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Math Help - Linear algebra: Prove the matrix having elements a_ij is invertible.

  1. #1
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    Linear algebra: Prove the matrix having elements a_ij is invertible.

    Prove the following theorem:

    If t_1, t_2, ..., t_n are distinct points in an interval  [a,b] , and if distinct points s_1, s_2, ..., s_n are outside that interval, then the matrix having elements a_{ij} = (t_i - s_j)^{-1} is invertible.
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  2. #2
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    Quote Originally Posted by mathwizard View Post
    Prove the following theorem:

    If t_1, t_2, ..., t_n are distinct points in an interval  [a,b] , and if distinct points s_1, s_2, ..., s_n are outside that interval, then the matrix having elements a_{ij} = (t_i - s_j)^{-1} 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-zero

    determinant. so suppose I is the determinant of your matrix. first see that:

    I=\frac{J(t_1, ... , t_n, s_1, ... , s_n)}{\prod_{i,j}(t_i-s_j)}, where J(t_1,...,t_n,s_1,...,s_n)=\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 that J(t_1,...,t_n,s_1,...,s_n) \neq 0. the proof is by induction on n. if n=2, then:

    J(t_1,t_2,s_1,s_2)=\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 that J(x_1, ..., x_k, y_1, ..., y_k) \neq 0, whenever x_i \neq x_j, \ y_i \neq y_j, \ \forall i \neq j, and k \leq n-1. now define:

    p(x)=\begin{vmatrix} \prod_{j \neq 1}(x-s_j) & \prod_{j \neq 2} (x-s_j) & . & . & . & \prod_{j \neq n} (x-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}. clearly p(x) is a polynomial of degree n-1 and

    p(t_1)=J(t_1,...,t_n,s_1,...,s_n). it's also evident that p(t_j)=0, \ j = 2, ..., n. thus p(x)=\alpha \prod_{j=2}^n(x-t_j), where \alpha is a constant.

    so we only need to prove that \alpha \neq 0. since \alpha=\frac{p(s_1)}{\prod_{j=2}^n(s_1-t_j)}, we only need to prove that p(s_1) \neq 0. this is easy to prove

    because from the definition of p(x), it's easy to see that: p(s_1)=\left(\prod_{j=2}^n(t_j-s_1)(s_1-s_j)\right)J(t_2,...,t_n,s_2,...,s_n). but by

    induction hypothesis J(t_2,...,t_n,s_2,...,s_n) \neq 0. thus \alpha \neq 0. \ \ \ \square
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