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Homework Statement
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)
Homework 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
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
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