Calculating Volume of a Parallelepiped in R^n

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The area of a parallelogram formed by two vectors in R^3 is calculated using the cross product, |v_1 × v_2|. For three vectors in R^4, arranging them in a 4 by 3 matrix allows for the use of Givens rotations to simplify the matrix, leading to a product of diagonal elements that represents the volume of a 3-D parallelepiped in the hyperplane spanned by those vectors. This concept can be extended to m vectors in R^n, where the product becomes the absolute value of the determinant when m equals n. The discussion suggests that this product may relate to the integral over an m-dimensional hyper-surface in R^n, contributing to the volume element dV. The connection to exterior algebra is noted, although the terminology remains unclear.
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Say you have two vectors v_1 and v_2 in \Re^3. Then the area of the parallelogram generated by these two vectors is \left|v_1\times v_2\right|.

Now consider the case where you have three vectors in \Re^4. You can place these vectors as columns in a 4 by 3 matrix. Now if you simultaneously rotate the three matrices together using Givins rotations you can introduce zeros in all entries below the main diagonal and non-negative values in all the diagonal elements. If the three matrices are linearly independent then all the diagonal elements will be positive. Now the product of the diagonal elements gives the area of a 3-D parallelepiped embedded in the hyperplane spanned by the vectors in \Re^4. What would you call this type of "triple" product?

This can be generalized to any m vectors in \Re^n as long as m\leq n. When m=n this product is just the absolute value of the determinant of the matrix formed by the vectors as columns.

It seems that in general this type of product will appear under the integral when integrating over some m dimensional hyper-surface that's embedded in \Re^n. It would be used for calculating the volume element dV on an m dimensional hyper-surface.

Hopefully this makes sense.
 
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I think the thing I'm thinking of is related to the exterior algebra, though I don't really understand the notation/terminogy.
 
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