Volumes of Polytopes

Question

I recently have come across the concept of the Gramian. For some set of vectors $\{v_1,\dots v_n\}\subset\mathbb{C}^m$, $n\leq m$, we have that $$ \operatorname{Vol}(v_1,\dots,v_n)=\sqrt{\det A} $$ where $A_{ij}=v_i\cdot v_j$. The value $\det A$ is then given the name of "the Gramian," and can be used to generalize volumes in higher dimensional spaces.

I wanted to prove that this is the case for myself. The proof for when $n=m$ is rather trivial. We can write that if $B_i=v_i$, then $A=B^\dagger B$, which means that $\det A=\det B^\dagger\det B=(\det B)^2$, and it is well known that $\det B=\det(v_1,\dots,v_m)=\operatorname{Vol}(v_1,\dots,v_n)$.

Where I start to struggle is for when $n<m$. Does anyone have any hints on how to proceed?

Answer 1

First I would say this, we have $V = (v_1,\dots,v_n)$ of size $m\times n$. Assume the $v_i$ are free I can have a weak equivalent of the polar decomposition: there exists $S \in S_n^{++}(\mathbb{R})$ and $U$ of size $m \times n$ right orthogonal, i.e $$U^TU = I_n $$ And $$V = US$$ For this just consider $S = \sqrt{V^TV}$ the only square root symmetric definite positive, and $U =V S^{-1}$. Since $U$ is right orthogonal, its restriction on $F = \text{Span}(v_i)_i$ will preserve the induced lebesgue measure.

Why? It turns out that we could extend $U$ to $F^\bot$ in $\tilde{U} \in O_m(\mathbb{R})$ and $\tilde{U}$ would send $F$ to $\mathbb{R}^n \times \{0_{m-n}\}$, on which an induced lebesgue measure is well defined.

So if we set $$P = [0,v_1] \times \cdots \times [0,v_n]\subset F$$ $$Q = [0,1]^n \subset \mathbb{R}^n$$ With the change of variable defined by $V$ which sends $F$ on $\mathbb{R}^n$ we would have $$\lambda_F (P) = |\det S| \lambda_{\mathbb{R}^n} (Q) = \det S = \sqrt{\det V^T V} = \text{Vol} (v_i)_i$$

Answer 2

I came up with a different solution, using just elementary calculus results. Suppose that we have a (not necessarily orthonormal) basis $\{v_1,\dots,v_n\}$ for $\mathbb{R}^n$. We will design an orthogonal basis $\{b_1,\dots,b_n\}$ from this basis by using a recursive process akin to the Gram-Schmidt process.

Let $b_i\in\operatorname{Span}\{v_1,\dots,v_{i-1}\}^\perp\cap\operatorname{Span}\{v_1,\dots,v_i\}$. In other words, there exists $\beta_1,\dots,\beta_i\in\mathbb{R}$ such that $$ b_i=\sum_{j=1}^i\beta_jv_j $$ Note we can write the following system of $(i-1)$ equations: $$ \begin{pmatrix} \sum_{j=1}^i\beta_j(v_j\cdot v_1)\\ \vdots\\ \sum_{j=1}^i\beta_j(v_j\cdot v_{i-1}) \end{pmatrix}= \begin{pmatrix} 0\\\vdots\\0 \end{pmatrix} $$ Now, if we add just one more (trivial) equation, $\beta_i=\beta_i$, then we can write a system of $i$ equations with $i$ unknowns: $$ \begin{pmatrix} v_1\cdot v_1&\dots&v_i\cdot v_1\\ \vdots&\ddots&\vdots\\ v_1\cdot v_{i-1}&\dots&v_i\cdot v_{i-1}\\ 0&\dots&1 \end{pmatrix} \begin{pmatrix} \beta_1\\\vdots\\\beta_i \end{pmatrix} = \begin{pmatrix} 0\\\vdots\\0\\\beta_i \end{pmatrix} $$ Let the matrix be the matrix $B$, and using Cramer's rule, we can claim $$ \det(B)\beta_j=\left\vert \begin{matrix} v_1\cdot v_1&\dots&v_{j-1}\cdot v_1&0&v_{j+1}\cdot v_1&\dots&v_i\cdot v_1\\ \vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ v_1\cdot v_{i-1}&\dots&v_{j-1}\cdot v_{i-1}&0&v_{j+1}\cdot v_{i-1}&\dots&v_i\cdot v_{i-1}\\ 0&\dots&0&\beta_i&0&\dots&1 \end{matrix} \right\vert $$ Notice that we can simplify this a bit to obtain $$ \det(B)\beta_j=\left\vert \begin{matrix} v_1\cdot v_1&\dots&v_{j-1}\cdot v_1&0&v_{j+1}\cdot v_1&\dots&v_i\cdot v_1\\ \vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ v_1\cdot v_{i-1}&\dots&v_{j-1}\cdot v_{i-1}&0&v_{j+1}\cdot v_{i-1}&\dots&v_i\cdot v_{i-1}\\ 0&\dots&0&1&0&\dots&1 \end{matrix} \right\vert\beta_i $$ If $\det(B)=\beta_i$, it is easy to confirm that $$ \beta_j=(-1)^{i+j}\det(v_k\cdot v_l),~k\in\{1,\dots,i-1\},~l\in\{1,\dots,i\}\setminus\{j\} $$ It is easy to then confirm that we can now say $$ b_i=\sum_{j=1}^i(-1)^{i+j}\det(v_k\cdot v_l) $$ where $k$ and $l$ are indexed by the same $k$ and $l$ as above. So, we can abuse a bit of notation and write $$ b_i=\left\vert \begin{matrix} v_1\cdot v_1&\dots&v_i\cdot v_1\\ \vdots&\ddots&\vdots\\ v_1\cdot v_{i-1}&\dots&v_i\cdot v_{i-1}\\ v_1&\dots&v_i \end{matrix} \right\vert $$ And, it follows that $\{b_1,\dots,b_n\}$ is an orthonormal basis for $\mathbb{R}^n$.

