## 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 .

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