# A Pythagorean Theorem for volume

Fredric D. Ancel  Department of Mathematical Sciences, University of
Wisconsin-Milwaukee<EMAIL_ADDRESS>
(Date: May 13, 2023)

###### Abstract.

Lebesgue measurable subsets $A$ and $B$ of parallel or identical
$k$-dimensional affine subspaces of Euclidean $n$-space $\mathbb{E}^{n}$
satisfy

_The Product Formula for Volume:_

$Vol_{k}(A)Vol_{k}(B)=\sum_{J\in\mathcal{S}(n,k)}Vol_{k}(\pi_{J}(A))Vol_{k}(\pi_{J}(B))$.

Here $Vol_{k}$ denotes $k$-dimensional Lebesgue measure; $\mathcal{S}(n,k)$
denotes the set of all $k$-element subsets of $\\{1,2,\cdots,n\\}$; and for
$J\in\mathcal{S}(n,k)$, $\mathbb{E}^{J}=$
$\\{(x_{1},x_{2},\cdots,x_{n})\in\mathbb{E}^{n}:x_{i}=0\text{ for all }i\notin
J\\}$ and $\pi_{J}:\mathbb{E}^{n}\rightarrow\mathbb{E}^{J}$ is the projection
that sends the $i^{th}$ coordinate of a point of $\mathbb{E}^{n}$ to $0$
whenever $i\notin J$. Setting $B=A$, we obtain the corollary:

_The Pythagorean Theorem for Volume:_

$Vol_{k}(A)^{2}=\sum_{J\in\mathcal{S}(n,k)}(Vol_{k}(\pi_{J}(A)))^{2}$.

##  1\. Introduction

If a vector space $V$ is equipped with an inner product $\langle\;,\;\rangle$,
then this inner product satisfies the following equation.

_The Product Formula for Inner Products:_

$\langle x,y\rangle=\sum_{1\leq i\leq n}\langle x,u_{i}\rangle\langle
y,u_{i}\rangle$

whenever $u_{1},u_{2},\cdots,u_{n}$ is an orthonormal basis for $V$. The fact
that the dot product on Euclidean $n$-space $\mathbb{E}^{n}$ satisfies this
equation with respect to the standard orthonormal basis for $\mathbb{E}^{n}$
allows us to deduce

_The Product Formula for Length:_

$Length(A)Length(B)=\sum_{1\leq j\leq n}Length(\pi_{j}(A))Length(\pi_{j}(B))$

provided that $A$ and $B$ are parallel or identical line segments in
$\mathbb{E}^{n}$. Here
$\mathbb{E}^{\\{j\\}}=\\{(x_{1},x_{2},\cdots,x_{n})\in\mathbb{E}^{n}:x_{i}=0\text{
for all }i\neq j\\}$ and
$\pi_{j}:\mathbb{E}^{n}\rightarrow\mathbb{E}^{\\{j\\}}$ is the projection that
sends the $i^{th}$ coordinate of a point of $\mathbb{E}^{n}$ to $0$ whenever
$i\neq j$. Setting $B=A$ then yields

_The Pythagorean Theorem for Length:_

$Length(A)^{2}=\sum_{1\leq j\leq n}(Length(\pi_{j}(A)))^{2}$

The object of this article is to exhibit an inner product on a vector space
that allows the measurement of - not the length of a line segment in
$\mathbb{E}^{n}$ \- but the volume of a subset of a k-dimensional affine
subspace of $\mathbb{E}^{n}$. Then, just as the dot product on
$\mathbb{E}^{n}$ leads to the Product Formula and the Pythagorean Theorem for
Length, this inner product yields the Product Formula and Pythagorean Theorem
for $k$-dimensional Volume stated in the Abstract. The emphasis in this
article is conceptual rather than computational: the Product Formula and
Pythagorean Theorem for Volume follow logically once Euclidean space with its
dot product have been generalized to the appropriate inner product space. The
inner product space which is the appropriate generalization of
$\mathbb{E}^{n}$ is a well-known algebraic object that is typically denoted
$\Lambda_{k}(\mathbb{E}^{n})$. It is a vector subspace of the exterior algebra
of $\mathbb{E}^{n}$ which is usually denoted $\Lambda(\mathbb{E}^{n})$. In
this article, we do not assume the readers are familiar with
$\Lambda_{k}(\mathbb{E}^{n})$ or its inner product. We will provide a complete
description of these objects.

The Product Formula and Pythagorean Theorem for Volume were new to the author
when he first observed them. Because of the remarkable way that the
Pythagorean Theorem for Volume echoes the universally renowned formula for
length, the author was surprised that the results for volume are not better
known. However, the author’s subsequent search of the literature revealed that
the Pythag-orean Theorem for Volume has been discovered, rediscovered, proved
and reproved repeatedly. (The Product Formula for Volume, on the other hand,
is not explicitly stated in any source found by the author; this article may
be its first formulation.) Apparently, the earliest statement of the
Pythagorean Theorem for the area of a triangle in $\mathbb{E}^{3}$ is found in
the 1220 book _Practica geometriae_ by L. Fibonacci (1172-1250) ([3] page
232). A typical application of this result would tell us that for a triangle
$T$ in $\mathbb{E}^{3}$ with vertices $P=(p,0,0)$, $Q=(0,q,0)$ and $R=(0,0,r)$
where $p,\;q\text{ and }r>0$,
$(Area(T))^{2}=(\frac{1}{2}qr)^{2}+(\frac{1}{2}pr)^{2}+(\frac{1}{2}pq)^{2}$,
because the projections of $T$ into the $YZ$-plane, the $XZ$-plane and the
$XY$-plane are triangles that have areas $\frac{1}{2}qr$, $\frac{1}{2}pr$ and
$\frac{1}{2}pq$, respectively.111To corroborate this formula for $Area(T)$,
note that $Area(T)=\frac{1}{2}Area(Z)$ where $Z$ is the parallelogram with
vertices $P$, $Q$, $R$ and $Q+R-P$ and $Area(Z)$ equals the norm of the cross
product $(Q-P)\times(R-P)$. According to page 37 of [11] and various Wikipedia
articles, following its appearance in Fibonacci’s book, this result (for a
triangle in $\mathbb{E}^{3}$) was familiar to R. Descartes (1596-1650) and J.
Faulhaber (1580-1635). Two independent proofs of this result were published in
the late $18^{th}$ century, one in 1772 by C. de Tinseau (1748-1822) and the
other in 1783 by J. P. de Gua de Malves (1712-1785). Ironically, the
Pythagorean Theorem for the area of a triangle in $\mathbb{E}^{3}$ then
acquired the name _de Gua’s Theorem_. A statement and careful proof of the
Pythagorean Theorem for the area of a planar polygonal figure in
$\mathbb{E}^{3}$ appeared in the influential 1813 memoir [13] by J. Hachette
(1769-1834) and G. Monge (1746-1818) (See _Theoreme 66_ on page 100 of [13],
also cited on page 436 of [3]). According to [16], in about 1812, J. Binet
(1786-1856) and A-L. Cauchy (1789-1857) independently discovered a formula
(named both _Binet-Cauchy_ and _Cauchy-Binet_) for the determinant of the
product of two non-square matrices whose product is a square matrix. (Binet
was the inventor of matrix multiplication.) This formula, if interpreted
correctly, is equivalent to the Product Formula for the inner product on
$\Lambda_{k}(\mathbb{E}^{n})$. Of the nineteen articles cited in the
references, eleven of them - [1], [2], [5], [6], [7], [8], [9], [12], [15],
[17] and [18] \- are recapitulations of the Pythagorean Theorem for Volume, in
some instances specialized to parallelograms in $\mathbb{E}^{3}$, and in other
instances covering the more general case of $k$-dimensional parallelopipeds in
$\mathbb{E}^{n}$. Of these, references [1] and [15] contain informative
historical remarks. Paper [7] is the only one found by the author that makes
the observation that the Pythagorean Theorem for Volume applies to all
Lebesgue measurable subsets of an affine subspace of $\mathbb{E}^{n}$. The
author was unaware of the existence of [7] when he discovered the results
stated in the Abstract; and the methods of this article are more linear
algebraic and less analytic in character than the methods of [7]. Since the
author’s search for papers on the Pythagorean Theorem for Volume was more
haphazard than exhaustive, there are no doubt many such articles that are
unintentionally omitted from the references.

