = integer = 1!*2!*3!*...*(n-1)!
a1^2 + a2^2 + ... an^2 = integer a1^3 + a2^3 + ... an^3 = integer ... a1^n + a2^n + ... an^n = integer
b) let V be the vandermonde matrix, S1 the lower-triangular stirling-matrix kind 1,
- which starts with 1 -1 1 1 -3/2 1/2 -1 11/6 -1 1/6 ... (the rows are scaled by the reciprocal factorials)
P the lower triangular binomial-matrix, S1~ the transpose od S1, then
V*S1~ = P
--Gotti 17:17, 8 January 2007 (UTC)
Definitely needs a citation for this being "more commonly known": 4 hits on Google Books, one of which disagrees completely with this definition and one of which differs in a more subtle way! I believe that the likely correct definition of Moore matrix, per a paper of Noam Elkies is that over GF(q) the matrix M has . Richard Pinch ( talk) 18:29, 16 July 2008 (UTC)
This matrix is of fundamental importance in the quantum hall effect, and forms part of the ansatz used as the wavefunction for quasi-particles in 2D. Perhaps a mention? —Preceding unsigned comment added by 137.195.250.2 ( talk) 10:20, 19 July 2009 (UTC)
Is there a way for this article to appear in the Wikipedia search if someone searches for "van der monde" instead of "vandermonde?" I did a wikipedia search for "van der monde" (thinking this was how it was spelled) and got no hits in the search. Brian Maurizi 72.51.124.202 ( talk) 20:10, 21 April 2010 (UTC)
The sentence
"Note that the Vandermonde determinant is alternating in the entries, meaning that permuting the αi by an odd permutation changes the sign, while permuting them by an even permutation does not change the value of the determinant."
is absolutely correct. However, it is a little misleading, because in fact this property is true for the determinant in general and has nothing to do with the Vandermonde matrix in particular. This should probably be mentioned. Brian Maurizi 72.51.124.202 ( talk) 20:13, 21 April 2010 (UTC)
The nice properties of the Vandermonde matrix only hold if the entries of the matrix commute. I think this should be explicitly mentioned.
Doetoe ( talk) 15:44, 5 July 2011 (UTC)
When you say "entries", do you mean the field over which the matrix space is constructed? For example, if the entries are non-commutative themselves. Isn't it assumed though that this matrix is only dealing with entries in R (or C more generally), with ordinary multiplication (which is always commutative). — Preceding unsigned comment added by 144.32.53.52 ( talk) 16:56, 3 January 2012 (UTC)
Although the article is careful to note the determinant formula and sequelae are valid only in the square case, essentially nothing is presented about the non-square case. Just as the square case for distinct arguments gives us the matrix for polynomial interpolation, the general case gives the normal equations for least square fit of polynomials (when multiplied by the transpose). This is an important application and gives an occasion to remark on rank and condition number, properties that are intrinsic to the non-square case. Hardmath ( talk) 21:01, 2 August 2011 (UTC)
I have to read through a whole page of dense mathematical definitions and anecdotes before I am told in the Vandermonde matrix#Applications what it's all about: evaluating a polynomial at multiple points. — Preceding unsigned comment added by 2A01:E35:8B11:FA90:1E6F:65FF:FE3E:10A9 ( talk) 23:42, 6 November 2017 (UTC)
I have removed the new proof recently added by Phill Schultz for the following reasons.
D.Lazard ( talk) 09:59, 30 June 2020 (UTC)
@ D.Lazard: can you go more in depth why you reverted my removal here. The reason I removed that source is because it was a wiki and wikis aren’t ideal sources for Wikipedia. CycoMa1 ( talk) 14:48, 1 December 2021 (UTC)
The introductory section contains this sentence:
"The identical term Vandermonde matrix was used for the transpose of the above matrix by Macon and Spitzbart (1958)."
This information is far too obscure and trivial for it to belong in the introduction. It is not even clear that this information belongs in the article at all. 2601:200:C000:1A0:E0B5:CA14:5879:A6C9 ( talk) 16:59, 18 June 2022 (UTC)
Who is the audience of this article? It is a list of features with no apparent order that could only be understood by someone who already has expertise in this subject. The article seems to have no utility for the expert who would already know this information. It also seems to have no utility for someone who wants to learn the subject, because the beginning has unnecessary details that could not be understood by someone who is not an expert.
I have used the Vandermonde matrix many times in my work as an engineer doing spatial computation. I never really needed to know the information the author thought belonged in the beginning of this article.
