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I am going back through old problems, and I still can't figure this one out.

Find an orthonormal basis for the orthogonal complement of the linear span of the vectors $(1,0,1,0)$ and $(1,0,-1,0)$ in $\mathbb{R}^4$.

So I understand what a orthonormal basis is, but how do you take the orthogonal complement of the linear span? Is it the linear span of the orthogonal complements of the components? i.e. $(1,0,1,0)^\perp$ and $(1,0,-1,0)^\perp$

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  • $\begingroup$ It's the vector space where each vector is orthogonal to the vector of the linear span of the vector given in the question, but you can think as the vector that are both orthogonal to $(1,0,1,0)$ and $(1,0,-1,0)$. $\endgroup$ – DiegoMath Oct 26 '14 at 16:21
  • $\begingroup$ So I'm correct? I just need to find a vector for $(1,0,1,0)^\perp$ and $(1,0,-1,0)^\perp$? I'm not quite sure what you mean by "the vector of the linear span of the vector". $\endgroup$ – Blackeyes Oct 26 '14 at 16:26
  • $\begingroup$ @DiegoMath - Your edit makes sense. Just to confirm, I need to find vectors are that perp to both of the vectors in the linear span. Would $(0,1,0,1),(0,-1,0,-1)$ satisfy the question? I just want to make sure that I fully grasp the solution. $\endgroup$ – Blackeyes Oct 26 '14 at 16:38
  • $\begingroup$ There are more simpler vector to take instead... Think a little, you want vectors such that $\langle(x,y,z,w),(1,0,\pm1,0)\rangle=0$, so, what are the conditions over $x,y,z,w$? $\endgroup$ – DiegoMath Oct 26 '14 at 16:56
  • $\begingroup$ So I could also take any two vectors such that $x=-z$ and $x=z$? $\endgroup$ – Blackeyes Oct 26 '14 at 16:59

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