##  2\. Linear algebraic preliminaries

In this article, we restrict ourselves to vector spaces and matrices over the
field $\mathbb{R}$ of real numbers.

Let $V$ be a vector space. For $p\in V$, define the _translation_
$T_{p}:V\rightarrow V$ by $T_{p}(x)=x+p$. Note that for $p$, $q\in V$,
$T_{q}\circ T_{p}=T_{p+q}$ and $T_{p}^{-1}=T_{-p}$. A subset $W$ of $V$ is a
_$k$ -dimensional affine subspace_ if there is a $k$-dimensional vector
subspace $W_{0}$ of $V$ and a $p\in V$ such that $W=T_{p}(W_{0})$. Two
$k$-dimensional affine subspaces $W$ and $W^{\prime}$ of $V$ are _identical or
parallel_ if there is a $q\in V$ such that $W^{\prime}=T_{q}(W)$. Observe that
if $W_{0}$ is a vector subspace of $V$ and $p$, $q\in V$, then
$T_{p}(W_{0})=T_{q}(W_{0})$ if and only if $p-q\in W_{0}$.

######  Proposition 2.1.

If $L:V\rightarrow V^{\prime}$ is a linear function between vector space, then
for every $p\in V$, $L\circ T_{p}=T^{\prime}_{L(p)}\circ L$. (Here, for $p\in
V$ and $p^{\prime}\in V^{\prime}$, $T_{p}:V\rightarrow V$ and
$T^{\prime}_{p^{\prime}}:V^{\prime}\rightarrow V^{\prime}$ denote
translations.)

###### Proof.

For every $x\in V$, $L\circ T_{p}(x)=L(x+p)=L(x)+L(p)=T^{\prime}_{L(p)}\circ
L(x).$ ∎

We call a vector space an _inner product space_ if it is equipped with an
inner product. If $V$ is an inner product space equipped with the inner
product $\langle$ , $\rangle_{V}$, then define the associated norm $||$
$||_{V}:V\rightarrow[0,\infty)$ by $||x||_{V}=(\langle
x,x\rangle_{V})^{\frac{1}{2}}$ and define the associated metric
$\rho_{V}:V\times V\rightarrow[0,\infty)$ by $\rho_{V}(x,y)=||x-y||_{V}$. The
metric $\rho_{V}$ determines a topology on $V$. If $V$ and $W$ are inner
product spaces, then a function $f:V\rightarrow W$ is an _isometry_ if
$f(V)=W$ and $\rho_{W}(f(x),f(y))=\rho_{V}(x,y)$ for all $x,y\in V$. Note that
the latter condition is equivalent to the statement
$||f(x)-f(y)||_{W}=||x-y||_{V}$ for all $x,y\in V$. Observe that if $V$ is an
inner product space and $p\in V$, then the translation $T_{p}:V\rightarrow V$
is an isometry.

For a vector space $V$ and an integer $k\geq 1$, let
$V^{k}=\\{(v_{1},v_{2},\cdots,v_{k}):v_{i}\in V$ for $1\leq i\leq k\\}$. Let
$V$ and $W$ be vector spaces, let $k\geq 1$ and let $f:V^{k}\rightarrow W$ be
a function. f is _multilinear_ if

$f(v_{1},\cdots,v_{i-1},av_{i}+bw_{i},v_{i+1},\cdots,v_{k})=$
$af(v_{1},\cdots,v_{i-1},v_{i},v_{i+1},\cdots,v_{k})+bf(v_{1},\cdots,v_{i-1},w_{i},v_{i+1},\cdots,v_{k})$

for all $1\leq i\leq k$, all
$v_{1},\cdots,v_{i-1},v_{i},w_{i},v_{i+1},\cdots,v_{k}\in V$ and all
$a,b\in\mathbb{R}$. When $k=2$ and $f$ is multilinear, then we say that $f$ is
_bilinear_. f is _alternating_ if

$f(v_{1},\cdots,v_{i-1},v_{j},v_{i+1},\cdots,v_{j-1},v_{i},v_{j+1},\cdots,v_{k})=$
$-f(v_{1},\cdots,v_{i-1},v_{i},v_{i+1},\cdots,v_{j-1},v_{j},v_{j+1},\cdots,v_{k})$

for all $1\leq i<j\leq k$ and all
$v_{1},\cdots,v_{i-1},v_{i},v_{i+1},\cdots,v_{j-1},v_{j},v_{j+1},\cdots,v_{k}\in
V$. (Equivalently, f is alternating if

$f(v_{1},\cdots,v_{i-1},v_{i},v_{i+1},\cdots,v_{j-1},v_{j},v_{j+1},\cdots,v_{k})=0$

whenever $v_{i}=v_{j}$.)

Let $V$ be a vector space and let $k\geq 1$ be an integer. Then there is a
vector space $\Lambda_{k}(V)$ and a multilinear map
$\textit{i}:V^{k}\rightarrow\Lambda_{k}(V)$ with the following _universality
property_. If $f:V^{k}\rightarrow W$ is a multilinear alternating map to a
vector space $W$, then there is a unique linear map
$\Phi:\Lambda_{k}(V)\rightarrow W$ such that $\Phi\circ\textit{i}=f$. It is
common practice to denote the image $\textit{i}(v_{1},v_{2},\cdots,v_{k})$ by
$v_{1}\wedge v_{2}\wedge\cdots\wedge v_{k}$ and to call this point a _$k$
-vector_ or a _wedge product_ of $k$ elements of $V$.

For a textbook treatment of $\Lambda_{k}(V)$, we refer the reader to pages
446-449 of [10] and to [21]. In those pages, it is revealed that
$\Lambda_{k}(V)$ is a vector subspace of a larger vector space $\Lambda(V)$
called the _exterior algebra_ of $V$. $\Lambda(V)$ is the direct sum of its
subspaces $\\{\Lambda_{k}(V):k\geq 0\\}$, and $\Lambda(V)$ is called an
_algebra_ because there is a product $(x,y)\mapsto x\wedge
y:\Lambda(V)\times\Lambda(V)\rightarrow\Lambda(V)$ which maps
$\Lambda_{k}(V)\times\Lambda_{m}(V)$ into $\Lambda_{k+m}(V)$. We will now list
the facts about $\Lambda_{k}(V)$ that we will need in the sequel, and we will
justify those facts when the justifications aren’t obvious. The reader can
consult [10] and [21] for a broader and more detailed overview of this topic.

The existence of $\Lambda_{k}(V)$ can be proved by identifying it with the
quotient space $\Phi(V^{k})/K$ where $\Phi(V^{k})$ is a vector space with
basis $V^{k}$ and $K$ is the vector subspace of $\Phi(V^{k})$ generated by all
elements of $\Phi(V^{k})$ of the form

$(v_{1},\cdots,v_{i-1},av_{i}+bw_{i},v_{i+1},\cdots,v_{k})$
$-a(v_{1},\cdots,v_{i-1},v_{i},v_{i+1},\cdots,v_{k})-b(v_{1},\cdots,v_{i-1},w_{i},v_{i+1},\cdots,v_{k})$

where $1\leq i\leq k$,
$v_{1},\cdots,v_{i-1},v_{i},w_{i},v_{i+1},\cdots,v_{k}\in V$ and
$a,b\in\mathbb{R}$ together with all elements of $\Phi(V^{k})$ of the form

$(v_{1},\cdots,v_{i},\cdots,v_{j},\cdots,v_{k})$

where $1\leq i<j\leq k$, $v_{1},\cdots,v_{i},\cdots,v_{j},\cdots,v_{k}\in V$
and $v_{i}=v_{j}$.222For a set $X$, a vector space $\Phi(X)$ with basis $X$
can be obtained by setting $\Phi(X)$ equal to the set of all functions
$\phi:X\rightarrow\mathbb{R}$ such that $\\{x\in X:\phi(x)\neq 0\\}$ is
finite, and endowing $\Phi(X)$ with the operations of pointwise addition and
pointwise scalar multiplication. For each $x\in X$, define
$\phi_{x}\in\Phi(X)$ by $\phi_{x}(x)=1$ and $\phi_{x}(y)=0$ for $y\in
X-\\{x\\}$. Then $\\{\phi_{x}:x\in X\\}$ is a basis for $\Phi(X)$. By
identifying each $x\in X$ with $\phi_{x}\in\Phi(X)$, we make $X$ a basis for
$\Phi(X)$.

