Definition talk:Determinant

• the summation $\sum_{\left({\lambda_1, \ldots, \lambda_n}\right)}$ goes over all the $n!$ permutations of $\left({1, 2, \ldots, n}\right\}$.

Should this really have a parenthesis on one side and a curly brace on the other? I assume it's just a typo, but I wanted to check. --Cynic 00:28, 3 December 2008 (UTC)

Good call, it was late and I was rushing. --Matt Westwood 06:16, 3 December 2008 (UTC)

Is it a good idea to define this concept (in a trivial manner, for now; not using exterior algebra and the like) for a linear transformation as well? It saves trouble in referring (one would need to invoke that $\det$ is invariant under change of basis of course). --Lord_Farin 08:05, 12 May 2012 (EDT)

I would say yes. --prime mover 08:43, 12 May 2012 (EDT)

Hmm, I just noticed some trouble appearing, in that one needs an inner product or so to define 'orthonormal basis' (or the determinant would be ill-defined). Of course, one can always pick an inner product on a finite dimensional vector space, but this is still allows for rescaling. So it will need to be defined with reference to a fixed inner product. --Lord_Farin 08:55, 12 May 2012 (EDT)