The following is a short proof showing that the moment for the balanced least squares fit is zero. We start with the equation for the line and let δ' be the distance to a point on the line and then determine the value of t for the minimum distance. It is equal to the projection of Δx onto the line.

δ is the same as before. e and e' are eigenvectors of the covariance matrix and the product of it with each of them results in a vector in the same direction whose magnitude is its eigenvalue. Since the eigenvectors are perpendicular to each other the product, M, is zero.

Supplemental: The moments are proportional to the change in the variance. One might define them as dV/dθ where θ represents the angle through which the direction of the line and one of the other eigenvectors are rotated through their common plane. One could use the product of the variance and the direction of the line, Ve', to define a surface about the center of the distribution. Since d(Ve') = (dV)e' + V(de') = (dV)e' + Vedθ both the direction and magnitude of Ve' will change with the rotation. Assuming all the eigenvalues are different, the change is zero for all rotations if and only if the direction of the line, e', is one of the eigenvectors of the covariance matrix.

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