I'm trying to find the basis of the subspace of R4 consisting of all vectors perpendicular to v where v is a 4 dimensional vector.

Question

I know that I have to use a variation of the dot product but I am not sure how to do that since I only have one vector?

Answer 1

I think I have figured it out. When you take v and one of it's orthogonal vectors the dot product will give you zero, therefore let the orthogonal vector = (a,b,c,d) and let v = (e,f,g,h), then ae+bf+cg+dh = 0 then if you isolate a you can get a = ((-f)b+(-g)c+(-h)d)/e, you can then replace a in your orthogonal vector giving

((-f)b+(-g)c+(-h)d)/e , b , c , d)/ which can the be decomposed into

b ((-f)/e , 1 , 0 , 0) + c ((-g)/e , 0 , 1 , 0) + d ((-h)/e , 0 , 0 , 1)

P.S. Sorry for the lack of notation and structure I'm not quite used to proofs yet.

Answer 2

Just take any three independent vectors $\vec v_1,\vec v_2,\vec v_3$, also independent of $\vec v$. You could use the standard basis vectors ([1,0,0,0] etc.). Then project them away from $\vec v$.

$$\vec b_1 = \vec v_1 - \left(\frac{\vec v_1\cdot\vec v}{\vec v\cdot\vec v}\right)\vec v$$ $$\vec b_2 = \vec v_2 - \left(\frac{\vec v_2\cdot\vec v}{\vec v\cdot\vec v}\right)\vec v$$ $$\vec b_3 = \vec v_3 - \left(\frac{\vec v_3\cdot\vec v}{\vec v\cdot\vec v}\right)\vec v$$

You could use the Gram-Schmidt process instead, if you want an orthonormal basis.

Answer 3

Let $v=<v_1,v_2,v_3,v_4>$

You want to find three linearly independent vectors, which are perpendicular to $v$

For example you may consider

$$a=<v_2, -v_1,0,0 >\\ b=<0,v_3, -v_2,0>\\ c=<0,0,v_4,-v_3> $$

in case of $v_i \ne 0$ for $i=1,2,3,4$