## Linear Algebra and Its Applications, Exercise 3.3.13

Exercise 3.3.13. Using least squares, find the line that is the best fit to the following measurements:

$b = 4$ at $t = -2$

$b = 3$ at $t = -1$

$b = 1$ at $t = 0$

$b = 0$ at $= 2$

Also, given the matrix

$A = \begin{bmatrix} 1&-2 \\ 1&-1 \\ 1&0 \\ 1&2 \end{bmatrix}$

find the projection of $b = (4, 3, 1, 0)$ onto the column space $\mathcal{R}(A)$.

Answer: Assuming that the line in question has the form $C + Dt$ the problem can be expressed as that of finding a solution to the system

$\begin{bmatrix} 1&-2 \\ 1&-1 \\ 1&0 \\ 1&2 \end{bmatrix} \begin{bmatrix} C \\ D \end{bmatrix} = \begin{bmatrix} 4 \\ 3 \\ 1 \\ 0 \end{bmatrix}$

or $Ax = b$ where $x = (C, D)$ is the exact solution.

In this case there is no exact solution, so we look for the least squares solution $\bar{x} = (\bar{C}, \bar{D})$ that minimizes the error vector $b-A\bar{x}$. The error vector is minimized when it is orthogonal to the column space of $A$ and is therefore in the left nullspace of $A$. We then have $A^T(b - A\bar{x}) = 0$ so that $\bar{x}$ is a solution to the system $A^TA\bar{x} = A^Tb$.

We have

$A^TA = \begin{bmatrix} 1&1&1&1 \\ -2&-1&0&2 \end{bmatrix} \begin{bmatrix} 1&-2 \\ 1&-1 \\ 1&0 \\ 1&2 \end{bmatrix} = \begin{bmatrix} 4&-1 \\ -1&9 \end{bmatrix}$

and

$A^Tb = \begin{bmatrix} 1&1&1&1 \\ -2&-1&0&2 \end{bmatrix} \begin{bmatrix} 4 \\ 3 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 8 \\ -11 \end{bmatrix}$

so that the system $A^TA\bar{x} = A^Tb$ reduces to

$\begin{bmatrix} 4&-1 \\ -1&9 \end{bmatrix} \begin{bmatrix} \bar{C} \\ \bar{D} \end{bmatrix} = \begin{bmatrix} 8 \\ -11 \end{bmatrix}$

or

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcl}4\bar{C}&-&\bar{D}&=&8 \\ -\bar{C}&+&9\bar{D}&=&-11 \end{array}$

expressed as a system of equations.

Multiplying the first equation by $\frac{1}{4}$ and adding it to the second equation produces the system

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcl}4\bar{C}&-&\bar{D}&=&8 \\ &&\frac{35}{4}\bar{D}&=&-9 \end{array}$

From the second equation we have $\bar{D} = -\frac{36}{35}$. Substituting that value into the first equation we have $4\bar{C} + \frac{36}{35} = 8$ or $\bar{C} = \frac{1}{4}(8 - \frac{36}{35}) = \frac{61}{35}$.

The line of best fit is therefore $\frac{61}{35} - \frac{36}{35}t$.

Given the matrix

$A = \begin{bmatrix} 1&-2 \\ 1&-1 \\ 1&0 \\ 1&2 \end{bmatrix}$

the projection matrix $P$ onto the column space of $A$ can be computed as

$P = A(A^TA)^{-1}A^T$

From above we have

$A^TA = \begin{bmatrix} 4&-1 \\ -1&9 \end{bmatrix}$

so that its inverse is

$(A^TA)^{-1} = \frac{1}{4 \cdot 9 - (-1)(-1)} \begin{bmatrix} 9&-(-1) \\ -(-1)&4 \end{bmatrix}$

$= \frac{1}{35} \begin{bmatrix} 9&1 \\ 1&4 \end{bmatrix}$

We then have

$P = A(A^TA)^{-1}A^T$

$= \begin{bmatrix} 1&-2 \\ 1&-1 \\ 1&0 \\ 1&2 \end{bmatrix} \frac{1}{35} \begin{bmatrix} 9&1 \\ 1&4 \end{bmatrix} \begin{bmatrix} 1&1&1&1 \\ -2&-1&0&2 \end{bmatrix}$

$= \frac{1}{35} \begin{bmatrix} 1&-2 \\ 1&-1 \\ 1&0 \\ 1&2 \end{bmatrix} \begin{bmatrix} 7&8&9&11 \\ -7&-3&1&9 \end{bmatrix}$

$= \frac{1}{35} \begin{bmatrix} 21&14&7&-7 \\ 14&11&8&2 \\ 7&8&9&11 \\ -7&2&11&29 \end{bmatrix}$

The projection of the vector $b = (4, 3, 1, 0)$ onto the column space of $A$ is then

$Pb = \frac{1}{35} \begin{bmatrix} 21&14&7&-7 \\ 14&11&8&2 \\ 7&8&9&11 \\ -7&2&11&29 \end{bmatrix} \begin{bmatrix} 4 \\ 3 \\ 1 \\ 0 \end{bmatrix}$

$= \frac{1}{35} \begin{bmatrix} 133 \\ 97 \\ 61 \\ -11 \end{bmatrix} = \begin{bmatrix} \frac{133}{35} \\ \frac{97}{35} \\ \frac{61}{35} \\ -\frac{11}{35} \end{bmatrix}$

The vector $Pb$ corresponds to the points on the least squares line of best fit $\bar{C} + \bar{D}t$ for the times $t = (-2, 1, 0, 2)$:

$\frac{61}{35} - \frac{36}{35} (-2) = \frac{133}{35}$

$\frac{61}{35} - \frac{36}{35} (-1) = \frac{97}{35}$

$\frac{61}{35} - \frac{36}{35} (0) = \frac{61}{35}$

$\frac{61}{35} - \frac{36}{35} (2) = \frac{-11}{35}$

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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