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Anybody know how I can build a orthogonal base using only a vector? I have a vector in the form $v_1 = [a, b, -a, -b]$, where $a$ and $b$ are real numbers. I did try build in the "adhoc way" but, nothing, I only got two orthogonal vectors:

$v_1 = [a, b, -a, -b], \text{ } v_2 = [a, -b, a, -b]$

I need more two vectors to complete the orthogonal basis $\{v_1, v_2, v_3, v_4\}$. Anybody can help me?

Thanks...

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    How do you want to use the basis? Since you are working in $\mathbb R^4$ andthe basis spans the same space, the standard basis should do.2012-09-08

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It's a rather old question, but from all the answers I was missing the observation that the vectors orthogonal to $v_1$ lie in its null space. That is, for any orthogonal vector $v$ it holds that the dot product $v_1\cdot v=0$.

A simple way to find a set of vectors that span the null space by hand is to use Gauss-Jordan elimination. You can find a very nice illustration here: https://en.wikipedia.org/wiki/Kernel_(linear_algebra)#Illustration

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No need for choosing a basis a priori - you just need one starting vector. There is a straight-forward algorithm that achieves exactly what you asked for:

Pick a vector. WLOG, you chose $(x_1,x_2,x_3,x_4)$. Now write it as a quaternion: $x_1+ix_2+jx_3+kx_4$ Then, since multiplication by $i,j,k$ rotates this vector $90^0$ across the various axes of our 4D space, the following three vectors make your initial choice of vector into an orthonormal basis: $i(x_1+ix_2+jx_3+kx_4)=ix_1-x_2+kx_3-jx_4\mapsto (-x_2,x_1,-x_4,x_3)$ $j(x_1+ix_2+jx_3+kx_4)=jx_1-kx_2-x_3+ix_4\mapsto (-x_3,x_4,x_1,-x_2)$ $k(x_1+ix_2+jx_3+kx_4)=kx_1+jx_2-ix_3-x_4\mapsto (-x_4,-x_3,x_2,x_1)$

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    The operation on complex, quaternions, and octonions is linear in the coefficients of the vector. The Gram or Cholesky procedures mentioned elsewhere are not. Fascinating difference!2016-02-22
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May be $v_3=[b,a,b,a]\quad v_4=[-b,a,b,-a]$

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    Ok, thank's! :))2012-09-08
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It's easy to find a basis $\{v_1, w_2, w_3, w_4\}$ of $\mathbb{R}^4$. Then use Gram–Schmidt process.

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    @user901366 Yes, that's why I wrote that you first find an (ordinary) basis containing $v_1$.2012-09-08
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Assume your vector has the form $u=(u_1,u_2,u_3,u_4)$, then you have,

$ v.u = a u_1 + b u_2 - a u_3 - b u_4 = 0 \,$

Now, since number of variables is bigger than the number of equations, you will have an infinite number of solutions. Assume $ u_2, u_3, u_4 $ are the free variables which means you can choose them freely from the real numbers and then substitute in the above equation to find $u_1$. For example, choosing $ u_2=-a,u_3=0,u_4=0 $ and substituting in the above equation yields $ u_1=b $. So you get the vector $ v_2 = ( b,-a,0,0 ) $. Now, you can find the other two vectors.