# How to efficiently compute the generalized cross product?

It's possible to extend the well known cross product between two vectors in $\mathbb{R}^3$ to $n-1$ vectors in $\mathbb{R}^n$.

Let $\vec{v_1}, \vec{v_2}, \dots, \vec{v}_{n-1} \in \mathbb{R}^n$ and $\vec{e}_1 = (1, 0, 0, \dots, 0)^T, \vec{e}_2 = (0, 1, 0, \dots, 0)^T, \dots, \vec{e}_n = (0, 0, \dots, 0, 1)^T$ be the unit vectors of the standard basis in $\mathbb{R}^n$. Then we can define the a "cross product" in $\mathbb{R}^n$ by the following determinant:

$\vec{v}_1 \times \vec{v}_2 \times \dots \times \vec{v}_{n-1} = det\ \begin{pmatrix} v_{1,1} & v_{2,1} & \cdots & v_{n-1,1} & \vec{e}_1\cr v_{1,2} & v_{2,2} & \cdots & v_{n-1,2} & \vec{e}_2\cr \vdots & \vdots & \ddots & \vdots & \vdots\cr v_{1,n} & v_{2,n} & \cdots & v_{n-1,n} & \vec{e}_n \end{pmatrix}$

In theory it's easy to compute the determinant by cofactor expansion along the last column, but i'm wondering how one would do this in practice. Computing the $n$ minors on their own seems like a lot of overhead to me and i guess it's not very stable.

Edit: As turned out later, the following system of equations is wrong! Furthermore the approach is not used to compute the cross product but an orthogonal vector with arbitrary length.

The only approach i've found on this topic is in the code of qhull. But unfortunately it's quite obfuscated and there aren't any helpful comments. I figured out that there the problem is adressed by solving the following system of equations:

$A * \vec{x} = \vec{b}$
with $A = \begin{pmatrix} v_{1,1} & v_{1,2} & \dots & v_{1,n}\cr v_{2,1} & v_{2,2} & \dots & v_{2,n}\cr\vdots & \vdots & \ddots & \vdots\cr v_{n-1,1} & v_{n-1,2} & \dots & v_{n-1,n}\cr 0 &0& \cdots & 1 \\end{pmatrix}$, $\vec{b} = \begin{pmatrix}0\cr0\cr\vdots\cr 0 \cr sgn(det(A)) \end{pmatrix}$ and $sgn(x) := \begin{cases} +1 & x \geq 0 \cr -1 & x < 0 \end{cases}$

It's clear that the first $n-1$ rows are enforcing $\vec{x}$ to be orthogonal to $\vec{v}_1, \dots, \vec{v}_{n-1}$ but i can't get the origin or meaning of the last row. Furthermore i guess this approach is problematic if $det(A) = 0$ (e.g. if vectors are parallel to a standard unit vector).

So my questions are:

1. How is the last row deduced and what is its meaning?
2. Are there other approaches that are faster and/or more stable?
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The last row seems to be a step toward enforcing the $n$-dimensional analog of the familiar "right-hand rule" for 3 dimensions. – Andreas Blass Oct 17 '12 at 21:20
In many places in qhull determinants are not calculated just their signs. That speeds up hyperplane arrangement related algorithms. Can you tell me please where the particular code you need decoding in qhull is? – Rabee Tourky Oct 17 '12 at 22:25
The last row is indeed weird because it enforces $x_n = \pm 1$, which cannot be possible if, say, you pick $\vec v_1 = \vec e_2$, ... $\vec v_{n-1} = \vec e_n$, in which case the only nonzero component of $x$ is the first one. There has to be a way of enforcing the right-hand rule, i.e. $\det (\vec v_1,...,\vec v_{n-1}, \vec v_1\times ...\times \vec v_{n-1}) \ge 0$, which is indeed the missing constraint. However, writing this condition as I just wrote it would amount to computing the components of the cross-product by determinants ! – Francois Monard Oct 19 '12 at 1:08

Regarding your question 2: the approach I'd take is computing the first determinant from a RQ factorization of the leading $(n-1)\times(n-1)$ matrix, and then each other by replacing in turn each row of $R$ with $(last row)Q$ and re-orthogonalizing manually with $O(n)$ Givens transformations on the left. In this way you pay $O(n^3)$ for the first determinant and then $O(n^2)$ (instead of the usual $O(n^3)$) for each subsequent one.
Actually, I just realized that the second part can be done in $O(n)$ per determinant. Maybe the best way to explain it is like this: first reduce to upper triangular form the last $n-1$ rows. Then do Givens transformations involving the first row to kill each of its elements in turn, starting from the leftmost one. After each of these transformations, you'll realize that there is a $(n-1)\times(n-1)$ which is upper triangular, and so you can compute its determinant easily. – Federico Poloni Oct 19 '12 at 22:22