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Math Help - Proving when Gram's determinant is equal to zero

  1. #1
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    Proving when Gram's determinant is equal to zero

    Task: Prove that Gram's determinant G(x_1,\dots, x_n)=0 if and only if x_1, \dots, x_k are linearly dependent.

    So I know that

    G(x_1,\dots, x_n)=\det \begin{vmatrix} \xi( x_1,x_1) & \xi( x_1,x_2) &\dots & \xi( x_1,x_n)\\<br />
 \xi( x_2,x_1) & \xi( x_2,x_2) &\dots & \xi( x_2,x_n)\\<br />
\vdots&\vdots&\ddots&\vdots\\<br />
 \xi( x_n,x_1) & \xi( x_n,x_2) &\dots & \xi( x_n,x_n)\end{vmatrix}

    (where \xi denotes an inner product) which means that if G(x_1,\dots, x_n)=0, then the determinant has to be equal to 0 as well. Why does it happen only with x_1,\dots, x_k being linearly dependent, though?

    Also, sorry for the <br> tags in the matrix but TeX converts my \\ signs that way here. Weird...
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  2. #2
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    Re: Proving when Gram's determinant is equal to zero

    Hi I would try this way, just briefly:

    If x_1,\dots ,x_n are linearly dependent, then there exist scalars \alpha_1,\dots ,\alpha_n such that
    \alpha_1 x_1+\alpha_2 x_2 +\dots + \alpha_n x_n =0 and \sum_{k=1}^n |\alpha_k|^2>0.

    Denote G the matrix, to which determinant G(x_1,\dots ,x_n) belongs, (G)_j its j-th row and \alpha=(\bar{\alpha_1},\dots ,\bar{\alpha_n})^{T}.

    Then there holds G\alpha=0, since for each j\in\{1,\dots ,n\} there holds
    (G)_j\alpha=\sum_{k=1}^n \bar{\alpha_k}\xi (x_1,x_n)=\xi \Big(x_j,\sum_{k=1}^n \alpha_k x_k\Big)=\xi (x_j,0)=0

    so nonzero vector \alpha belongs to Ker(G) and thus G is singular matrix.

    The other way,
    if G is singular matrix, there exists nonzero vector \alpha=(\alpha_1,\dots ,\alpha_n) such that for all j\in\{1,\dots ,n\} there holds
    \xi (x_j,\beta)=0, where \beta=\sum_{k=1}^n\bar{\alpha_k} x_k. But then \xi (\beta,\beta)=0 so \beta=0
    and x_1,\dots ,x_n are linearly dependent
    Last edited by alteraus; June 9th 2013 at 05:00 PM.
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