We make two additional remarks about $\Lambda_{k}(V)$.

######  Remark 1.

The uniqueness assertion in the universality property stated in the definition
of $\Lambda_{k}(V)$ implies that the set $\textit{i}(V^{k})=\\{v_{1}\wedge
v_{2}\wedge\cdots\wedge v_{k}:(v_{1},v_{2},\cdots,v_{k})\in V^{k}\\}$
generates $\Lambda_{k}(V)$.

######  Remark 2.

Suppose $(u_{1},u_{2},\cdots,u_{n})$ is an ordered basis for $V$ and $1\leq
k\leq n$. For $J\in\mathcal{S}(n,k)$, let $1\leq j_{1}<j_{2}<\cdots<j_{k}\leq
n$ so that $J=\\{j_{1},j_{2},\cdots,j_{k}\\}$, let
$\textbf{u}_{J}=(u_{j_{1}},u_{j_{2}},\cdots,u_{j_{k}})$ and let
$\wedge_{J}\textbf{u}=\textit{i}(\textbf{u}_{J})=u_{j_{1}}\wedge
u_{j_{2}}\wedge\cdots\wedge u_{j_{k}}$. Then
$\\{\wedge_{J}\textbf{u}:J\in\mathcal{S}(n,k)\\}$ is a basis for
$\Lambda_{k}(V)$. Hence, if $1\leq k\leq n$ and $dim(V)=n$, then
$dim(\Lambda_{k}(V))=\binom{n}{k}$. The universality property is used to prove
that $\\{\wedge_{J}\textbf{u}:J\in\mathcal{S}(n,k)\\}$ is linearly
independent.

Now suppose that $V$ is an $n$-dimensional inner product space with inner
product $\langle\;,\;\rangle$ and $1\leq k\leq n$. We will define an inner
product $\langle\langle\;,\;\rangle\rangle$ on $\Lambda_{k}(V)$ with the
property that if $(u_{1},u_{2},\cdots,u_{n})$ is an ordered orthonormal basis
for $V$, then $\\{\wedge_{J}\textbf{u}:J\in\mathcal{S}(n,k)\\}$ is an
orthonormal basis for $\Lambda_{k}(V)$. Define $\prec\;,\;\succ:V^{k}\times
V^{k}\rightarrow\mathbb{R}$ by
$\prec(v_{1},v_{2},\cdots,v_{k}),(w_{1},w_{2},\cdots,w_{k})\succ\;=Det(\langle
v_{i},w_{j}\rangle)$. Observe that $\prec\;,\;\succ$ is multilinear in the
sense that it is linear in each of the variables
$v_{1},v_{2},\cdots,v_{k},w_{1},w_{2},\cdots,w_{k}$. Also observe that
$\prec\;,\;\succ$ is _bi-alternating_ ; in other words, $\prec\;,\;\succ$ is
alternating separately in the variables $v_{1},v_{2},\cdots,v_{k}$ and in the
variables $w_{1},w_{2},\cdots,w_{k}$; that is,
$\prec(v_{1},v_{2},\cdots,v_{k}),(w_{1},w_{2},\cdots,w_{k})\succ\;=0$ whenever
either $v_{i}=v_{j}$ or $w_{i}=w_{j}$ for $1\leq i<j\leq k$. Further observe
that if $(u_{1},u_{2},\cdots,u_{n})$ is an ordered basis for $V$, then for
$J,K\in\mathcal{S}(n,k)$: $\prec\textbf{u}_{J},\textbf{u}_{K}\succ\;=1$ if
$J=K$ and $\prec\textbf{u}_{J},\textbf{u}_{K}\succ\;=0$ if $J\neq K$. To
“lift” $\prec\;,\;\succ$ to an inner product on $\Lambda_{k}(V)$, we require
the following generalization of the universality property.

######  Lemma 2.2.

If $\alpha:V^{k}\times V^{k}\rightarrow W$ is a multilinear bi-alternating map
to a vector space W, then there is a unique bilinear map
$\beta:\Lambda_{k}(V)\times\Lambda_{k}(V)\rightarrow W$ such that
$\beta\circ(\textit{i}\times\textit{i})=\alpha$.

###### Proof.

For each $\textbf{w}\in V^{k}$, define $\alpha_{\textbf{w}}:V_{k}\rightarrow
W$ by $\alpha_{\textbf{w}}(\textbf{v})=\alpha(\textbf{v},\textbf{w})$. Then
each $\alpha_{\textbf{w}}$ is a multilinear alternating map. Hence, for each
$\textbf{w}\in V^{k}$, the universality property provides a unique linear map
$\phi_{\textbf{w}}:\Lambda_{k}(V)\rightarrow W$ such that
$\phi_{\textbf{w}}\circ\textit{i}=\alpha_{\textbf{w}}$. Let
$\mathcal{L}(\Lambda_{k}(V),W)$ denote the set of all linear maps from
$\Lambda_{k}(V)$ to $W$. Regard $\mathcal{L}(\Lambda_{k}(V),W)$ as a vector
space with respect to the operations of pointwise addition and pointwise
scalar multiplication. Define
$\Phi:V^{k}\rightarrow\mathcal{L}(\Lambda_{k}(V),W)$ by
$\Phi(\textbf{w})=\phi_{\textbf{w}}$. It is easy to verify that $\Phi$ is
multilinear and alternating. Therefore the universality property provides a
unique linear map
$\Psi:\Lambda_{k}(V)\rightarrow\mathcal{L}(\Lambda_{k}(V),W)$ such that
$\Psi\circ\textit{i}=\Phi$. Define
$\beta:\Lambda_{k}(V)\times\Lambda_{k}(V)\rightarrow W$ by
$\beta(x,y)=\Psi(y)(x)$. $\beta$ is clearly bilinear and clearly satisfies
$\beta\circ(\textit{i}\times\textit{i})=\alpha$. The uniqueness of $\beta$
follows from the fact that $\textit{i}(V^{k})$ generates $\Lambda_{k}(V).$ ∎

This lemma implies that there is a bilinear map
$\langle\langle\;,\;\rangle\rangle:\Lambda_{k}(V)\times\Lambda_{k}(V)\rightarrow\mathbb{R}$
such that
$\langle\langle\;,\;\rangle\rangle\circ(\textit{i}\times\textit{i})=\prec\;,\;\succ$.
Hence, if $(u_{1},u_{2},\cdots,u_{n})$ is an ordered orthonormal basis for
$V$, then for $J,K\in\mathcal{S}(n,k)$:
$\langle\langle\wedge_{J}\textbf{u},\wedge_{K}\textbf{u}\rangle\rangle=\;\prec\textbf{u}_{J},\textbf{u}_{K}\succ\;=1$
if $J=K$ and
$\langle\langle\wedge_{J}\textbf{u},\wedge_{K}\textbf{u}\rangle\rangle=\;\prec\textbf{u}_{J},\textbf{u}_{K}\succ\;=0$
if $J\neq K$. Thus, $\\{\wedge_{J}\textbf{u}:J\in\mathcal{S}(n,k)\\}$ is
orthonormal with respect to $\langle\langle\;,\;\rangle\rangle$. It remains to
prove that $\langle\langle\;,\;\rangle\rangle$ is positive definite. This
follows easily because $\\{\wedge_{J}\textbf{u}:J\in\mathcal{S}(n,k)\\}$ is a
basis for $\Lambda_{k}(V)$ that is orthonormal with respect to
$\langle\langle\;,\;\rangle\rangle$ and because
$\langle\langle\;,\;\rangle\rangle$ is bilinear. These observations combined
with the Remark 2 above tell us that
$\\{\wedge_{J}\textbf{u}:J\in\mathcal{S}(n,k)\\}$ is an orthonormal basis for
$\Lambda_{k}(V)$ with respect to the inner product
$\langle\langle\;,\;\rangle\rangle$.

