Talk:Vandermonde matrix

Add in proof of determinant formula

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)

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Transpose

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)

Now fixed. -- Jitse Niesen 13:36, 12 Jul 2004 (UTC)
Do we really want j<k vs. jk rather than jk vs. j>k in the case distinctions? Not knowing much of anything about these matrices, the latter seems more naturally to me, seeing that k=0 is the case of an ordinary Vandermonde matrix, and j=k gives a (-1)! in the denominator. AxelBoldt 16:17, 12 Jul 2004 (UTC)
You're absolutely right. Thanks for spotting this. -- Jitse Niesen 16:33, 12 Jul 2004 (UTC)

In the formula for the determinant, what is the set ${\displaystyle S_{n}}$? It is not defined or referenced. -- Grubber

This refers to the formula
${\displaystyle \det(V)=\prod _{1\leq i
Sn denotes the set of permutations of {1, 2, ..., n}, and sgn(σ) denotes the signature of the permutation σ. However, I now removed the second expression for the determinant, since it is just a straightforward application of the Leibniz formula for the determinant, and it does not seem very useful. Anyway, thanks for bringing this to our attention. -- Jitse Niesen 03:55, 15 Jun 2005 (UTC)
I object to removing this. Why? To you, its a "straightforward application of the Leibniz formula". However, I know that AxelBoldt is very smart, and clearly, it wasn't straightforward to him. If it had been obvious, he would not have needed to ask "what is S_n?" I know AxelBoldt is smarter than 98% of all other wikipedia math readers, so I think its a bit of a mistake that average readers will have the Leibniz formula pop into their heads. Please, the goal of WP hould not be to be terse and concise; it should be healpful and useful. We should put the formula back in, and explain what S_n is, and explain that its obvious. (I mean, it is obvious, but sometimes one can be tired and bleary when reading WP, and when one is tired, its helpful to have things pop up). (Besides, the vandermonde determinant is all about the representation theory of the symmetric group, it makes sense to have the symmetric group appear here.). linas 05:06, 15 Jun 2005 (UTC)
If someone wouldn't mind keeping it and providing a reference and a definition, I'd appreciate it. I ran across Vandermonde matrices in a coding paper I'm reading, and all the 'obvious' tricks and formulae you all know are helpful! -- Grubber 09:35 15 Jun 2005 (UTC)
Fair enough, I put the formula back in. My primary reason for removing it was not that it's a "straightforward application of the Leibniz formula", but that I did not see that it is useful. However, I know little representation theory, so I believe you when you say it is useful in that context. -- Jitse Niesen 08:49, 15 Jun 2005 (UTC) (via edit conflict with Grubber)

Order of material

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 thought about it, and decided against it. The way things are now, the polynomial interpolation is used as just a motivation for introducting confluent Vandermonde matrices, so people can ignore that motivational sentence. But I did clarify that polymial interpolation is defined a few sections below. Oleg Alexandrov (talk) 01:09, 25 May 2006 (UTC)

more properties

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)

Moore matrix

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 ${\displaystyle M_{i,j}=\alpha _{i}^{q^{j-1}}}$. Richard Pinch (talk) 18:29, 16 July 2008 (UTC)

The Quantum Hall effect / Anyons

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)

Vandermonde currently redirects to Alexandre-Théophile Vandermonde which mentions both Vandermonde matrices and Vandermonde's identity. One could create another redirect van der monde to Alexandre-Théophile Vandermonde. I think this would be not be inconsistent with the spirit of WP:REDIRECT#KEEP. I am going to create that redirect. — Tobias Bergemann (talk) 14:15, 22 April 2010 (UTC)

Comment on the Introduction

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)

Well, this is true about permuting rows or columns of the determinant. If I understand this correctly, this sentence speaks about permuting variables when we view the as a polynomial. (In the other words: I believe that the intention was to explain that "Vandermonde polynomial is an alternating polynomial".) --Kompik (talk) 08:36, 22 April 2010 (UTC)

Mention that it is assumed that the entries commute

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)

The non-square case

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)