June 26
Given n-1 vectors (guaranteed independent but not necessarily orthogonal) in n-space, how do I get the vector orthogonal to all of them when n>3? —Tamfang (talk) 20:53, 26 June 2012 (UTC)[reply]
- Don't you use the determinant trick? Say your vectors are (1,1,0,0), (1,0,1,0) and (0,0,1,1). You'd abuse notation and put them into a 4-by-4 matrix:
![{\displaystyle M:=\left[{\begin{array}{cccc}{\vec {i))&{\vec {j))&{\vec {k))&{\vec {l))\\1&1&0&0\\1&0&1&0\\0&0&1&1\end{array))\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1658813f97755aa4f8875220acf030536a8f83ea)
- Then det(M) gives a vector perpendicular to (1,1,0,0), (1,0,1,0) and (0,0,1,1):
— Fly by Night (talk) 21:20, 26 June 2012 (UTC)[reply]
- Thanks, I knew that trick for 3-space, didn't know it carried on up. —Tamfang (talk) 04:43, 27 June 2012 (UTC)[reply]
- You're welcome. I think it works in all dimensions. There will be some fancy way of viewing it it terms of tensors. In three dimensions, the cross product of u and v is the Hodge dual of the exterior product u ∧ v. — Fly by Night (talk) 16:10, 27 June 2012 (UTC)[reply]
Now, what's an easy way to do that in Numpy? —Tamfang (talk) 21:08, 1 July 2012 (UTC)[reply]
Further Question
Consider the matrix trick above. Let's say you have vectors
in an n-dimensional vector space. What do we get if we take the Hodge dual of the wedge product, i.e. what is
? Is it even a vector?! If it is, then is it orthogonal to all of the
? — Fly by Night (talk) 16:10, 27 June 2012 (UTC)[reply]
- I think
is a vector. So, is it orthogonal to all of the
? — Fly by Night (talk) 17:37, 27 June 2012 (UTC)[reply]
- Yes. Let
. Then
by definition, so for any vi,
. Rckrone (talk) 18:33, 27 June 2012 (UTC)[reply]
- Could you please explain, in detail, exactly how that follows from the definition? I am confused by your used of the double Hodge dual, i.e. you define w to be a Hodge dual, but then use the Hodge dual of w. It would be helpful if you could explain exactly what everything is, e.g. if it's a one-vector, or a bi-covector, etc. — Fly by Night (talk) 22:35, 27 June 2012 (UTC)[reply]
- In the Hodge dual article you'll find the identity
(which is used as the definition of the Hodge dual) as well as the fact that
equals
up to a scalar factor. Taking
to be any vector from the set
and
to be the vector
, you get
. Therefore
. -- BenRG (talk) 18:37, 29 June 2012 (UTC)[reply]
- That's great, thanks! It would be helpful if you could explain exactly what everything is. Let's take the example above: say your vectors are (1,1,0,0), (1,0,1,0) and (0,0,1,1). Can you follow your above argument through in this case, and explicitly say what everything is, please? — Fly by Night (talk) 19:13, 29 June 2012 (UTC)[reply]
- Those vectors are
,
, and
. Their product is
. You can find the Hodge star of that by hitting it on the left with
in turn to get its dot product with each basis vector, i.e., its components. For example
, so the second component is −1. Note that this is basically the same as computing the determinant of the 4x4 matrix. I can add more if that's not clear enough. -- BenRG (talk) 01:10, 30 June 2012 (UTC)[reply]