We finish this section by establishing some matrix notation that will be used
later. Suppose $\textbf{u}=(u_{1},u_{2},\cdots,u_{k})$ is an ordered basis for
a vector space $V$ and $\textbf{v}=(v_{1},v_{2},\cdots,v_{m})$ is an m-tuple
of elements of $V$. Let $C_{\textbf{u}}(\textbf{v})$ be the $m\times k$ matrix
$(c_{i,j})$ determined by the equations

$v_{i}=\sum_{1\leq j\leq k}c_{i,j}u_{j}$

for $1\leq i\leq m$, and call $C_{\textbf{u}}(\textbf{v})$ the _coordinate
matrix of v with respect to u_. (The $i^{th}$ row of
$C_{\textbf{u}}(\textbf{v})$ lists the coordinates of $v_{i}$ with respect to
u.) Note that since $u_{1},u_{2},\cdots,u_{k}$ are linearly independent, then
$C_{\textbf{u}}(\textbf{v})$ is uniquely determined by u and v. Observe that
$C_{\textbf{u}}(\textbf{u})$ is the $k\times k$ identity matrix.

Suppose $\textbf{u}=(u_{1},u_{2},\cdots,u_{k})$ and
$\textbf{v}=(v_{1},v_{2},\cdots,v_{m})$ are ordered bases for vector spaces
$U$ and $V$, respectively, and $L:U\rightarrow V$ is a linear map. Let
$C_{\textbf{u},\textbf{v}}(L)$ be the $k\times m$ matrix $(a_{i,j})$
determined by the equations

$L(u_{i})=\sum_{1\leq j\leq m}a_{i,j}v_{j}$

for $1\leq i\leq k$, and call $C_{\textbf{u},\textbf{v}}(L)$ the _coordinate
matrix of $L$ with respect to u and v_. (The $i^{th}$ row of
$C_{\textbf{u},\textbf{v}}(L)$ lists the coordinates of $L(\textbf{u}_{i})$
with respect to v.) As above, $C_{\textbf{u},\textbf{v}}(L)$ is uniquely
determined by u, v and $L$ because $v_{1},v_{2},\cdots,v_{m}$ are linearly
independent. We state without proof:

######  Proposition 2.3.

If $\textbf{u}=(u_{1},u_{2},\cdots,u_{k})$ and
$\textbf{v}=(v_{1},v_{2},\cdots,v_{m})$ are ordered bases for vector spaces
$U$ and $V$, respectively, $L:U\rightarrow V$ is a linear map,
$\textbf{w}=(w_{1},w_{2},\cdots,w_{r})$ is an $r$-tuple of elements of $U$,
and

$L(\textbf{w})=(L(w_{1}),L(w_{2}),\cdots,L(w_{r})),$

then
$C_{\textbf{v}}(L(\textbf{w}))=C_{\textbf{u}}(\textbf{w})C_{\textbf{u},\textbf{v}}(L)$.
In particular, if $V=U$ and $\textbf{w}=\textbf{v}=\textbf{u}$, then
$C_{\textbf{u}}(L(\textbf{u}))=C_{\textbf{u}}(\textbf{u})C_{\textbf{u},\textbf{u}}(L)=C_{\textbf{u},\textbf{u}}(L)$.
∎

If $A=(a_{i,j})$ is a $k\times m$ matrix, recall that the _transpose_ of $A$
is the $m\times k$ matrix $A^{T}=(a_{i,j}^{T})$ defined by
$a_{i,j}^{T}=a_{j,i}$ for $1\leq i\leq m,1\leq j\leq k$. Also recall that if
$A$ is a $k\times k$ matrix, then $Det(A^{T})=Det(A)$.

##  3\. Measure theoretic preliminaries

$k$-dimensional Lebesgue measure is typically developed for subsets of
$\mathbb{E}^{k}$. For the purposes of this article, it is more useful to
formulate $k$-dimensional Lebesgue measure in a slightly more general context
- for subsets of $k$-dimensional inner product spaces. Therefore, we will
state the basic definitions and relevant fundamental results of measure theory
in this more general setting. The proofs of the theorems about Lebesgue
measure stated in this section, at least in the setting of $\mathbb{E}^{k}$,
can be found in the standard texts on measure theory, and references to these
proofs are provided. The process of generalizing these proofs to the setting
of inner product spaces is, for the most part, routine and is left to the
reader.

A collection $\mathcal{M}$ of subsets of a set $X$ is a _$\sigma$ -algebra_ on
$X$ if it satisfies the following three properties: i)
$\emptyset\in\mathcal{M}$. ii) $\mathcal{M}$ is _closed under the formation of
countable unions;_ i.e., if $\mathcal{A}$ is a countable subcollection of
$\mathcal{M}$, then $\bigcup\mathcal{A}\in\mathcal{M}$. iii) $\mathcal{M}$ is
_closed under the formation of complements;_ i.e., if $A\in\mathcal{M}$, then
$X-A\in\mathcal{M}$. It follows easily that every $\sigma$-algebra
$\mathcal{M}$ on X also satisfies the following properties: iv) $\mathcal{M}$
is _closed under the formation of differences;_ i.e., if $A,B\in\mathcal{M}$,
then $A-B\in\mathcal{M}$. v) $\mathcal{M}$ is _closed under the formation of
countable intersections;_ i.e., if $\mathcal{A}$ is a countable subcollection
of $\mathcal{M}$, then $\bigcap\mathcal{A}\in\mathcal{M}$.

A _measure_ on a set $X$ is a function $\mu:\mathcal{M}\rightarrow[0,\infty]$
where $\mathcal{M}$ is a $\sigma$-algebra on $X$ and $\mu$ satisfies the
following two properties: i) $\mu(\emptyset)=0$. ii) $\mu$ is _countably
additive_ ; i.e., if $\mathcal{A}$ is a countable pairwise disjoint
subcollection of $\mathcal{M}$, then
$\mu(\bigcup\mathcal{A})=\sum_{A\in\mathcal{A}}\mu(A)$. It is easily proved
that every measure $\mu$ satisfies the following additional properties: iii)
$\mu$ is _monotone_ ; i.e., if $A,B\in\mathcal{M}$ and $A\subset B$, then
$\mu(A)\leq\mu(B)$. iv) $\mu$ is _countably subadditive_ ; i.e., if
$\mathcal{A}$ is a countable (not necessarily pairwise disjoint) subcollection
of $\mathcal{M}$, then
$\mu(\bigcup\mathcal{A})\leq\sum_{A\in\mathcal{A}}\mu(A)$. v) If $A_{1}\subset
A_{2}\subset A_{3}\subset\cdots$ is an increasing sequence of elements of
$\mathcal{M}$, then $\mu(\bigcup_{i\geq 1}A_{i})=sup\\{\mu(A_{i}):i\geq 1\\}$.
vi) If $A_{1}\supset A_{2}\supset A_{3}\supset\cdots$ is a decreasing sequence
of elements of $\mathcal{M}$ and $\mu(A_{1})<\infty$, then $\mu(\bigcap_{i\geq
1}A_{i})=inf\\{\mu(A_{i}):i\geq 1\\}$.

A measure $\mu:\mathcal{M}\rightarrow[0,\infty]$ on a set $X$ is _complete_ if
every subset of a measure 0 element of $\mathcal{M}$ is an element of
$\mathcal{M}$. A simple argument shows that if
$\mu:\mathcal{M}\rightarrow[0,\infty]$ is a complete measure on $X$, then for
$A\subset B\subset C\subset X$, if $A,C\in\mathcal{M}$ and $\mu(A)=\mu(C)$,
then $B\in\mathcal{M}$.

A $\sigma$-algebra $\mathcal{M}$ on a topological space $X$ is called a _Borel
$\sigma$-algebra_ if $\mathcal{M}$ contains every open subset of X. A measure
$\mu:\mathcal{M}\rightarrow[0,\infty]$ on a topological space is called a
_Borel measure_ if $\mathcal{M}$ us a Borel $\sigma$-algebra.

