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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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4  
You could take any basis containing $v_1$ and apply the Gram–Schmidt process. –  Trevor Wilson Sep 8 '12 at 2:16
    
...and to find a basis in the first place, you can just pick vectors randomly and then test to see if you have a basis. "Most" choices of four vectors will form a basis for $\mathbb{R}^4$. –  Trevor Wilson Sep 8 '12 at 2:44
    
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. –  Tpofofn Sep 8 '12 at 4:52

4 Answers 4

up vote 1 down vote accepted

May be $$v_3=[b,a,b,a]\quad v_4=[-b,a,b,-a]$$

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Yesss, you have build it testing or using any process? –  user901366 Sep 8 '12 at 2:32
    
By testing.. :) –  Tapu Sep 8 '12 at 2:33
    
Ok, thank's! :)) –  user901366 Sep 8 '12 at 2:37

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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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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To use the Gram-Schmidt process I need four vectors to build four new orthogonal vectors, or not? –  user901366 Sep 8 '12 at 2:25
    
@user901366 Yes, that's why I wrote that you first find an (ordinary) basis containing $v_1$. –  Makoto Kato Sep 8 '12 at 2:30

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.

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