Linear Algebra and Its Applications, Exercise 3.4.14

Exercise 3.4.14. Given the vectors

$a = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \quad b = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \quad c = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$

find the corresponding orthonormal vectors $q_1$ , $q_2$ , and $q_3$ .

Answer: We first choose $a' = a$ . We then have

$b' = b - \frac{(a')^Tb}{(a')^Ta'}a' = b - \frac{1 \cdot 1 + 1 \cdot 0 + 0 \cdot 1}{1^2 + 1^2 + 0^2}a' = b - \frac{1}{2}a'$

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

We then have

$c' = c - \frac{(a')^Tc}{(a')^Ta'}a' - \frac{(b')^Tc}{(b')^Tb'}b' = c - \frac{1 \cdot 0 + 1 \cdot 1 + 0 \cdot 1}{1^2 + 1^2 + 0^2}a' - \frac{\frac{1}{2} \cdot 0 + (-\frac{1}{2}) \cdot 1 + 1 \cdot 1}{(\frac{1}{2})^2 + (-\frac{1}{2})^2+ 1^2}b'$

$= c' - \frac{1}{2}a' - \frac{\frac{1}{2}}{\frac{3}{2}}b' = c - \frac{1}{2}a' - \frac{1}{3}b'$

$= \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} - \frac{1}{2} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} - \frac{1}{3} \begin{bmatrix} \frac{1}{2} \\ -\frac{1}{2} \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} - \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ 0 \end{bmatrix} - \begin{bmatrix} \frac{1}{6} \\ -\frac{1}{6} \\ \frac{1}{3} \end{bmatrix}$

$= \begin{bmatrix} -\frac{2}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{bmatrix}$

Now that we have calculated the orthogonal vectors $a'$ , $b'$ , and $c'$ , we can normalize them to create the orthonormal vectors $q_1$ , $q_2$ , and $q_3$ . We have

$\|a'\| = \sqrt{1^2+1^2 + 0^2} = \sqrt{2}$

$\|b'\| = \sqrt{(\frac{1}{2})^2 + (-\frac{1}{2})^2 + 1^2} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}}$

$\|c'\| = \sqrt{(-\frac{2}{3})^2 + (\frac{2}{3})^2 + (\frac{2}{3})^2} = \sqrt{\frac{12}{9}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$

so that

$q_1 = a' / \|a'\| = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{bmatrix}$

$q_2 = b' / \|b'\| = \frac{\sqrt{2}}{\sqrt{3}} \begin{bmatrix} \frac{1}{2} \\ -\frac{1}{2} \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}\sqrt{3}} \\ -\frac{1}{\sqrt{2}\sqrt{3}} \\ \frac{2}{\sqrt{2}\sqrt{3}} \end{bmatrix}$

$q_3 = c' / \|c'\| = \frac{\sqrt{3}}{2} \begin{bmatrix} -\frac{2}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{bmatrix} = \begin{bmatrix} -\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{bmatrix}$

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

Linear Algebra and Its Applications, Exercise 3.4.14

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