Question

Apply the Gram-Schmidt orthogonalisation process to find an orthonormal basis for the subspace of R4 generated by the vectors
{(-1,1,0,1),(1,0 ,-1,0),(1,0,2,-1)}

Answer 1

Answer on Question #66803–Math–Linear Algebra

Question

Apply the Gram-Schmidt orthogonalization process to find an orthonormal basis for the subspace of $\mathbb{R}^4$ generated by the vectors $\{(-1,1,0,1),(1,0,-1,0),(1,0,2,-1)\}$.

Solution

Let $u_{1} = (-1,1,0,1)$, $u_{2} = (1,0, -1,0)$, $u_{3} = (1,0,2, -1)$.

By the Gram-Schmidt orthogonalization process $\{\{1\}, \{2, \text{pp } 544, 558\}\}$ consider $v_{1} = (-1,1,0,1)$ and $v_{2} = u_{2} - \frac{u_{2} \cdot v_{1}}{v_{1} \cdot v_{1}} v_{1}$;

Note that

$$\begin{array}{l} u_{2} \cdot v_{1} = 1 \cdot (-1) + 0 \cdot 1 + (-1) \cdot 0 + 0 \cdot 1 = -1; \\ v_{1} \cdot v_{1} = \|v_{1}\| = (-1) \cdot (-1) + 1 \cdot 1 + 0 \cdot 0 + 1 \cdot 1 = 3. \\ v_{2} = u_{2} - \frac{u_{2} \cdot v_{1}}{v_{1} \cdot v_{1}} v_{1} = (1, 0, -1, 0) - \frac{(-1)}{3} (-1, 1, 0, 1) = (1, 0, -1, 0) + \frac{1}{3} (-1, 1, 0, 1) = \frac{1}{3} (2, 1, -3, 1), \end{array}$$

we can to consider $v_{2} = (2,1, -3,1)$.

Next calculate $v_{3} = u_{3} - \frac{u_{3} \cdot v_{1}}{v_{1} \cdot v_{1}} v_{1} - \frac{u_{3} \cdot v_{2}}{v_{2} \cdot v_{2}} v_{2}$;

$$\begin{array}{l} u_{3} \cdot v_{1} = 1 \cdot (-1) + 0 \cdot 1 + 2 \cdot 0 + (-1) \cdot 1 = -2; \\ u_{3} \cdot v_{2} = 1 \cdot 2 + 0 \cdot 1 + 2 \cdot (-3) + (-1) \cdot 1 = -5; \\ v_{2} \cdot v_{2} = 2 \cdot 2 + 1 \cdot 1 + (-3) \cdot (-3) + 1 \cdot 1 = 15. \end{array}$$

We get $v_{3} = u_{3} - \frac{u_{3} \cdot v_{1}}{v_{1} \cdot v_{1}} v_{1} - \frac{u_{3} \cdot v_{2}}{v_{2} \cdot v_{2}} v_{2} = (1, 0, 2, -1) - \frac{(-2)}{3} (-1, 1, 0, 1) - \frac{(-5)}{15} (2, 1, -3, 1) =$

$$= (1, 0, 2, -1) + \frac{2}{3} (-1, 1, 0, 1) + \frac{1}{3} (2, 1, -3, 1) = (1, 1, 1, 0), \text{ hence, } v_{3} = (1, 1, 1, 0).$$

So we get an orthogonal basis $v_{1} = (-1,1,0,1)$, $v_{2} = (2,1, -3,1)$, $v_{3} = (1,1,1,0)$.

Normalizing the vectors:

$$\begin{array}{l} \|v_{1}\| = \sqrt{v_{1} \cdot v_{1}} = \sqrt{3}; \quad e_{1} = \frac{v_{1}}{\|v_{1}\|} = \frac{1}{\sqrt{3}} (-1, 1, 0, 1). \\ \|v_{2}\| = \sqrt{v_{2} \cdot v_{2}} = \sqrt{15}; \quad e_{2} = \frac{v_{2}}{\|v_{2}\|} = \frac{1}{\sqrt{15}} (2, 1, -3, 1). \\ \|v_{3}\| = \sqrt{v_{3} \cdot v_{3}} = \sqrt{1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 + 0 \cdot 0} = \sqrt{3}; \quad e_{3} = \frac{v_{3}}{\|v_{3}\|} = \frac{1}{\sqrt{3}} (1, 1, 1, 0). \end{array}$$

Answer: $\frac{1}{\sqrt{3}} (-1, 1, 0, 1); \frac{1}{\sqrt{15}} (2, 1, -3, 1) \cdot \frac{1}{\sqrt{3}} (1, 1, 1, 0)$.

References:

1. Hazewinkel, Michel (2001) Orthogonalization. Encyclopedia of Mathematics, Springer ISBN 978-1-55608-010-4

https://www.encyclopediaofmath.org/index.php/Orthogonalization

2. E W. Cheney, D. Ronald (2009) Linear Algebra: Theory and Applications, Sudbury Massachusetts, 740 p. ISBN 978-0-7637-5020-6.

https://books.google.com/books?id=Gg3Uj1GkHK8C&pg=PA544&redir_esc=y#v=onepage&q&f=false

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