First off, I'm not a mathematician and I'm getting a bit over my head here, so please have mercy on me.
If I have n-1 linearly independent vectors in n-dimensionas, I want to calculate an nth vector that is linearly independent to each of the others. In 3 dimensions I would use the cross product, which can be defined as the determinant of a 3x3 matrix where the first row is the unit vectors and the remaining rows are the row vectors of the 2 input vectors. Can this be translated to n-dimensions, where the result is the determinant of an n x n matrix constructed in the same way using n-1 vectors.
I've been searching the internet but dont really know what to search for and havent found anything. I wrote the following but don't know if it is valid:
We start with the following formula (from here) for the determinant of an NxN matrix.
$\mathrm{det}(\mathbf{A}) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma_i} $
$\mathrm{det}(\mathbf{A}) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) a_{1,\sigma_1} \prod_{i=2}^n a_{i,\sigma_i}$
If the first row of the matrix is composed of the unit vectors of N-Space
$a_{1,i} = \mathbf{e}_i$
then we can write
$\mathrm{det}(\mathbf{A}) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \mathbf{e}_{\sigma_1} \prod_{i=2}^n a_{i,\sigma_i} $
EDIT
This may be a duplicate of Generalized Cross Product. However, I do not yet understand the math in that answer, though I will work on that. The formula above is certainly in a different form, and I would like to know if it is valid.
