Linear Algebra and Its Applications, Exercise 3.3.5

Exercise 3.3.5. Given the system

$Ax = \begin{bmatrix} 1&-1 \\ 1&0 \\ 1&1 \end{bmatrix} \begin{bmatrix} C \\ D \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \\ 9 \end{bmatrix} = b$

with no solution, provide a graph of a straight line that minimizes

$(C - D -4)^2 + (C - 5)^2 + (C + D - 9)^2$

and solve for the equation of the line. What is the result of projecting the vector $b$ onto the column space of $A$ ?

Answer: We have

$(C - D -4)^2 + (C - 5)^2 + (C + D - 9)^2 = \| Ax - b \|^2$

In other words, this problem is essentially that of finding $\bar{x} = \begin{bmatrix} \bar{C} \\ \bar{D} \end{bmatrix}$ , the least squares solution to $Ax = b$ that minimizes the error vector $E^2 = \| Ax - b \|^2$ .

From 3L on page 156 we have $\bar{x} = (A^TA)^{-1}A^Tb$ . In this case we have

$A^TA = \begin{bmatrix} 1&1&1 \\ -1&0&1 \end{bmatrix} \begin{bmatrix} 1&-1 \\ 1&0 \\ 1&1 \end{bmatrix} = \begin{bmatrix} 3&0 \\ 0&2 \end{bmatrix}$

and thus

$(A^TA)^{-1} = \frac{1}{6} \begin{bmatrix} 2&0 \\ 0&3 \end{bmatrix}$

so that

$\bar{x} = (A^TA)^{-1}A^Tb = \frac{1}{6} \begin{bmatrix} 2&0 \\ 0&3 \end{bmatrix} \begin{bmatrix} 1&-1 \\ 1&0 \\ 1&1 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \\ 9 \end{bmatrix}$

$= \frac{1}{6} \begin{bmatrix} 2&0 \\ 0&3 \end{bmatrix} \begin{bmatrix} 18 \\ 5 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 36 \\ 15 \end{bmatrix} = \begin{bmatrix} 6 \\ \frac{5}{2} \end{bmatrix}$

Thus the line that minimizes

$E^2 = \|Ax -b\|^2$

$= (C - D - 4)^2 + (C - 5)^2 + (C + D - 9)^2$

is $\bar{C} + \bar{D}t$ with slope $\bar{D} = \frac{5}{2}$ and intercept $\bar{C} = 6$ .

Also from 3L on page 156 the projection of $b$ onto the column space of $A$ is

$p = A\bar{x} = \begin{bmatrix} 1&-1 \\ 1&0 \\ 1&1 \end{bmatrix} \begin{bmatrix} 6 \\ \frac{5}{2} \end{bmatrix} = \begin{bmatrix} \frac{7}{2} \\ 6 \\ \frac{17}{2} \end{bmatrix}$

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, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

Linear Algebra and Its Applications, Exercise 3.3.5

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Linear Algebra and Its Applications, Exercise 3.3.5

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