A measure $\mu:\mathcal{M}\rightarrow[0,\infty]$ on a topological space $X$ is
_regular_ if $\mathcal{M}$ is a Borel $\sigma$-algebra and for every
$A\in\mathcal{M}$,

$\mu(A)=inf\\{\mu(U):U\text{ is an open subset of }X\text{ and }A\subset U\\}$
$=sup\\{\mu(C):C\text{ is a closed subset of }X\text{ and }C\subset A\\}.$

A measure $\mu:\mathcal{M}\rightarrow[0,\infty]$ on a vector space $V$ is
_translation invariant_ if for every $A\in\mathcal{M}$ and every $p\in V$,
$\mu(T_{p}(A))=\mu(A)$. (Recall that $T_{p}:V\rightarrow V$ is the translation
specified by $T_{p}(x)=x+p$.)

A _pseudometric_ on a set $X$ is a function $\rho:X\times
X\rightarrow[0,\infty]$ which for all $x,y\text{ and }z\in X$ has the
following three properties: i) $\rho(x,x)=0$. ii) $\rho(x,y)=\rho(y,x)$. iii)
$\rho$ satisfies the _triangle inequality_ ; i.e.,
$\rho(x,z)\leq\rho(x,y)+\rho(y,z)$. If $\rho$ is a pseudometric on $X$, then
for every $x\in X$ and every $\epsilon>0$, let
$\mathcal{N}_{\rho}(x,\epsilon)=\\{y\in X:\rho(x,y)<\epsilon\\}$. Then
$\\{\mathcal{N}_{\rho}(x,\epsilon):x\in X\text{ and }\epsilon>0\\}$ is a basis
for a topology on $X$ called the _topology on $X$ determined by $\rho$_. If
$\rho$ is a pseudometric on a set $X$, then the pair $(X,\rho)$ is called a
_pseudometric space_.

For two sets $A$ and $B$, the _symmetric difference of $A$ and $B$_ is the set
$A\Delta B=(A-B)\cup(B-A)$.

Let $\mu:\mathcal{M}\rightarrow[0,\infty]$ be a measure on a set $X$. A
pseudometric $\rho$ on $\mathcal{M}$, called the _symmetric difference
pseudometric associated with $\mu$_, is defined by $\rho(A,B)=\mu(A\Delta B)$.
(The verification that $\rho$ is a pseudometric on $\mathcal{M}$ is
straightforward. The verification of the triangle inequality depends on the
fact that for $A,B,C\in\mathcal{M},A\Delta C\subset(A\Delta B)\cup(B\cup C)$.)
We note that $\rho$ satifies the inequality $|\mu(A)-\mu(B)|\leq\rho(A\Delta
B)$. (_Proof._ $|\mu(A)-\mu(B)|=|(\mu(A-B)+\mu(A\cap B))-(\mu(B-A)+\mu(A\cap
B))|=|\mu(A-B)-\mu(B-A)|\leq\mu(A-B)+\mu(B-A)=\rho(A,B)$. ∎) It follows that
if $\mathcal{M}$ is equipped with the topology determined by $\rho$, then
$\mu:\mathcal{M}\rightarrow[0,\infty]$ is continuous.

Let $V$ be a $k$-dimensional inner product space, let
$\textbf{u}=(u_{1},u_{2},\cdots,u_{k})$ be an ordered orthonormal basis for
$V$, and let $[\textbf{u}]=[u_{1},u_{2},\cdots,u_{k}]=\\{\sum_{1\leq i\leq
k}t_{i}u_{i}:0\leq t_{i}\leq 1\text{ for }1\leq i\leq k\\}$. A measure
$\mu:\mathcal{M}(V,\textbf{u})\rightarrow[0,\infty]$ is called a _$k$
-dimensional Lebesgue u-measure on V_ if $\mu$ is a complete, Borel, regular,
translation invariant measure such that $\mu([\textbf{u}])=1$.

A measure $\mu:\mathcal{M}(V,\textbf{u})\rightarrow[0,\infty]$ on a set X is
_$\sigma$ -finite_ if there is a countable subset $\mathcal{A}$ of
$\mathcal{M}$ such that $\bigcup\mathcal{A}=X$ and $\mu(A)<\infty$ for every
$A\in\mathcal{A}$. If $V$ is a $k$-dimensional inner product space and u is an
ordered orthonormal basis for $V$, then every $k$-dimensional Lebesgue
u-measure $\mu$ on $V$ is $\sigma$-finite. Indeed, if $P=\\{\sum_{1\leq i\leq
k}n_{i}u_{i}:n_{i}\text{ is an integer for }1\leq i\leq k\\}$, then $V$ is the
union of the countable collection $\\{T_{p}([\textbf{u}]):p\in P\\}$ and
$\mu(T_{p}([\textbf{u}]))=\mu([\textbf{u}])=1$ for each $p\in P$.

######  Existence Theorem 3.1.

If $V$ is a $k$-dimensional inner product space and u is an ordered
orthonormal basis for $V$, then a $k$-dimensional Lebesgue u-measure on $V$
exists.

The statement and a proof of the Existence Theorem can be cobbled together
from Theorems 20.1.11, 20.1.13 and Problem 20.2.20 on pages 426-427, 427-428
and 435, respectively, of [19].

######  Uniqueness Theorem 3.2.

Suppose $V$ is a $k$-dimensional inner product space, u is an ordered
orthonormal bases for $V$ and $\mu:\mathcal{M}\rightarrow[0,\infty]$ is a
$k$-dimensional Lebesgue u-measure on $V$. If
$\mu^{\prime}:\mathcal{M}^{\prime}\rightarrow[0,\infty]$ is a Borel,
translation invariant measure on V such that $\mu^{\prime}(U_{0})<\infty$ for
some non-empty open subset $U_{0}$ of $V$, then either $\mu^{\prime}=0$ or
there is a $\kappa\in(0,\infty)$ such that
$\mu^{\prime}|\mathcal{M}\cap\mathcal{M}^{\prime}=\kappa\mu|\mathcal{M}\cap\mathcal{M}^{\prime}$.
If, in addition, $\mu^{\prime}$ is non-zero, complete and regular, then
$\mathcal{M}^{\prime}=\mathcal{M}$.

A proof of the Uniqueness Theorem 3.2 can be derived, with some effort, from a
result known as the Carathéodory-Hahn Theorem (pages 356-357 of [19]).

We note two important consequences of the Uniqueness Theorem. First, if u is
an ordered orthonormal basis for a $k$-dimensional inner product space $V$ and
$\mu:\mathcal{M}\rightarrow[0,\infty]$ and
$\mu^{\prime}:\mathcal{M}^{\prime}\rightarrow[0,\infty]$ are both
$k$-dimensional Lebesgue u-measures on $V$, then
$\mathcal{M}=\mathcal{M}^{\prime}$ and $\mu=\mu^{\prime}$. Indeed, the
Uniqueness Theorem implies $\mathcal{M}^{\prime}=\mathcal{M}$ and
$\mu^{\prime}=\kappa\mu$ for some $\kappa\in(0,\infty)$. However,
$\kappa=\kappa\mu([\textbf{u}])=\mu^{\prime}([\textbf{u}])=1$. In other words,
for every ordered orthonormal basis u for V, there is _one and only one_
$k$-dimensional Lebesgue u-measure on $V$. Second, if u and
$\textbf{u}^{\prime}$ are ordered orthonormal bases for the $k$-dimensional
inner product space $V$, $\mu:\mathcal{M}\rightarrow[0,\infty]$ is a
$k$-dimensional Lebesgue u-measure on $V$ and
$\mu^{\prime}:\mathcal{M}^{\prime}\rightarrow[0,\infty]$ is a $k$-dimensional
Lebesgue u′-measure on $V$, then the Uniqueness Theorem implies
$\mathcal{M}=\mathcal{M}^{\prime}$. In other words, there is _one and only
one_ $\sigma$-algebra that can serve as the domain of a $k$-dimensional
Lebesgue u-measure on $V$ independent of the choice of the ordered orthonormal
basis u. This observation justifies the following terminology. If the
$\sigma$-algebra $\mathcal{M}$ on $V$ is the domain of a $k$-dimensional
Lebesgue u-measure on V for some orthonormal basis u for $V$, then we call the
elements of $\mathcal{M}$ the _Lebesgue measurable subsets_ of $V$ and we
denote $\mathcal{M}$ by $\mathcal{M}(V)$.

