Let x1,x2,…,xmx_1,x_2,\dots,x_m be distinct real numbers and n=m−1n=m-1. Define matrix A=(aij)A=(a_{ij}) where

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

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1 Answer

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You can write A=VVTA = VV^T, where VV is the Vandermonde matrix with entries vik=xi−1kv_{ik} = x_k^{i-1}. Now rank(V)=n\renewcommand\rank{\operatorname{rank}} \rank(V) = n because n \le mn \le m and the x_ix_i are distinct, so \rank(A) = \rank(VV^T) = n\rank(A) = \rank(VV^T) = n and AA is nonsingular.

Note: You could even let n = mn = m and AA would still be nonsingular.