Wikipedia:Reference desk/Archives/Mathematics/2012 June 26

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June 26

hypercross

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)

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:

$M:=\left[{\begin{array}{cccc}{\vec {i}}&{\vec {j}}&{\vec {k}}&{\vec {l}}\\1&1&0&0\\1&0&1&0\\0&0&1&1\end{array}}\right].$

Then det(M) gives a vector perpendicular to (1,1,0,0), (1,0,1,0) and (0,0,1,1): $\det(M)={\vec {i}}-{\vec {j}}-{\vec {k}}+{\vec {l}}\equiv (1,-1,-1,1)\,.$ — Fly by Night (talk) 21:20, 26 June 2012 (UTC)

Thanks, I knew that trick for 3-space, didn't know it carried on up. —Tamfang (talk) 04:43, 27 June 2012 (UTC)

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)

Now, what's an easy way to do that in Numpy? —Tamfang (talk) 21:08, 1 July 2012 (UTC)

Further Question

Consider the matrix trick above. Let's say you have vectors ${\vec {v}}_{1},\ldots ,{\vec {v}}_{n-1}$ in an n-dimensional vector space. What do we get if we take the Hodge dual of the wedge product, i.e. what is $*({\vec {v}}_{1}\wedge \ldots \wedge {\vec {v}}_{n-1})$? Is it even a vector?! If it is, then is it orthogonal to all of the ${\vec {v}}_{i}$? — Fly by Night (talk) 16:10, 27 June 2012 (UTC)

I think $*({\vec {v}}_{1}\wedge \ldots \wedge {\vec {v}}_{n-1})$ is a vector. So, is it orthogonal to all of the ${\vec {v}}_{i}$? — Fly by Night (talk) 17:37, 27 June 2012 (UTC)

Yes. Let $w=*(v_{1}\wedge \ldots \wedge v_{n-1})$. Then $v\wedge (*w)=\langle v,w\rangle e_{1}\wedge \ldots \wedge e_{n}$ by definition, so for any v_i, $\langle v_{i},w\rangle =0$. Rckrone (talk) 18:33, 27 June 2012 (UTC)

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)

In the Hodge dual article you'll find the identity ${\vec {v}}\wedge (*{\vec {w}})=\langle {\vec {v}},{\vec {w}}\rangle {\vec {e}}_{1}\wedge \ldots \wedge {\vec {e}}_{n}$ (which is used as the definition of the Hodge dual) as well as the fact that $*{*\eta }$ equals $\eta$ up to a scalar factor. Taking ${\vec {v}}$ to be any vector from the set $\{{\vec {v}}_{1},\ldots ,{\vec {v}}_{n-1}\}$ and ${\vec {w}}$ to be the vector $*({\vec {v}}_{1}\wedge \ldots \wedge {\vec {v}}_{n-1})$, you get $\langle {\vec {v}},{\vec {w}}\rangle {\vec {e}}_{1}\wedge \ldots \wedge {\vec {e}}_{n}={\vec {v}}\wedge (*{\vec {w}})={\vec {v}}\wedge (**({\vec {v}}_{1}\wedge \ldots \wedge {\vec {v}}_{n-1}))\propto {\vec {v}}\wedge {\vec {v}}_{1}\wedge \ldots \wedge {\vec {v}}_{n-1}=0$. Therefore $\langle {\vec {v}},{\vec {w}}\rangle =0$. -- BenRG (talk) 18:37, 29 June 2012 (UTC)

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)

Those vectors are $e_{1}+e_{2}$, $e_{1}+e_{3}$, and $e_{3}+e_{4}$. Their product is $(e_{1}+e_{2})\wedge (e_{1}+e_{3})\wedge (e_{3}+e_{4})=e_{1}\wedge e_{3}\wedge e_{4}+e_{2}\wedge e_{1}\wedge e_{3}+e_{2}\wedge e_{1}\wedge e_{4}+e_{2}\wedge e_{3}\wedge e_{4}$. You can find the Hodge star of that by hitting it on the left with $e_{1},e_{2},e_{3},e_{4}$ in turn to get its dot product with each basis vector, i.e., its components. For example $e_{2}\wedge (e_{1}\wedge e_{3}\wedge e_{4}+e_{2}\wedge e_{1}\wedge e_{3}+e_{2}\wedge e_{1}\wedge e_{4}+e_{2}\wedge e_{3}\wedge e_{4})=e_{2}\wedge (e_{1}\wedge e_{3}\wedge e_{4})=-e_{1}\wedge e_{2}\wedge e_{3}\wedge e_{4}$, 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)