Reputation: 4733
As per the title, is the best way to calculate the n-dimensional cross product just using the determinant definition and using the LU Decomposition method of doing as such or could you guys suggest a better one?
Thanks
Edit: for clarity I mean http://en.wikipedia.org/wiki/Cross_product and not the Cartesian Product
Edit: It also seems that using the Leibniz Formula might help - though I don't know how that compares to LU Decomp. at the moment.
Upvotes: 3
Views: 2166
Reputation: 60988
From your comment, it seems like you are looking for an operation which takes n −1 vectors as input and computes a single vector as its result, which will be orthogonal to all the input vectors and perhaps have a well-defined length as well.
You can characterize the 3-dimensional cross product v =a ×b using the identity v ∙w =det(a,b,w). In other words, taking the cross product of the input vectors and then computing the dot product with any other vector w is the same as plugging the input vectors and that other vector into a matrix and computing its determinant.
This definition can be generalized to arbitrary dimensions. Due to the way a determinant can be computed using Laplace expansion along the last column, the resulting coordinates of that cross product will be the values of all (n −1)×(n −1) sub-determinants you can form from the input vectors, with alternating signs. So yes, Leibniz might be useful in theory, although it is hardly suitable for real-world computations. In practice, you'll soon have to figure out ways to avoid repeating computationswhile computing these n determinants. But wait for the last section of this answer…
Most applications however can do with a weaker requirement. They don't care about the length of the resulting vector, but only about its direction. In that case, what you are asking for is the kernel of the (n −1)×n matrix you can form by taking the input vectors as rows. Any element of that kernel will be orthogonal to the input vectors, and since computing kernels is a common task, you can build on a lot of existing implementations, e.g. Lapack. Details might depend on the language you are using.
You can even combine the two approaches above: compute one element of the kernel, and for a non-zero entry of that vector, also compute the corresponding (n −1)×(n −1) determinant which would give you that single coordinate using the first approach. You can then simply scale the vector so that the selected coordinate reaches the computed value, and all the other coordinates will match that one.
Upvotes: 4