Calculus Integration Based Proof that the Volume of a Parallelepiped spanned by $\{v_1, \ldots, v_n\}$ is $\text{det}(v_1, \ldots, v_n)$?

Solution 1:

This is just Cavalieri's principle in action, but I get it that you want to write down the integral. So, $$E^{\alpha\beta}(c)Q = \{x\in\Bbb R^N: 0\le x_i\le 1, i\ne\alpha,\ cx_\beta\le x_\alpha\le cx_\beta + 1\}.$$ By Fubini's Theorem, do the $\alpha\beta$ integral and then the rest over the usual cube in $\Bbb R^{N-2}$. Thus, the volume is $$\int_0^1\int_{cx_\beta}^{cx_\beta+1}dx_\alpha\,dx_\beta = \int_0^1 x_\alpha\Big|_{cx_\beta}^{cx_\beta+1}\,dx_\beta = \int_0^1 1\,dx_\beta = 1,$$ of course, as required. (Drawing the two-dimensional parallelogram is, of course, helpful, but not necessary.)

Solution 2:

We can actually make further reductions. Suppose $T:\Bbb{R}^n\to\Bbb{R}^n$ is the linear transformation \begin{align} T(x_1,\dots, x_n)&=(x_1+x_2,x_2,\cdots, x_n) \end{align} In terms of matrices, we're taking the second row of the identity matrix and adding it to the first row. We only need to restrict attention to this particular one, because we can perform row swaps to ensure we're only looking at rows 1 and 2, and then perform a scalar multiplication to ensure we only deal with $c=1$. Now, \begin{align} T(Q)&=\{\xi\in\Bbb{R}^n\,:\, \xi_2\leq \xi_1\leq \xi_2+1\,\quad\text{and}\quad \xi_2,\dots, \xi_n\in [0,1]\} \end{align} (i.e just put $\xi_1=x_1+x_2$, and $\xi_j=x_j$ for $j\geq 2$, and manipulate the inequalities $x_i\in [0,1]$ for all $i$ in terms of $\xi$). Now, we have \begin{align} \text{vol}(T(Q))&=\int_{T(Q)}1\,dV\\ &=\int_{[0,1]^{n-2}}\int_0^1\int_{\xi_2}^{\xi_2+1}1\,d\xi_1\,d\xi_2\, d(\xi_3,\dots, \xi_n)\tag{by Fubini}\\ &=\int_0^1(\xi_2+1-\xi_2)\,d\xi_2\\ &=1. \end{align} Here, it's clear that the integral over the last $n-2$ coordinates is trivially $1$ (this is just the $(n-2)$-dimensional volume of the cube $[0,1]^{n-2}$).