Now, we will use the fact that $$ \int_{[0,1]^n}d^nx=(1)^n=1 $$ to find $$ \operatorname{Vol}(v_1,\dots,v_n)\equiv\int_{[0,v_1]\times\dots\times[0,v_n]}\prod_{i=1}^nd(x\cdot v_i) $$ Let $e_i=b_i/\left\vert b_i\right\vert$, where $\{b_1,\dots,b_n\}$ is the orthogonal basis found using the process used before. Notice that $$ d^nk=\prod_{i=1}^nd(k\cdot e_i) $$ This is true because $\{e_i\}$ is related to all other orthonormal bases by linear plane isometries, which all have Jacobian of $1$. So, by simple rescaling, we get $$ d^nk=\prod_{i=1}^n\frac{d(k\cdot b_i)}{\left\vert b_i\right\vert} $$ We can write that $$ b_i^2= \left\vert \begin{matrix} v_1\cdot v_1&\dots&v_i\cdot v_1\\ \vdots&\ddots&\vdots\\ v_1\cdot v_{i-1}&\dots&v_i\cdot v_{i-1}\\ b_i\cdot v_1&\dots&b_i\cdot v_i \end{matrix} \right\vert $$ Because by construction, $b_i\in\operatorname{Span}(v_1,\dots,v_{i-1})^\perp$, $b_i\cdot v_j=0$ for $j<i$. Now, notice that $$ b_i\cdot v_i= \left\vert \begin{matrix} v_1\cdot v_1&\dots&v_i\cdot v_1\\ \vdots&\ddots&\vdots\\ v_1\cdot v_{i-1}&\dots&v_i\cdot v_{i-1}\\ v_i\cdot v_1&\dots&v_i\cdot v_i \end{matrix} \right\vert $$ Or, in other words, $b_i=G(v_1,\dots,v_i)$, where $G$ is the Gramian. This means $$ b_i^2= \left\vert \begin{matrix} v_1\cdot v_1&\dots&v_i\cdot v_1\\ \vdots&\ddots&\vdots\\ v_1\cdot v_{i-1}&\dots&v_i\cdot v_{i-1}\\ 0&\dots&G(v_1,\dots,v_i) \end{matrix} \right\vert $$ Or, in other words, $$ b_i^2= G(v_1,\dots,v_i)G(v_1,\dots,v_{i-1}) $$ I will now define $G_i\equiv G(v_1,\dots,v_i)$. This means that $$ d^nk=\prod_{i=1}^n\frac{d(k\cdot b_i)}{\sqrt{G_{i-1}G_i}} $$ Notice that by reordering this product, we can simply write $$ d^nk=\sqrt{\frac{G_0}{G_n}}\prod_{i=1}^n\frac{d(k\cdot b_i)}{\sqrt{G_{i-1}^2}}=G_n^{-1/2}\prod_{i=1}^nd(k\cdot b_i) $$ Now, notice that $$ \frac{\partial(k\cdot b_i)}{\partial(k\cdot v_j)}=0 $$ if $j>i$ because then $k\cdot b_i$ is independent of $k\cdot v_j$. This means that our Jacobian matrix is upper triangular. So, this means that the Jacobian determinant is just going to be $$ \det(J)=\prod_{i=1}^n\frac{\partial(k\cdot b_i)}{\partial(k\cdot v_i)} $$ Notice that $k\cdot b_i=(k\cdot v_i)G_{i-1}+C$, where $C$ is a function that does not depend on $k\cdot v_i$, so we can simply write $$ \det(J)=\prod_{i=1}^nG_{i-1} $$ So, we can finally write that $$ d^nk=G_n^{-1/2}\prod_{i=1}^n\frac{G_{i-1}}{G_{i-1}}d(v_i\cdot k) $$ Or, in other words, $$ \int_{[0,1]^n}d^nk=G^{-1/2}\int_{[0,v_1]\times\dots\times[0,v_n]}\prod_{i=1}^nd(v_i\cdot k) $$ And, finally, $$ \operatorname{Vol}(v_1,\dots,v_n)=G(v_1,\dots,v_n)^{1/2} $$ I know the proof can be made shorter, but this was what was most intuitive for me!