The Uniqueness Theorem alone does not tell us that all Lebesgue measures on a
finite dimensional inner product space are identical. More precisely, the
Uniqueness Theorem alone does not tell us that $\mu=\mu^{\prime}$ in the case
that $\mu:\mathcal{M}(V)\rightarrow[0,\infty]$ is a $k$-dimensional Lebesgue
u-measure on V and $\mu^{\prime}:\mathcal{M}(V)\rightarrow[0,\infty]$ is a
$k$-dimensional Lebesgue u′-measure on V where u and u′ are _distinct_ ordered
orthonormal bases for $V$. To reach that conclusion, we also need the
following result.

######  Theorem 3.3.

Suppose $V$ and $W$ are $k$-dimensional inner product spaces with ordered
orthonormal bases v and w, respectively,
$\lambda:\mathcal{M}(V)\rightarrow[0,\infty]$ is a $k$-dimensional Lebesgue
v-measure on V and $\mu:\mathcal{M}(W)\rightarrow[0,\infty]$ is a
$k$-dimensional Lebesgue w-measure on W. If $f:V\rightarrow W$ is an isometry,
then for every $A\in\mathcal{M}(V)$, $f(A)\in\mathcal{M}(W)$ and
$\mu(f(A))=\lambda(A)$.

A version of this result for isometries of $\mathbb{E}^{n}$ appears as
Corollary 20.2.24 on page 435 of [19]. The proof of this theorem has two
parts. First, the statement that an isometry sends elements of
$\mathcal{M}(V)$ to elements of $\mathcal{M}(W)$ can be deduced from the facts
that isometries satisfy a Lipschitz condition (defined on page 216 of [19])
and that a function from $V$ to $W$ which satisfies a Lipschitz condition
sends elements of $\mathcal{M}(V)$ to elements of $\mathcal{M}(W)$. (See
Proposition 20.2.20 on page 432 of [19].) Second, the assertion that an
isometry preserves measure ($\mu(f(A))=\lambda(A)$ for all
$A\in\mathcal{M}(V)$) is proved in [19] in a way that relies on properties of
the Lebesgue integral. We will present a different proof here that avoids
invoking the Lebesgue integral and instead depends on the Uniqueness Theorem
3.2.

###### Proof of Theorem 3.3.

First, consider the case in which $W=V$, $\textbf{w}=\textbf{v}$ and
$f:V\rightarrow V$ is an isometry such that $f(0_{V})=0_{V}$ where $0_{V}$ is
the additive identity of $V$. An elementary argument shows that any isometry
$f:V\rightarrow W$ which sends $0_{V}$ to $0_{W}$ is linear. (First observe
that $||f(x)||_{W}=||x||_{V}$ for all $x\in V$. Next, show that $\langle
f(x),f(y)\rangle_{W}=\langle x,y\rangle_{V}$ for all $x,y\in V$ by expanding
both sides of the equation $(||f(x)-f(y)||_{W})^{2}=(||x-y||_{V})^{2}$. Third,
show that $f(ax+by)-af(x)-bf(y)=0$ for all $x,y\in V$ and all
$a,b\in\mathbb{R}$ by expanding $(||f(ax+by)-af(x)-bf(y)||_{W})^{2}$.) Define
the function $\lambda^{\prime}:\mathcal{M}(V)\rightarrow[0,\infty]$ by
$\lambda^{\prime}(A)=\lambda(f(A))$. Using that fact that $f$ is a linear
function and a homeomorphism, one verifies that $\lambda^{\prime}$ is a non-
zero translation invariant measure. (Proposition 2.1 is helpful in proving
that $\lambda^{\prime}$ is translation invariant.) Hence, the Uniqueness
Theorem provides a $\kappa\in(0,\infty)$ such that
$\lambda^{\prime}=\kappa\lambda$. If $U=\\{x\in V:||x||_{V}<k\\}$, then
$f(U)=U$ because f is an isometry that fixes $0_{V}$, and $\lambda(U)>0$
because $[\textbf{v}]\subset U$. Hence,
$\kappa\lambda(U)=\lambda^{\prime}(U)=\lambda(f(U))=\lambda(U)$. Consequently,
$\kappa=1$. Therefore, $\lambda^{\prime}=\lambda$. We conclude that
$\lambda(f(A))=\lambda(A)$ for all $A\in\mathcal{M}(V)$.

Second, consider the case in which $W=V,\textbf{w}=\textbf{v},\mu=\lambda$ and
$f:V\rightarrow V$ is an isometry such that $f(0_{V})=p\neq 0_{V}$. Let
$T_{-p}:V\rightarrow V$ denote the translation $T_{-p}(x)=x-p$. Let
$A\in\mathcal{M}(V)$. Since $\lambda$ is translation invariant, then
$\lambda(f(A))=\lambda(T_{-p}(f(A))$ and since $T_{-p}\circ f:V\rightarrow V$
is an isometry such that $T_{-p}\circ f(0_{V})=0_{V}$, then
$\lambda(T_{-p}\circ f(A))=\lambda(A)$. Hence, $\lambda(f(A))=\lambda(A)$.

Finally, consider the general case in which $W\neq V$. Suppose
$\textbf{v}=(v_{1},v_{2},\cdots,v_{k})$ and
$\textbf{w}=(w_{1},w_{2},\cdots,w_{k})$. Then a linear isometry
$g:W\rightarrow V$ is determined by the equation $g(\sum_{1\leq i\leq
k}t_{i}w_{i})=\sum_{1\leq i\leq k}t_{i}v_{i}$. Observe that
$g([\textbf{w}])=[\textbf{v}]$. Define the function
$\mu^{\prime}:\mathcal{M}(W)\rightarrow[0,\infty]$ by
$\mu^{\prime}(B)=\lambda(g(B))$. As in the first paragraph of this proof, one
can verify that $\mu^{\prime}$ is a non-zero translation invariant measure.
Hence, the Uniqueness Theorem provides a $\kappa\in(0,\infty)$ such that
$\mu^{\prime}=\kappa\mu$. Therefore,
$\kappa=\kappa\mu([\textbf{w}])=\mu^{\prime}([\textbf{w}])=\lambda(g([\textbf{w}]))=\lambda([\textbf{v}])=1$.
Thus, $\mu^{\prime}=\mu$. Consequently, for $A\in\mathcal{M}(V)$,
$\mu(f(A))=\mu^{\prime}(f(A))=\lambda(g\circ f(A))$. Since $g\circ
f:V\rightarrow V$ is an isometry, then $\lambda(g\circ f(A))=\lambda(A)$. We
conclude that $\mu(f(A))=\lambda(A)$. ∎

Theorem 3.3 allows us to settle the issue of the relationship between two
$k$-dimensional Lebesgue measures arising from different ordered orthonormal
bases for the same inner product space.

######  Corollary 3.4.

If u and u′ are ordered orthonormal bases for a $k$-dimensional inner product
space $V$, $\mu:\mathcal{M}(V)\rightarrow[0,\infty]$ is a $k$-dimensional
Lebesgue u-measure on $V$ and
$\mu^{\prime}:\mathcal{M}(V)\rightarrow[0,\infty]$ is a $k$-dimensional
Lebesgue $\textbf{u}^{\prime}$-measure on V, then $\mu=\mu^{\prime}$.

###### Proof.

Since the identity function from $V$ to itself is an isometry, then Theorem
3.3 implies $\mu^{\prime}(A)=\mu(A)$ for every $A\in\mathcal{M}(V)$. ∎

From Corollary 3.4 we conclude that there is _one and only one_
$k$-dimensional Lebesgue u-measure on a $k$-dimensional inner product space
$V$, independent of the choice of ordered orthonormal basis u. We will
subsequently call this measure _$k$ -dimensional Lebesgue measure on $V$_ and
denote it by $Vol_{k}:\mathcal{M}(V)\rightarrow[0,\infty]$.

