How to build a orthogonal basis from a vector? 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...
 A: 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:
$$i(x_1+ix_2+jx_3+kx_4)=ix_1-x_2+kx_3-jx_4\mapsto (-x_2,x_1,-x_4,x_3)$$
$$j(x_1+ix_2+jx_3+kx_4)=jx_1-kx_2-x_3+ix_4\mapsto (-x_3,x_4,x_1,-x_2)$$
$$k(x_1+ix_2+jx_3+kx_4)=kx_1+jx_2-ix_3-x_4\mapsto (-x_4,-x_3,x_2,x_1)$$
A: It's easy to find a basis $\{v_1, w_2, w_3, w_4\}$ of $\mathbb{R}^4$.
Then use Gram–Schmidt process.
A: May be $$v_3=[b,a,b,a]\quad v_4=[-b,a,b,-a]$$
A: 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.  
A: 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
