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

    If $\displaystyle t_1, t_2, ..., t_n$ are distinct points in an interval $\displaystyle [a,b] $, and if distinct points $\displaystyle s_1, s_2, ..., s_n$ are outside that interval, then the matrix having elements $\displaystyle a_{ij} = (t_i - s_j)^{-1}$ 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 $\displaystyle I$ is the determinant of your matrix. first see that:

    $\displaystyle I=\frac{J(t_1, ... , t_n, s_1, ... , s_n)}{\prod_{i,j}(t_i-s_j)},$ where $\displaystyle J(t_1,...,t_n,s_1,...,s_n)=\begin{vmatrix} \prod_{j \neq 1}(t_1-s_j) & \prod_{j \neq 2} (t_1-s_j) & . & . & . & \prod_{j \neq n} (t_1-s_j) \\ \prod_{j \neq 1} (t_2 - s_j) & \prod_{j \neq 2}(t_2-s_j) & . & . & . & \prod_{j \neq n} (t_2 - s_j) \\ . & . & . & . & . & . & \\
    . & . & . & . & . & . \\ . & . & . & . & . & . \\ \prod_{j \neq 1}(t_n-s_j) & \prod_{j \neq 2} (t_n - s_j) & . & . & . & \prod_{j \neq n}(t_n-s_j) \end{vmatrix}.$

    so we just need to prove that $\displaystyle J(t_1,...,t_n,s_1,...,s_n) \neq 0.$ the proof is by induction on $\displaystyle n.$ if $\displaystyle n=2,$ then:

    $\displaystyle J(t_1,t_2,s_1,s_2)=\begin{vmatrix}t_1-s_2 & t_1-s_1 \\ t_2-s_2 & t_2-s_1 \end{vmatrix}=(t_1-t_2)(s_2-s_1) \neq 0.$

    now suppose that $\displaystyle J(x_1, ..., x_k, y_1, ..., y_k) \neq 0,$ whenever $\displaystyle x_i \neq x_j, \ y_i \neq y_j, \ \forall i \neq j$, and $\displaystyle k \leq n-1.$ now define:

    $\displaystyle p(x)=\begin{vmatrix} \prod_{j \neq 1}(x-s_j) & \prod_{j \neq 2} (x-s_j) & . & . & . & \prod_{j \neq n} (x-s_j) \\ \prod_{j \neq 1} (t_2 - s_j) & \prod_{j \neq 2}(t_2-s_j) & . & . & . & \prod_{j \neq n} (t_2 - s_j) \\ . & . & . & . & . & . & \\
    . & . & . & . & . & . \\ . & . & . & . & . & . \\ \prod_{j \neq 1}(t_n-s_j) & \prod_{j \neq 2} (t_n - s_j) & . & . & . & \prod_{j \neq n}(t_n-s_j) \end{vmatrix}.$ clearly $\displaystyle p(x)$ is a polynomial of degree $\displaystyle n-1$ and

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

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

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

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