# find orthonormal basis of span of 2 vectors in r4?

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? – straykiwi Oct 25 '12 at 1:25
$w$ is 2-dimensional, so you do not need 4 vectors. Have you read this article? en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process – wj32 Oct 25 '12 at 1:26
Yeah I know how to use gram schmidt. w is 2 dimensional, but the vectors are 4 dimensional, does that matter? – straykiwi Oct 25 '12 at 1:29
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?) – straykiwi Oct 25 '12 at 1:31
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. – wj32 Oct 25 '12 at 1:41

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