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Finding determinant of Vandermonde matrix

  1. Jun 19, 2007 #1
    Can any one help with finding the determinant of vandermonde matrix or the way in which I can start the problem. I did with 4X4 matrix but I would have to show that the generalisation that I get works for nXn matrix. Please anyone
  2. jcsd
  3. Jun 19, 2007 #2


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    It's a bit too messy to write down right now, but try to eliminate a row/column up to a single non-zero member and use Laplace expansion on that row/column and see where it gets you.
  4. Jun 29, 2007 #3
    This Wikipedia article might help.


    The formula is given, but a proof is not given.

    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

    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.

    Hope that was clear. I didn't write the solution super carefully, but the ideas are all there... Let me know if it made sense...
    Last edited: Jun 30, 2007
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