## Linear Algebra and Its Applications, Exercise 3.3.23

Exercise 3.3.23. Given measurements $y_1, y_2, \dots, y_m$ show that the best least squares fit to the horizontal line $y = C$ is given by $C = (y_1 + y_2 + \cdots + y_m)/m$

Answer: This corresponds to the system $Ax = b$ where $A$ is an $m$ by 1 matrix with all entries equal to 1 and $b = (y_1, y_2, \cdots, y_m)$. To find the least squares solution we form the system $A^TA\bar{x} = A^Tb$. We have $A^TA = \begin{bmatrix} 1&1&\cdots&1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = \sum_{i=1}^m 1 = m$

and $A^Tb = \begin{bmatrix} 1&1&\cdots&1 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \end{bmatrix} = \sum_{i=1}^m y_i$

The system $A^TA\bar{x} = A^Tb$ thus reduces to $m\bar{C} = \sum_{i=1}^m y_i$ so that $\bar{C} = (y_1 + y_2 + \cdots + y_m)/m$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the accompanying free online course, and Dr Strang’s other books .

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