Timeline for How to efficiently compute the generalized cross product?
Current License: CC BY-SA 3.0
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| Oct 19, 2012 at 22:22 | comment | added | Federico Poloni | 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. | |
| Oct 19, 2012 at 17:01 | vote | accept | aegirxx | ||
| Oct 18, 2012 at 9:29 | history | answered | Federico Poloni | CC BY-SA 3.0 |