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I have been trying to solve this problem (Fraleigh - Linear Algebra (3rd Ed.)):

6.2.17. Find an orthonormal basis for $\mathbb{R^4}$ that contains an orthonormal basis for the subspace $sp([1, 0, 1, 0], [0, 1, 1, 0])$.

The first thing I thought was the usage of the standard linearly independent vectors for $\mathbb{R^4}$, but that did not work.

Ideas, suggestions (none solution) to solve this problem?

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    $\begingroup$ Use the Gram-Schmidt process. $\endgroup$
    – Ege Erdil
    May 6, 2016 at 21:45
  • $\begingroup$ You want to read about the Gram Schmidt process. You can read on Wikipedia, or search math.se. $\endgroup$
    – fonini
    May 6, 2016 at 21:45
  • $\begingroup$ Answered down below, showcasing the Gram Schmidt process to you. Make sure you ask me if you have any questions and if the answer fits you, then make sure you approve it so that the thread goes down as asnwered. $\endgroup$
    – Rebellos
    May 6, 2016 at 21:48

1 Answer 1

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Let's say that : $ x_1 = [1,0,1,0] $ and $x_2 = [0,1,1,0]$

Let $v_1 = x_1$.

$y = proj_{v_1}x_2 = \frac{x_2v_1}{v_1v_1}v_1 $

and

$v_2 = x_2 - y = x_2 - \frac{x_2v_1}{v_1v_1}v_1 $

This is also called as "The Gram-Schmidt" process. Plug in the correct vectors and you will have your component which will be orthogonal to $x_1$ and then you'll be able to form your basis.

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  • $\begingroup$ I have found the first two vectors for the orthonormal basis: $\{\frac{1}{\sqrt{2}}[1, 0, 1, 0], \frac{1}{\sqrt{6}}[-1, 2, 1, 0]\}$, but the answer for this exercise includes the previous vectors and two more. Why are they needed? Thanks $\endgroup$
    – InfZero
    May 7, 2016 at 19:38
  • $\begingroup$ I have solved it. What I did to solve it was the addition of two other linearly independent vectors $\{[0,0,0,1], [0, 1, 0, 0]\}$, and then apply the Gram-Schmidt algorithm to these ones. $\endgroup$
    – InfZero
    May 7, 2016 at 20:20

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