Re: what does 'determinant Δ' mean


young wrote:

please give an explanation as detailed as possible
any reply will be appreciated

If u and v are vectors in an n-dimesional vector space, define the
wedge product by the relations v ⋀ v = 0; which entails u ⋀ v = - v
⋀ u. These now define vectors in an n(n-1)/2
dimensional vector space, and we can keep on going, using the same
rules. Then an n-fold wedge product v1 ⋀ v2 ⋀ ... ⋀ vn is vector
in a 1-dimensional vector space, so it is basically a scalar. The
scalar quantity is the determinant of the square matrix whose rows are
the vectors v1, ..., vn. From the defining properties of a wedge
product it easily follows that the determininat is zero iff v1 ... vn
are linearly dependent, and that the determinant is antisymmetric with
respect to swapping roles. Transposing, this is also true of columns.

.

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