tarjetaroja2t

Answered

2022-09-25

Gram Determinant equals volume?

I have been trying to solve this problem of finding the 'n-volume' of a paralleletope spanned by m vectors, where clearly $m\le n$. In general, for computational purposes, what I have managed to do is define volume as the product of absolute values of vectors obtained by gram-schmidt orthogonalizationn. (Makes sense right? That's the natural interpretation when we say volume)

I had to do two things, firstly to show that this definition of volume is a well defined one (i.e. any set of orthogonal vectors obtained by the process will give the same volume), and secondly to find a quick way to do this. I managed to prove the first one by induction, but the second part is a little bit of a problem. I managed to obtain formulae for small dimensions as 2,3 or even 4 but this process is impractical for any bigger dimensions as the substitutions for smaller dimensions into the formula for the next dimension becomes exponentially complicated

How does one prove that the gram determinant is equal to the volume of a paralleletope spanned by a set of vectors?

Answer & Explanation

Marnovdk

Expert

2022-09-26Added 6 answers

Step 1

Let d vectors ${\mathbf{a}}_{i}\in V$ $\phantom{\rule{mediummathspace}{0ex}}(1\le i\le d)$ be given, where V is an n-dimensional euclidean space, e.g., $V={\mathbb{R}}^{n}$ with the usual scalar product. Then there is a d-dimensional subspace $U\subset V$ containing the ${\mathbf{a}}_{i}\phantom{\rule{thinmathspace}{0ex}}$, and U inherits the euclidean structure from V. Let $({\mathbf{e}}_{j}{)}_{1\le j\le d}$ be an orthonormal basis of U. Then the d-dimensional volume of the parallelotope P spanned by the ${\mathbf{a}}_{i}$ is given by

${\mathrm{v}\mathrm{o}\mathrm{l}}_{d}(P)=|det(A)|\text{},$

where A is the $d\times d$- matrix containing the coordinates $({a}_{i.1},{a}_{i.2},\dots ,{a}_{i.d})$ of the ${\mathbf{a}}_{i}$ in terms of the ${\mathbf{e}}_{j}$ in its columns. (This is the d-dimensional analogue of the formula

${\mathrm{v}\mathrm{o}\mathrm{l}}_{3}(P)=|\phantom{\rule{mediummathspace}{0ex}}det\left[\begin{array}{ccc}{a}_{1.1}& {a}_{2.1}& {a}_{3.1}\\ {a}_{1.2}& {a}_{2.2}& {a}_{3.2}\\ {a}_{1.3}& {a}_{2.3}& {a}_{3.3}\end{array}\right]\phantom{\rule{mediummathspace}{0ex}}|$

valid in 3-space.) It follows that

$\begin{array}{}\text{(1)}& {({\mathrm{v}\mathrm{o}\mathrm{l}}_{d}(P))}^{2}={\textstyle (}det(A){{\textstyle )}}^{2}=det({A}^{\prime}\phantom{\rule{mediummathspace}{0ex}}A)=det(G)\text{}.\end{array}$

Step 2

The elements ${g}_{ik}$ of the $d\times d$-matrix $G:={A}^{\prime}\phantom{\rule{thinmathspace}{0ex}}A$ are given by

${g}_{ik}:={\mathrm{r}\mathrm{o}\mathrm{w}}_{i}({A}^{\prime})\cdot {\mathrm{c}\mathrm{o}\mathrm{l}}_{k}(A)={\mathbf{a}}_{i}\cdot {\mathbf{a}}_{k}\phantom{\rule{2em}{0ex}}(1\le i\le d,\text{}1\le k\le d)\text{}.$

It turns out that G is the matrix of the scalar products ${\mathbf{a}}_{i}\cdot {\mathbf{a}}_{k}\phantom{\rule{thinmathspace}{0ex}}$ and is independent of the basis chosen for U. This Gram matrix G depends only on the given ${\mathbf{a}}_{i}$ and the euclidean structure in V.

Taking the square root in (1) gives the final formula ${\mathrm{v}\mathrm{o}\mathrm{l}}_{d}(P)=\sqrt{det(G)}\text{}.$

pilinyir1

Expert

2022-09-27Added 2 answers

Step 1

You put the vectors into a matrix A, then apply one of the three methods for QR-factorization to it, $A=QR$, and read off the volume as the product (or absolute value thereof) of the diagonal entries of R.

Step 2

The parallelotope is $P=A[0,1{]}^{m}$, and applying ${Q}^{T}$ to it, ${Q}^{T}P=R[0,1{]}^{m}$, transforms it into a position where it is obtained by axis-parallel, volume -preserving shearing from a rectangular box with side-lengths equal to the diagonal entries of R.

Step 3

Since $G={A}^{T}A={R}^{T}{Q}^{T}QR={R}^{T}R$, one has $\sqrt{detG}=|detR|$

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