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

= 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&gt;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 vi, $$\langle v_i,w \rangle = 0$$. 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)