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About LinkBacksProve the following theorem:
Ifare distinct points in an interval
, and if distinct points
are outside that interval, then the matrix having elements
is invertible.
for n = 1, there's nothing to prove. so, from now on i'll assume that n > 1. we want to prove that your matrix has a non-zeroOriginally Posted by mathwizard
Prove the following theorem:
If
determinant. so supposeis the determinant of your matrix. first see that:
where
=\begin{vmatrix} \prod_{j \neq 1}(t_1-s_j) & \prod_{j \neq 2} (t_1-s_j) & . & . & . & \prod_{j \neq n} (t_1-s_j) \\ \prod_{j \neq 1} (t_2 - s_j) & \prod_{j \neq 2}(t_2-s_j) & . & . & . & \prod_{j \neq n} (t_2 - s_j) \\ . & . & . & . & . & . & \\<br />
. & . & . & . & . & . \\ . & . & . & . & . & . \\ \prod_{j \neq 1}(t_n-s_j) & \prod_{j \neq 2} (t_n - s_j) & . & . & . & \prod_{j \neq n}(t_n-s_j) \end{vmatrix}.)
so we just need to prove thatthe proof is by induction on
if
then:
=\begin{vmatrix}t_1-s_2 & t_1-s_1 \\ t_2-s_2 & t_2-s_1 \end{vmatrix}=(t_1-t_2)(s_2-s_1) \neq 0.)
now suppose thatwhenever
, and
now define:
clearly
is a polynomial of degree
and
it's also evident that
thus
where
is a constant.
so we only need to prove thatsince
we only need to prove that
this is easy to prove
because from the definition ofit's easy to see that:
but by
induction hypothesisthus
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