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Below is a proof found in Gilbert Strang's book and here Why is the determinant the volume of a parallelepiped in any dimensions? that the determinant equals the volume of a box :

To find the volume of a box whose edges are given by a set of vectors $\{v_{1},\ldots,v_{n}\}$ we apply Gram-Schmidt orthogonalization to $\{v_{1},\ldots,v_{n}\}$, so that \begin{eqnarray*} v_{1} & = & v_{1}\\ v_{2} & = & c_{12}v_{1}+v_{2}^{\perp}\\ v_{3} & = & c_{13}v_{1}+c_{23}v_{2}+v_{3}^{\perp}\\ & \vdots \end{eqnarray*} where $v_{2}^{\perp}$ is orthogonal to $v_{1}$; and $v_{3}^{\perp}$ is orthogonal to $span\left\{ v_{1},v_{2}\right\} $, etc. Since determinant is multilinear, anti-symmetric, then \begin{eqnarray*} \det\left(v_{1},v_{2},v_{3},\ldots,v_{n}\right) & = & \det\left(v_{1},c_{12}v_{1}+v_{2}^{\perp},c_{13}v_{1}+c_{23}v_{2}+v_{3}^{\perp},\ldots\right)\\ & = & \det\left(v_{1},v_{2}^{\perp},v_{3}^{\perp},\ldots,v_{n}^{\perp}\right)\\ & =?& \mbox{signed volume}\left(v_{1},\ldots,v_{n}\right) \end{eqnarray*}

Question: This only proves that the determinant of the original edges equals the volume of the new created box.

That is we don't show $\det(v_1,...,v_n)$ = signed volume $(v_1,...v_n)$. What we do prove is $\det(v_1,...,v_n)$ = Volume of the box with orthogonal edges.

Martin Sleziak
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Kashmiri
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    The volume of a parallelepiped is the height dropped from a vertex times the $n-1$ volume of the $n-1$ parallelepiped the height is dropped on. So by induction you can show the Gram Schmidt procedure gives you the volume of the parallelepiped you started with (up to sign because the sign depends on orientation) –  May 29 '21 at 05:40
  • When you apply orthogonalization, you're turning the parallelepiped into a cuboid of equal volume - think about how you can 'stretch' a parallelogram into a rectangle and vice-versa without changing the area. Now the matrix of these now-orthogonal vectors, that make up the sides of this cuboid is basically $diag(l_1,l_2,\ldots,l_n)$ where $l_i$ are the lengths of the sides. The determinant of this diagonal matrix is just the product of the lengths, i.e. the volume. – Raad Shaikh May 29 '21 at 05:41
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    Shear transformations preserve volume. https://en.wikipedia.org/wiki/Shear_mapping – CyclotomicField May 29 '21 at 05:42
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    Yes, this messed me up for a long time. I think the answer @CyclotomicField gives is almost enough, but still missing something. Basically, the proof uses the fact that the determinant is invariant to shears of the vertices $v_i$. However, just showing that some quantity is invariant to shear is NOT the same as saying that quantity is a volume. – user3180 May 29 '21 at 05:49
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    The only way to rigorously prove this to my satisfaction, would be to derive a formula for volume of a paralleliped which is independent of the determinant, and then showing that using the determinant gives the same answer as that formula. To date I have not seen such a proof yet. – user3180 May 29 '21 at 05:50

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A sketch: Consider a parallelepided spanned by vectors $(v_1,v_2,...,v_n)$: $$ K(v_1,v_2,...,v_n)= \{ \sum_{k=1}^n t_k v_k : 0 \leq t_k\leq 1, 1\leq k\leq n\}$$

Given $c\in [0,1]$ consider an elementary shear of this solid in the 1-2 direction: $K'=K(v_1,v_2-cv_1,v_3,...,v_n)$. Note that the directions $v_3,...,v_n$ are left untouched. $K'$ may be cut into two pieces $K'=(K'\cap K) \cup (K'\setminus K)=: K_1\cup K_2$ (make a drawing). If you translate $K_2$ by the vector $-v_1$ you almost get $K\setminus K'$. True for the interior but not for the borders which, however, are of zero volume. $K$ is in this way reconstructed from $K'$ by cutting the latter into two pieces and translating one of them (up to zero volume corrections). Thus, the volume of $K$ and $K'$ are the same and as you see by inspection, the determinants as well. Repeating the operation and also using the inverse you see that the same holds true for all $c\in {\Bbb R}$.

The transformation described by Strang may be decomposed into $n(n-1)/2$ such operations.

H. H. Rugh
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