Skew Symmetry: Inner Product of Rows & Determinant

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I am entranced. I mean There are so many thing which seem to deal with skew symmetry, like the skew anti-symmetric matrices of Electromagnetic 4-tensor, I have this thing, a question. And it is stated as: " if you take the inner product of the rows of a skew symmetric matrix would it be equal to 0" because I remember hearing, or reading some where that the determinant of a matrix is its diagonals
 
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No but the determinant itself is an alternating form meaning the determinant of a matrix with two identical columns will vanish; you might have seen in say relativity textbooks the determinant is defined using the levi-civita symbol so the above property of the determinant is immediately apparent when cast in such form (in LA books like Friedberg you would see the determinant characterized as the unique n-linear alternating map that sends the identity matrix to 1).
 
That actually helps! But, the levi-civita symbols are a little confusing @_@ I mean just in general I don't understand why that is a reason for skew anti-symmetry
 
Why what is a reason?
 
D: maybe I asked the question wrong, I was asking, or trying to ask why does the determinant function hold to be skew anti-symmetric n-linear function phi from one vector space to another. In the book the determinant function is defined as ## \Delta = \sum_{\sigma} \epsilon_{\sigma}\left(\sigma\Phi\right) ## and ##\forall\sigma\in S_{n}## which is a set of permutations and epsilon is a homomorphism from ##S_{n}## to the multiplicative set {1,-1} and when ##\epsilon_{\sigma}## is even (when ##\epsilon_{\sigma}## is odd respectively) < that right there is similar to the levi-civita but I don't understand why it makes it skew anti-symmetric, is it because phi is skew symmetric?
 
D: but I am not looking for a basis free definition, thank you for the resource but that hasn't settled this problem D:
 
bump D:
please help?
 
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