Re: derivatives and determinant
- From: quasi <quasi@xxxxxxxx>
- Date: Tue, 14 Mar 2006 09:35:56 -0500
On Mon, 13 Mar 2006 05:53:57 -0500, quasi <quasi@xxxxxxxx> wrote:
On Mon, 13 Mar 2006 05:23:52 EST, eugene <jane1806@xxxxxxx> wrote:
It was a typo in my first message
Let f(x)=(x-x_1)(x-x_2)...(x-x_n)where the numbers x_1,x_2,...,x_n-pairwise distinct and a_{ii}=f''(x_i), a_{ij}=(f'(x_i)-f'(x_j))/(x_i-x_j). i \neq j.
1<=i,j<=n.
Now, A mustn't be always diagonal..
And with that correction, det(A)=0 now appears to be true (but I don't
have time to look at it right now).
quasi
Ok, I think my prior attempt can be repaired.
Here is my rough sketch of a proof strategy, version 2 ...
The first 7 steps are the same as in version 1.
(1) What happens to A as x_i --> x_j ?
Fix x_1 and let x_2 approach x_1. In the limit, rows 1 and 2 become
equal and opposite. The same for columns 1 and 2. Hence A becomes
singular as x1 --> x2. Likewise for any pair i,j with i <> j.
(2) Simplify the entries of A to polynomials.
(3) Let f(x_1,...,x_n) = det(A).
Clearly f is a polynomial in x_1,...,x_n. Our goal is to show f=0.
(4) Let p = product (x_i - x_j) for all pair i,j with i<>j.
(5) f is a multiple of p.
From step (1), for distinct i,j, f is 0 if x_i = x_j. Hence f is amultiple of x_i - x_j for all pairs i,j with i<>j. It follows that f
is a multiple of p.
(6) f is a symmetric polynomial in x_1,...,x_n.
Again from step (1), any transposition of a pair x_i, x_j with i<>j
swaps both rows i,j (with a sign change) and also columns i,j (also
with a sign change). The row swap possibly negates det(A) but if so,
the column swap negates it again so det(A) remains unchanged.
Since det(A) is unchanged by transpositions, it's must be unchanged
for any permutation of x_1,...,x_n, hence f is symmetric.
(7) f is a multiple of p^2
By step (5), f is a multiple of (x_i - x_j) but then since f is
symmetric, f must be a multiple of (x_i - x_j)^2 (otherwise there
would be a sign change if x_i and x_j were swapped). Hence, f is a
multiple of (x_i - x_j)^2 for all pairs i,j with i<>j. It follows that
f is a multiple of p^2.
(8) Each entry of A is a homogeneous polynomial (in the variables
x_1,...,x_n), of total degree n-2 (assuming n>1).
(9) f is a homogeneous polynomial of total degree at most n*(n-2).
This is based on step (8) together with viewing det(A) as a signed sum
of the product of generalized diagonals.
(10) p^2 is a homogeneous polynomial of total degree n*(n-1).
(11) det(A)=0
p^2 divides f but the total degree of p^2 exceeds the total degree of
f. It follows that f=0, and hence, det(A)=0, as required.
quasi
.
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