Wikipedia:Reference desk/Archives/Mathematics/2012 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:

::

:Then det(M) gives a vector perpendicular to (1,1,0,0), (1,0,1,0) and (0,0,1,1): $\backslash det(M)\; =\; \backslash vec\{i\}\; -\; \backslash vec\{j\}\; -\; \backslash vec\{k\}\; +\; \backslash vec\{l\}\; \backslash equiv\; (1,-1,-1,1)\; \backslash ,\; .$ — 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 {{nowrap|1=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)

Consider the matrix trick above. Let's say you have vectors $\backslash vec\{v\}\_1,\backslash ldots,\backslash vec\{v\}\_\{n-1\}$ in an n-dimensional vector space. What do we get if we take the Hodge dual of the wedge product, i.e. what is $*(\backslash vec\{v\}\_1\; \backslash wedge\; \backslash ldots\; \backslash wedge\; \backslash vec\{v\}\_\{n-1\})$? Is it even a vector?! If it is, then is it orthogonal to all of the $\backslash vec\{v\}\_i$? — Fly by Night (talk) 16:10, 27 June 2012 (UTC)

:I think $*(\backslash vec\{v\}\_1\; \backslash wedge\; \backslash ldots\; \backslash wedge\; \backslash vec\{v\}\_\{n-1\})$ is a vector. So, is it orthogonal to all of the $\backslash vec\{v\}\_i$? — Fly by Night (talk) 17:37, 27 June 2012 (UTC)

::Yes. Let $w\; =\; *(v\_1\; \backslash wedge\; \backslash ldots\; \backslash wedge\; v\_\{n-1\})$. Then $v\; \backslash wedge\; (*w)\; =\; \backslash langle\; v,w\; \backslash rangle\; e\_1\backslash wedge\backslash ldots\backslash wedge\; e\_n$ by definition, so for any v_i, $\backslash langle\; v\_i,w\; \backslash rangle\; =\; 0$. 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 $\backslash vec\; v\; \backslash wedge\; (*\backslash vec\; w)\; =\; \backslash langle\; \backslash vec\; v,\backslash vec\; w\; \backslash rangle\; \backslash vec\; e\_1\backslash wedge\backslash ldots\backslash wedge\; \backslash vec\; e\_n$ (which is used as the definition of the Hodge dual) as well as the fact that $*\{*\backslash eta\}$ equals $\backslash eta$ up to a scalar factor. Taking $\backslash vec\; v$ to be any vector from the set $\backslash \{\; \backslash vec\; v\_1,\; \backslash ldots,\; \backslash vec\; v\_\{n-1\}\; \backslash \}$ and $\backslash vec\; w$ to be the vector $*(\backslash vec\{v\}\_1\; \backslash wedge\; \backslash ldots\; \backslash wedge\; \backslash vec\{v\}\_\{n-1\})$, you get $\backslash langle\; \backslash vec\; v,\backslash vec\; w\; \backslash rangle\; \backslash vec\; e\_1\backslash wedge\backslash ldots\backslash wedge\; \backslash vec\; e\_n\; =\; \backslash vec\; v\; \backslash wedge\; (*\backslash vec\; w)\; =\; \backslash vec\; v\; \backslash wedge\; (**(\backslash vec\; v\_1\; \backslash wedge\backslash ldots\backslash wedge\; \backslash vec\; v\_\{n-1\}))\; \backslash propto\; \backslash vec\; v\; \backslash wedge\; \backslash vec\; v\_1\; \backslash wedge\backslash ldots\backslash wedge\; \backslash vec\; v\_\{n-1\}\; =\; 0$. Therefore $\backslash langle\; \backslash vec\; v,\backslash vec\; w\; \backslash rangle\; =\; 0$. -- 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 $(e\_1+e\_2)\backslash wedge(e\_1+e\_3)\backslash wedge(e\_3+e\_4)\; =\; e\_1\backslash wedge\; e\_3\backslash wedge\; e\_4\; +\; e\_2\backslash wedge\; e\_1\backslash wedge\; e\_3\; +\; e\_2\backslash wedge\; e\_1\backslash wedge\; e\_4\; +\; e\_2\backslash wedge\; e\_3\backslash wedge\; e\_4$. 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 $e\_2\; \backslash wedge\; (e\_1\backslash wedge\; e\_3\backslash wedge\; e\_4\; +\; e\_2\backslash wedge\; e\_1\backslash wedge\; e\_3\; +\; e\_2\backslash wedge\; e\_1\backslash wedge\; e\_4\; +\; e\_2\backslash wedge\; e\_3\backslash wedge\; e\_4)\; =\; e\_2\; \backslash wedge\; (e\_1\backslash wedge\; e\_3\backslash wedge\; e\_4)\; =\; -e\_1\backslash wedge\; e\_2\backslash wedge\; e\_3\backslash wedge\; e\_4$, 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)