How do you know that the least squares regression line is unique and corresponds to a minimum

We already know that using the criterion of either

  1. minimizing sum of residuals OR
  2. minimizing sum of the absolute value of residuals

is BAD as either of the criteria do not give a unique line. Visit these notes for an example where these criteria are shown to be inadequate.

So we use minimizing the sum of the squares of the residuals as the criterion. How can we show that this criterion gives a unique line?

The proof is given below as image files because the proof is equation intensive. I made a better resolution pdf file also.


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Author: Autar Kaw

Autar Kaw ( is a Professor of Mechanical Engineering at the University of South Florida. He has been at USF since 1987, the same year in which he received his Ph. D. in Engineering Mechanics from Clemson University. He is a recipient of the 2012 U.S. Professor of the Year Award. With major funding from NSF, he is the principal and managing contributor in developing the multiple award-winning online open courseware for an undergraduate course in Numerical Methods. The OpenCourseWare ( annually receives 1,000,000+ page views, 1,000,000+ views of the YouTube audiovisual lectures, and 150,000+ page views at the NumericalMethodsGuy blog. His current research interests include engineering education research methods, adaptive learning, open courseware, massive open online courses, flipped classrooms, and learning strategies. He has written four textbooks and 80 refereed technical papers, and his opinion editorials have appeared in the St. Petersburg Times and Tampa Tribune.

One thought on “How do you know that the least squares regression line is unique and corresponds to a minimum”

  1. Nice. Note that (13) is positive because it’s a multiple of the variance of the x_i’s. Of course, if all the x_i’s are equal, the variance is zero, so having the x_i’s not all equal is necessary and sufficient for uniqueness.

    But a single critical point which is a local minimum need not be a global minimum, see . Here I think you’re OK since the Hessian is positive everywhere.


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