## Linear Algebra and Its Applications, Exercise 3.4.17

Exercise 3.4.17. Given the matrix $A = \begin{bmatrix} 1&1 \\ 2&3 \\ 2&1 \end{bmatrix}$

from the previous exercise and the vector $b = (1, 1, 1)$, solve $Ax = b$ by least squares using the factorization $A = QR$.

Answer: From the previous exercise we have $Q = \begin{bmatrix} \frac{1}{3}&0 \\ \frac{2}{3}&\frac{1}{\sqrt{2}} \\ \frac{2}{3}&-\frac{1}{\sqrt{2}} \end{bmatrix}$ $R = \begin{bmatrix} 3&3 \\ 0&\sqrt{2} \end{bmatrix}$

To find the least squares solution $\bar{x}$ to $Ax = b$ where $b = (1, 1, 1)$, we take advantage of the fact that $R\bar{x} = Q^Tb$.

On the right side of the equation we have $Q^Tb = \begin{bmatrix} \frac{1}{3}&\frac{2}{3}&\frac{2}{3} \\ 0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{5}{3} \\ 0 \end{bmatrix}$

so that the system $R\bar{x} = Q^Tb$ is then $\begin{bmatrix} 3&3 \\ 0&\sqrt{2} \end{bmatrix} \begin{bmatrix} \bar{x_1} \\ \bar{x_2} \end{bmatrix} = \begin{bmatrix} \frac{5}{3} \\ 0 \end{bmatrix}$

From the second equation we have $\bar{x_2} = 0$. Substituting into the first equation we have $3\bar{x_1} + 3\bar{x_2} = 3\bar{x_1} + 3 \cdot 0 = \frac{5}{3}$ so that $3\bar{x_1} = \frac{5}{3}$ or $\bar{x_1} = \frac{5}{9}$.

The least squares solution to $Ax = b$ with $b = (1, 1, 1)$ is therefore $\bar{x} = (\frac{5}{9}, 0)$.

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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