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And please, do not use the determinant, because I am trying to figure this out so that I can use it as a step in my proof that the determinant is the signed volume of a parallelepiped.

I am trying to rigorously prove this line: "This corresponds to a skew translation of the parallelepiped, which does not affect its volume." for the answer from Hagen von Eitzen

A similar idea is used in James answer when he equates $det(v_1, v_2^\perp, ... v_n^\perp) = det(v_1, v_2, ... v_n)$

for this question here: why determinant is volume of parallelepiped in any dimensions

user3180
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1 Answers1

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Say you have a skew transformation: $$\mathbf{r}_1 = \mathbf{r}+(\mathbf r\mathbf u)\mathbf{v},$$ where $\mathbf{uv} = 0$, $|\mathbf u|=1$.

Consider chopping the initial parallelepiped by planes $\mathbf{ru}=c=\mathrm{const}$. Then the volume of parallelepiped is: $$ V=\int_{c_\min}^{c_\max}S(c)dc, $$ where $S(c)$ is the area of cross-section.

Then for every cross-section of a plane, the transformation is a translation $\mathbf{r}_1 = \mathbf{r}+c\mathbf{v}$. So the shape of cross-section doesn't change: $S(c)=S'(c)$ and thus the volume stays the same.

Vasily Mitch
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