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Let $x_1,x_2,\dots,x_m$ be distinct real numbers and $n=m-1$. Define matrix $A=(a_{ij})$ where

$$a_{ij}=\sum_{k=1}^m x_k^{i+j-2}$$ $i,j=1,2,\dots,n$. This matrix is the left hand side matrix of normal equation from least square method using polynomial function. Is $A$ non singular? From its definition we know that $A$ is symmetric and its diagonal components are positive numbers.

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You can write $A = VV^T$, where $V$ is the Vandermonde matrix with entries $v_{ik} = x_k^{i-1}$. Now $\renewcommand\rank{\operatorname{rank}} \rank(V) = n$ because $n \le m$ and the $x_i$ are distinct, so $\rank(A) = \rank(VV^T) = n$ and $A$ is nonsingular.

Note: You could even let $n = m$ and $A$ would still be nonsingular.

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