Linear Algebra and Its Applications, Exercise 3.4.14

Posted on December 27, 2017 by hecker

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.

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