Theorem 3.3 reveals that isometries between $k$-dimensional inner product
spaces preserve Lebesgue measure. We need one other result that describes how
functions between $k$-dimensional inner product spaces affect Lebesgue
measure.

######  Theorem 3.5.

Suppose $V$ and $W$ are $k$-dimensional inner product spaces and
$Vol_{k}:\mathcal{M}(V)\rightarrow[0,\infty]$ and
$Vol_{k}:\mathcal{M}(W)\rightarrow[0,\infty]$ are the $k$-dimensional Lebesgue
measures on $V$ and $W$, respectively. If $L:V\rightarrow W$ is a linear
function, then for every $A\in\mathcal{M}(V)$, $L(A)\in\mathcal{M}(W)$.
Furthermore, if v and w are ordered orthonormal bases for $V$ and $W$,
respectively, then for every $A\in\mathcal{M}(V)$,

$Vol_{k}(L(A))=|Det(C_{\textbf{v},\textbf{w}}(L))|\;Vol_{k}(A).$

Since linear functions between finite dimensional inner product spaces satisfy
a Lipschitz condition (Proposition 20.2.17 on page 430 of [19]) and since a
function which satisfies a Lipschitz condition sends Lebesgue measurable sets
to Lebesgue measurable sets (Proposition 20.2.18 on page 431 of [19]), then
every linear function between finite dimensional inner product spaces $V$ and
$W$ sends elements of $\mathcal{M}(V)$ to elements of $\mathcal{M}(W)$. A
statement and proof of the equation
$Vol_{k}(L(A))=|Det(C_{\textbf{v},\textbf{w}}(L))|\;Vol_{k}(A)$ for linear
functions from $\mathbb{E}^{n}$ to itself appears as Corollary 20.2.23 on page
434 of [19]. (Again the proof of this result in [19] relies on properties of
the Lebesgue integral. A simple proof which avoids the Lebesgue integral can
be given using a linear algebra theorem known as the _Polar Decomposition
Theorem_. For linear functions between finite dimensional inner product
spaces, this theorem has a proof which, though simple, is beyond the scope of
this article.

Suppose $\mathcal{M}$ is a $\sigma$-algebra on a set $X$, $\mathcal{N}$ is a
$\sigma$-algebra on a set $Y$ and $f:X\rightarrow Y$ is a function such that
$f(A)\in\mathcal{N}$ for all $A\in\mathcal{M}$. Then $f$ determines a function
from $\mathcal{M}$ to $\mathcal{N}$ which we denote
$f_{\ast}:\mathcal{M}\rightarrow\mathcal{N}$ and specify by the equation
$f_{\ast}(A)=f(A)$ for all $A\in\mathcal{M}$.

Suppose $(X,\rho)$ and $(Y,\sigma)$ are pseudometric spaces and
$f:X\rightarrow Y$ is a function. $f$ is an _isometry_ if $f(X)=Y$ and
$\sigma(f(x),f(x^{\prime}))=\rho(x,x^{\prime})$ for all $x,x^{\prime}\in X$.
$f$ is a _dilation with dilation constant $\delta$_ if $\delta\in[0,\infty)$
and $\sigma(f(x),f(x^{\prime}))=\delta\rho(x,x^{\prime})$ for all
$x,x^{\prime}\in X$. Note that isometries and dilations are continuous.

Observe that if $f:X\rightarrow Y$ is a function and $A$ and $B$ are subsets
of $X$, then $f(A)\Delta f(B)\subset f(A\Delta B)$, with equality holding if
$f$ is injective. Theorems 3.3 and 3.5 together with this observation yield
the following conclusions.

######  Corollary 3.6.

Suppose $V$ and $W$ are $k$-dimensional inner product spaces,
$Vol_{k}:\mathcal{M}(V)\rightarrow[0,\infty]$ and
$Vol_{k}:\mathcal{M}(W)\rightarrow[0,\infty]$ are the $k$-dimensional Lebesgue
measures on $V$ and $W$, respectively, and $\mathcal{M}(V)$ and
$\mathcal{M}(W)$ are assigned the symmetric difference pseudometrics
associated with these Lebesgue measures.

i) If $f:V\rightarrow W$ is an isometry, then
$f_{\ast}:\mathcal{M}(V)\rightarrow\mathcal{M}(W)$ is also an isometry.

ii) If $L:V\rightarrow W$ is a linear function and v and w are ordered
orthonormal bases for $V$ and $W$, respectively, then
$L_{\ast}:\mathcal{M}(V)\rightarrow\mathcal{M}(W)$ is a dilation with dilation
constant $|Det(C_{\textbf{v},\textbf{w}}(L)|$. ∎

Next we provide a technique for calculating $k$-dimensional Lebesgue measure
in a $k$-dimensional inner product space $V$. Suppose u =
$(u_{1},u_{2},\cdots,u_{k})$ is an ordered orthonormal basis for $V$. For real
numbers $b_{i}<b_{i}^{\prime}$ for $1\leq i\leq k$, we call any set of the
form

$B=\Biggl{\\{}\sum_{1\leq i\leq k}t_{i}u_{i}:b_{i}\leq t_{i}\leq
b_{i}^{\prime}\text{ for }1\leq i\leq k\Biggr{\\}}$

a _u -box_ and we define the _u -volume_ of $B$ to be

$vol_{\textbf{u}}(B)=\prod_{1\leq i\leq k}(b_{i}^{\prime}-b_{i}).$

Also let

$int(B)=\Biggl{\\{}\sum_{1\leq i\leq
k}t_{i}u_{i}:b_{i}<t_{i}<b_{i}^{\prime}\text{ for }1\leq i\leq k\Biggr{\\}}$

and note that $int(B)$ in non-empty. Let $\mathbb{B}(V,\textbf{u})$ denote the
set of all countable collections u-boxes in $V$. If
$\mathcal{B}\in\mathbb{B}(V,\textbf{u})$, we define the _u -volume sum of_
$\mathcal{B}$ to be

$vol_{\textbf{u}}sum(\mathcal{B})=\sum_{B\in\mathcal{B}}vol_{\textbf{u}}(B).$

We stipulate that $\mathbb{B}(V,\textbf{u})$ contains the empty collection
$\emptyset$ and that $vol_{\textbf{u}}sum(\emptyset)=0$. Observe that since
$\mathcal{M}(V)$ is a Borel $\sigma$-algebra and since all u-boxes are closed
subsets of $V$, then all u-boxes are elements of $\mathcal{M}(V)$ and
$\bigcup\mathcal{B}\in\mathcal{M}(V)$ for all
$\mathcal{B}\in\mathbb{B}(V,\textbf{u})$. The connection between countable
collections of u-boxes in $V$ and $k$-dimensional Lebesgue measure on $V$ is
the following result.

######  Theorem 3.7.

If u is an ordered orthonormal basis for a $k$-dimensional inner product space
$V$, then

$Vol_{k}(A)=inf\\{vol_{\textbf{u}}sum(\mathcal{B}):\mathcal{B}\in\mathbb{B}(V,\textbf{u})\text{
and }A\subset\bigcup\mathcal{B}\\}$.

The usual approach to constructing Lebesgue measure on a finite dimensional
inner product space $V$ is to define a function
$\mu^{\ast}:\mathcal{P}(V)\rightarrow[0,\infty]$ by the formula

$\mu^{\ast}(A)=inf\\{vol_{\textbf{u}}sum(\mathcal{B}):\mathcal{B}\in\mathbb{B}(V,\textbf{u})\text{
and }A\subset\bigcup\mathcal{B}\\}$.

