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[Matrix Algebra] Special matrix, columns as a first order derivatives

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Let a_1,...,a_n be Real. Then define a 2n x 2n matrix A as follows.

    The following are the first 4 and the last 2 columns;
    note that the (2k) column equals (a_k)(derivative of the (2k-1) column

    A[,1] = a_1^(2n),a_1^(2n-1),a_1^(2n-2),...,a_1^(2), a_1

    A[,2] = (2n)a_1^(2n),(2n-1)a_1^(2n-1),(2n-2)a_1^(2n-2),...,2a_1^(
    2), a_1

    A[,3] = a_2^(2n),a_2^(2n-1),a_2^(2n-2),...,a_2^(2), a_2

    A[,4] = (2n)a_2^(2n),(2n-1)a_2^(2n-1),(2n-2)a_2^(2n-2),...,2a_2^(2), a_2
    A[,2n-1] = a_n^(2n),a_n^(2n-1),a_n^(2n-2),...,a_n^(2), a_n

    A[,2n] = (2n)a_n^(2n),(2n-1)a_n^(2n-1),(2n-2)a_n^(2n-2),...,2a_n^(2), a_n

    Calulate det(A)

    2. Relevant equations

    I'd like to need to know if there's a special name for this kind of matrix? (It seems to be a special case, which can be generalized to any order derivative)

    Also, I outlined my idea below, and I'm curious if there's an easier way of solving this problem

    3. The attempt at a solution
    I partition A into 4 square matrices, say P,Q,R, and S. Then proceed inductively, with the top-left partition P at step k+1 being equal to the whole A matrix from the previous kth step. This way at each step the dimensions are as follows

    top-left P: (2k-2)x(2k-2)
    top-right Q: (2k-2)x2
    bottom-left R: 2x(2k-2)
    bottom-right S: 2x2

    Next, at each step, I use det(A)=det(P)det(S-RP^(-1)Q)

    This is where I'm getting confused. Can I somehow use the dependence between the columns without going through all the gory calculations?

    Appreciate any help
    Last edited: Jan 24, 2009
  2. jcsd
  3. Jan 24, 2009 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    There seem to be some obvious row and column operations to fiddle around with, that would simplify your matrix.... Did any of them do anything useful?
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