The article really needs a simple example somewhere near the first paragraph. It would be great if the example was taken from history. The example should also have a figure that really demonstrates the basic ideas. The reader can hold this image in their mind as they try to understand the more advanced ideas of the subject. Openmir ( talk) 17:23, 1 May 2023 (UTC)
The 1st proof uses the factor theorem, which Wikipedia only states for univariate or bivariate polynomials. As stated, it does not apply to the kind of multivariate polynomial that the Vandermonde determinant is. I think you could manually imitate the proof of the Factor theorem by substitutions. Maybe this is really an application of Hilbert's Nullstellensatz. -- Svennik ( talk) 16:10, 15 October 2023 (UTC)
The first proof still depends on knowing that multivariate polynomials over a field form a UFD. The proof of this is usually absent from most linear algebra courses - especially at an elementary level - and is in fact rather advanced. I've made a note that it isn't appropriate for introductory linear algebra classes. It remains a popular problem to set students. -- Svennik ( talk) 11:27, 18 October 2023 (UTC)
The proof is OK now. -- Svennik ( talk) 17:16, 18 October 2023 (UTC)
![]() | This article is rated C-class on Wikipedia's
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Here is a proof, taken from physicsforums.com. If this does not violate copyright, perhaps someone should clean it up and add it to the Wikipedia page.
"To prove the formula given in the article, replace the ith row by
1 x x^2 ... x^(n-1)
Then, take the determinant. The determinant is a function of x. Lets call it V(x).
It is a polynomial in x of degree (n-1). Hence, it has (n-1) roots. Furthermore, V(a_i) is the value of the vandermonde determinant, and you know that letting x=a_j for any j between 1 and n(excluding i) makes the determinant 0. Hence, a_i is a root of the polynomial V(x). Thus, (a_i - a_j) is a factor in the expansion of V(a_i). Repeating this with each of the i rows tells us that that
det(V)= C product (a_j - a_i), where the product is taken over 1<=i<j<=n. C is a constant. The fact that C=1 follows from induction. To show that C=1, just consider the cofactor expansion along the last column and examine the coefficient of the highest power of a_n. This is again a vandermonde determinant. Hence, C is the same constant as the smaller Vandermonde determinant. Of course, you need to check that in the case n=2 that C=1." 68.51.75.84 ( talk) 03:13, 30 November 2009 (UTC)particle25
I have added a link to a proof (not the same as yours, I considered the one I found simpler). 84.226.97.134 ( talk) 15:39, 6 August 2010 (UTC)
I switched to the transpose, since the equation for polynomial interpolation was incorrect and the transposed matrix works better in companion matrix as well.
Something is wrong with the formula for the confluent Vandermonde matrix; the index surely cannot be 0? AxelBoldt 18:05, 11 Jul 2004 (UTC)
In the formula for the determinant, what is the set ? It is not defined or referenced. -- Grubber
COMMENT BY ANOTHER PERSON: In its current incarnation, "the polynomial interpolation problem is ill-posed" comes before the mention of polynomial interpolation as an application. This needs to be switched around. --anon
I came across some more properties. Do you think they're woth to be included?
a) the eigenvalues a1,a2,...an of a vandermonde-matrix of dimension n have the following property:
all eigenvalues are irrational (roots of order n)
a1 + a2 + a3 + ... + an = integer a1*a2 + a1*a3 + ... + a2*a3 + a2*a4 + ... +a(n-1) *an = integer a1*a2*a3 + ... + a(n-2)*a(n-1)*an = integer ...
a1*a2*a3...*an = integer = 1!*2!*3!*...*(n-1)!
a1^2 + a2^2 + ... an^2 = integer a1^3 + a2^3 + ... an^3 = integer ... a1^n + a2^n + ... an^n = integer
b) let V be the vandermonde matrix, S1 the lower-triangular stirling-matrix kind 1,
- which starts with 1 -1 1 1 -3/2 1/2 -1 11/6 -1 1/6 ... (the rows are scaled by the reciprocal factorials)
P the lower triangular binomial-matrix, S1~ the transpose od S1, then
V*S1~ = P
--Gotti 17:17, 8 January 2007 (UTC)
Definitely needs a citation for this being "more commonly known": 4 hits on Google Books, one of which disagrees completely with this definition and one of which differs in a more subtle way! I believe that the likely correct definition of Moore matrix, per a paper of Noam Elkies is that over GF(q) the matrix M has . Richard Pinch ( talk) 18:29, 16 July 2008 (UTC)
This matrix is of fundamental importance in the quantum hall effect, and forms part of the ansatz used as the wavefunction for quasi-particles in 2D. Perhaps a mention? —Preceding unsigned comment added by 137.195.250.2 ( talk) 10:20, 19 July 2009 (UTC)
Is there a way for this article to appear in the Wikipedia search if someone searches for "van der monde" instead of "vandermonde?" I did a wikipedia search for "van der monde" (thinking this was how it was spelled) and got no hits in the search. Brian Maurizi 72.51.124.202 ( talk) 20:10, 21 April 2010 (UTC)
The sentence
"Note that the Vandermonde determinant is alternating in the entries, meaning that permuting the αi by an odd permutation changes the sign, while permuting them by an even permutation does not change the value of the determinant."
