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Let $w = \mbox{span} \{(1,1,-2,0)^T,(1,0,2,1)^T\}$

Find an orthonormal basis for $w$ using dot product

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    I'm a bit confused as to what I actually need to do. Do I use gram-schmidt? I only have 2 vectors, don't I need 4 to solve this? if so, whats an easy way to choose 2 more vectors that I know are lineraly independant?2012-10-25
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    $w$ is 2-dimensional, so you do not need 4 vectors. Have you read this article? http://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process2012-10-25
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    Yeah I know how to use gram schmidt. w is 2 dimensional, but the vectors are 4 dimensional, does that matter?2012-10-25
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    so v1 = (1,1,-2,0)trans and v2 = (1/2)*(3,1,2,2) Is what I got, is that correct? (next step would be to normalise v1 and v2?)2012-10-25
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    You are working in $\mathbb{R}^4$, but you are trying to find an orthonormal basis for $w$, which is a 2-dimensional subspace of $\mathbb{R}^4$. I referred you to that article not because I didn't think you knew how to follow the computational steps, but because I didn't think you understood what the Gram-Schmidt process actually *does*. And yes, that computation looks correct.2012-10-25
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    Thanks, so my final answer is (1/sqrt(6))*(1,1,-2,0)t and v2 = 1/(3*sqrt(2))* (3,1,2,2)t where t = transpose is this the final correct answer? Thanks for your help!2012-10-25
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    @straykiwi to see if it is correct, check the dot-product between your final gram-schmidt'd vectors. Do you get zero? If so then they are indeed orthogonal and then you just need to check for length one.2012-10-25
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    And when you are sure you understand, you can post a solution, and then later you can accept it.2012-10-25

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Just use Gram-Schmidt on $w$ which gives $$v_1 = \frac{1}{\sqrt{6}}(1,1,-2,0)$$ and $$v_2 = \frac{1}{3\sqrt{2}}(3,1,2,2).$$