In this question, the determinant of a matrix is explained to be a measure of the volume of a parallelepiped formed by using the columns in a matrix as vectors.

It is also noted that the determinant is linear in each row. We can represent this geometrically as the volume of a parallelpiped being the sum of the volumes of two other ones. Is there a geometric proof that the volume corresponding to these matrices is linear in this manner?

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  • The determinant, by cinstruction, ismultilinear in its rows (or its columns as well), independently of any geometric consideration. – Bernard Nov 21 '18 at 20:57
  • @Bernard: Sure, but it seems like there should be a geometric representation. Maybe I need to reword the question. – Casebash Nov 21 '18 at 21:16
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    I think there is a geometric representation. Note that the determinant does not change upon adding multiples of one row/column to another. This corresponds to a shear mapping of the paralelepiped, so the volume does not change under this operation. Then, you use the Gram-Schmidt process to express the original vectors which form the base,width and height (etc if there are more than 3 dimensions) in terms of a determinant which results from applying the operation of adding multiples of one row/column to another. – Matematleta Nov 21 '18 at 21:55

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