is absolutely correct. However, it is a little misleading, because in fact this property is true for the determinant in general and has nothing to do with the Vandermonde matrix in particular. This should probably be mentioned. Brian Maurizi 72.51.124.202 ( talk) 20:13, 21 April 2010 (UTC)
The nice properties of the Vandermonde matrix only hold if the entries of the matrix commute. I think this should be explicitly mentioned.
Doetoe ( talk) 15:44, 5 July 2011 (UTC)
When you say "entries", do you mean the field over which the matrix space is constructed? For example, if the entries are non-commutative themselves. Isn't it assumed though that this matrix is only dealing with entries in R (or C more generally), with ordinary multiplication (which is always commutative). — Preceding unsigned comment added by 144.32.53.52 ( talk) 16:56, 3 January 2012 (UTC)
Although the article is careful to note the determinant formula and sequelae are valid only in the square case, essentially nothing is presented about the non-square case. Just as the square case for distinct arguments gives us the matrix for polynomial interpolation, the general case gives the normal equations for least square fit of polynomials (when multiplied by the transpose). This is an important application and gives an occasion to remark on rank and condition number, properties that are intrinsic to the non-square case. Hardmath ( talk) 21:01, 2 August 2011 (UTC)
I have to read through a whole page of dense mathematical definitions and anecdotes before I am told in the Vandermonde matrix#Applications what it's all about: evaluating a polynomial at multiple points. — Preceding unsigned comment added by 2A01:E35:8B11:FA90:1E6F:65FF:FE3E:10A9 ( talk) 23:42, 6 November 2017 (UTC)
I have removed the new proof recently added by Phill Schultz for the following reasons.
D.Lazard ( talk) 09:59, 30 June 2020 (UTC)
@ D.Lazard: can you go more in depth why you reverted my removal here. The reason I removed that source is because it was a wiki and wikis aren’t ideal sources for Wikipedia. CycoMa1 ( talk) 14:48, 1 December 2021 (UTC)
The introductory section contains this sentence:
"The identical term Vandermonde matrix was used for the transpose of the above matrix by Macon and Spitzbart (1958)."
This information is far too obscure and trivial for it to belong in the introduction. It is not even clear that this information belongs in the article at all. 2601:200:C000:1A0:E0B5:CA14:5879:A6C9 ( talk) 16:59, 18 June 2022 (UTC)
Who is the audience of this article? It is a list of features with no apparent order that could only be understood by someone who already has expertise in this subject. The article seems to have no utility for the expert who would already know this information. It also seems to have no utility for someone who wants to learn the subject, because the beginning has unnecessary details that could not be understood by someone who is not an expert.
I have used the Vandermonde matrix many times in my work as an engineer doing spatial computation. I never really needed to know the information the author thought belonged in the beginning of this article.
The article really needs a simple example somewhere near the first paragraph. It would be great if the example was taken from history. The example should also have a figure that really demonstrates the basic ideas. The reader can hold this image in their mind as they try to understand the more advanced ideas of the subject. Openmir ( talk) 17:23, 1 May 2023 (UTC)
The 1st proof uses the factor theorem, which Wikipedia only states for univariate or bivariate polynomials. As stated, it does not apply to the kind of multivariate polynomial that the Vandermonde determinant is. I think you could manually imitate the proof of the Factor theorem by substitutions. Maybe this is really an application of Hilbert's Nullstellensatz. -- Svennik ( talk) 16:10, 15 October 2023 (UTC)
The first proof still depends on knowing that multivariate polynomials over a field form a UFD. The proof of this is usually absent from most linear algebra courses - especially at an elementary level - and is in fact rather advanced. I've made a note that it isn't appropriate for introductory linear algebra classes. It remains a popular problem to set students. -- Svennik ( talk) 11:27, 18 October 2023 (UTC)
The proof is OK now. -- Svennik ( talk) 17:16, 18 October 2023 (UTC)