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So I have this symmetric matrix: $$ S = \begin{bmatrix} n+1 & \sum\limits_{i=0}^n x_i & \sum\limits_{i=0}^n x_i^2 & \dots & \sum\limits_{i=0}^n x_i^n \\ \sum\limits_{i=0}^n x_i & \sum\limits_{i=0}^n x_i^2 & \dots & \dots & \sum\limits_{i=0}^n x_i^{n+1} \\ \sum\limits_{i=0}^n x_i^2 &\vdots &\ddots& & \vdots\\ \vdots & \vdots & &\ddots& \vdots \\ \sum\limits_{i=0}^n x_i^n & \sum\limits_{i=0}^n x_i^{n+1} & \dots & \dots & \sum\limits_{i=0}^n x_i^{2n} \end{bmatrix} $$ The dimension of this matrix is $n+1$. The $x_i$ represent data points (coordinates). So every element in the matrix should be a real number. I'd like to show that this matrix is invertible for values $x_i \neq x_j$ for $i \neq j$.

I have to show that $$det(S) \neq 0.$$ $S$ is symmetric so the determinant is equal to the product of the eigenvalues of S: $$ det(S) = \prod_{i=0}^{n} \lambda_i.$$ My idea was to prove that $S$ has no zero-eigenvalues but I really have no idea on how to start doing this (due to complexity of the matrix).

If anyone has an idea how to prove this that would be appreciated! It is, of course, possible that $S$ is not invertible (I have no idea). If you can find a counter-example, please go along.

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In this post it is proved that your matrix is positive definite, since it can be written as a quadratic form $B^TB$ . Hence, it is also invertible.

This directly proves the claim. $\Box$

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  • $\begingroup$ Thanks, just what I needed! $\endgroup$ – WarreG Jan 23 '18 at 14:22
  • $\begingroup$ Why, exactly, is the matrix positive definite if it can be written as $B^TB$? $\endgroup$ – WarreG Jan 23 '18 at 14:29
  • $\begingroup$ Define $y=Bx$. Then for any nonzero vector $x$, $x^TS x = y^T y = \sum y_i^2 >0$. $\endgroup$ – Andreas Jan 23 '18 at 14:35

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