Here $\mathcal{P}(V)=\\{A:A\subset V\\}$ is the _power set_ of $V$. The
function $\mu^{\ast}$ is called _Lebesgue u-outer measure on $V$_. Then a
subcollection $\mathcal{M}(V,\textbf{u})$ of $\mathcal{P}(V)$ is specified by
the formula

$\mathcal{M}(V,\textbf{u})=\\{A\in\mathcal{P}(V):\mu^{\ast}(B\cap
A)+\mu^{\ast}(B-A)=\mu^{\ast}(B)\text{ for all }B\in\mathcal{P}(V)\\}.$

The elements of $\mathcal{M}(V,\textbf{u})$ are called _Lebesgue u-measurable
subsets of $V$_. Next Carathéodory’s Theorem (Theorem 17.4.8 on page 34 of
[19]) is invoked to reveal that $\mathcal{M}(V,\textbf{u})$ is a
$\sigma$-algebra on $V$ and the restriction
$\mu^{\ast}|\mathcal{M}(V,\textbf{u})$ is a complete measure on $V$. Finally,
one shows that $\mu^{\ast}|\mathcal{M}(V,\textbf{u})$ is Borel, regular,
translation invariant $\mu^{\ast}([\textbf{u}])=1$; in other words,
$\mu^{\ast}|\mathcal{M}(V,\textbf{u})$ is a $k$-dimensional Lebesgue u-measure
on $V$. (See Theorems 20.2.11 and 20.2.13 and Problem 20.2.20 on pages
426-427, 427-428 and 435, respectively, of [19].) Then the Uniqueness Theorem
3.2 and Corollary 3.4 imply $\mathcal{M}(V,\textbf{u})=\mathcal{M}(V)$ and
$\mu^{\ast}|\mathcal{M}(V,\textbf{u})=Vol_{k}$, thereby yielding Theorem 3.7.
This approach to the construction of Lebesgue measure was pioneered in [4]. In
addition to [19], we mention [14] as a historically influential comprehensive
source for information about measure theory.

We again consider a $k$-dimensional inner product space $V$ and let u be an
ordered orthonormal basis for $V$. We introduce a special subcollection of
$\mathbb{B}(V,\textbf{u})$ that is quite useful in both the development of
Lebesgue measure and in the remainder of this paper. A countable collection
$\mathcal{B}$ of u-boxes in $V$ is _almost disjoint_ if
$\\{int(B):B\in\mathcal{B}\\}$ is pairwise disjoint. Let

$\mathbb{B}_{0}(V,\textbf{u})=\\{\mathcal{B}\in\mathbb{B}(V,\textbf{u}):\mathcal{B}\text{
is almost disjoint}\\}.$

The following lemmas state two basic properties of almost disjoint collections
of u-boxes.

######  Lemma 3.8.

Suppose $V$ is a $k$-dimensional inner product space and u is an ordered
orthonormal basis for $V$. If $\mathcal{B}\in\mathbb{B}_{0}(V,\textbf{u})$,
then
$Vol_{k}(\bigcup\mathcal{B})=vol_{\textbf{u}}sum(\mathcal{B})=\sum_{B\in\mathcal{B}}Vol_{k}(B)$.

######  Lemma 3.9.

Suppose $V$ is a $k$-dimensional inner product space and u is an ordered
orthonormal basis for $V$. For every open subset $U$ of $V$, there is a
$\mathcal{B}\in\mathbb{B}_{0}(V,\textbf{u})$ such that $\bigcup\mathcal{B}=U$.

These results appear in [20] as Lemmas 1.2.9 and 1.2.11 on pages 23-25.

The next result hints at the value of the concept of almost disjoint
collections of boxes. This theorem plays a key role in our proof of the
Product Formula for Volume.

######  Theorem 3.10.

Suppose $V$ is a $k$-dimensional inner product space and u is an ordered
orthonormal basis for $V$. Let $\rho$ be the symmetric difference pseudometric
on $\mathcal{M}(V)$ associated with $k$-dimensional Lebesgue measure
$Vol_{k}$. Then
$\\{\bigcup\mathcal{B}:\mathcal{B}\in\mathbb{B}_{0}(V,\textbf{u})\\}$ is a
dense subset of $\mathcal{M}(V)$ in the topology on $\mathcal{M}(V)$
determined by $\rho$.

###### Proof.

Lemma 3.9 tells us that it suffices to prove that the set of all open subsets
of $V$ is dense in $\mathcal{M}(V)$. To this end, let $A\in\mathcal{M}(V)$ and
let $\epsilon>0$. Since $Vol_{k}$ is $\sigma$-finite, there is a countable
subset $\\{A_{i}:i\geq 1\\}$ of $\mathcal{M}(V)$ such that $A=\bigcup_{i\geq
1}A_{i}$ and $Vol_{k}(A_{i})<\infty$ for each $i\geq 1$. Since $Vol_{k}$ is
regular, then for each $i\geq 1$, there is an open subset $U_{i}$ of $V$ such
that $A_{i}\subset U_{i}$ and $Vol_{k}(U_{i})<Vol_{k}(A_{i})+\epsilon/2^{i}$.
Hence, for each $i\geq 1$,
$Vol_{k}(U_{i}-A_{i})=Vol_{k}(U_{i})-Vol_{k}(A_{i})<\epsilon/2^{i}$. Let
$U=\bigcup_{i\geq 1}U_{i}$. Then $U$ is an open subset of $V$ such that
$A\subset U$. Furthermore, $U-A=\bigcup_{i\geq
1}(U_{i}-A)\subset\bigcup_{i\geq 1}(U_{i}-A_{i})$. Hence,

$\rho(A,U)=Vol_{k}(U-A)\leq\sum_{i\geq 1}Vol_{k}(U_{i}-A_{i})<\epsilon$. ∎

Suppose $V_{0}$ is a $k$-dimensional vector subspace of $\mathbb{E}^{n}$.
Clearly, the dot product on $\mathbb{E}^{n}$ restricts to an inner product on
$V_{0}$. Hence, there is an associated unique Borel $\sigma$-algebra
$\mathcal{M}(V_{0})$ of Lebesgue measurable subsets of $V_{0}$ and there is a
unique $k$-dimensional Lebesgue measure
$Vol_{k}:\mathcal{M}(V_{0})\rightarrow[0,\infty]$ on $V_{0}$. Now consider a
$k$-dimensional affine subspace $V$ of $\mathbb{E}^{n}$ such that
$V=T_{p}(V_{0})$ where $p\in\mathbb{E}^{n}$. If we restrict the Euclidean
metric on $\mathbb{E}^{n}$ (i.e., the metric on $\mathbb{E}^{n}$ associated
with the dot product) to $V$, then $T_{p}|V_{0}:V_{0}\rightarrow V$ becomes an
isometry. Hence, we feel justified in calling the set
$\\{T_{p}(A):A\in\mathcal{M}(V_{0})\\}$ the _$\sigma$ -algebra of Lebesgue
measurable subsets of $V$_, denoting this set by $\mathcal{M}(V)$, and in
defining _$k$ -dimensional Lebesgue measure_
$Vol_{k}:\mathcal{M}(V)\rightarrow[0,\infty]$ on $V$ by
$Vol_{k}(T_{p}(A))=Vol_{k}(A)$ for $A\in\mathcal{M}(V_{0})$. In further
support of these choices, we observe that the definitions of $\mathcal{M}(V)$
and $Vol_{k}:\mathcal{M}(V)\rightarrow[0,\infty]$ are independent of the
choice of the point $p\in\mathbb{E}^{n}$ such that $T_{p}(V_{0})=V$. For
suppose $p$ and $q\in\mathbb{E}^{n}$ such that $T_{p}(V_{0})=T_{q}(V_{0})=V$.
Then $q-p\in V_{0}$. Therefore, $T_{q-p}$ restricts to a translation of
$V_{0}$ which sends $\mathcal{M}(V)$ onto itself. Hence,
$\\{T_{q}(A):A\in\mathcal{M}(V_{0})\\}=\\{T_{p}\circ
T_{q-p}(A):A\in\mathcal{M}(V_{0})\\}=\\{T_{p}(A):A\in\mathcal{M}(V_{0})\\}$.
It follows that the definition of $\mathcal{M}(V)$ is independent of the
choice of $p$. Also if $A,B\in\mathcal{M}(V_{0})$ such that
$T_{p}(A)=T_{q}(B)$, then $T_{q-p}(B)=A$. The translation invariance of
$Vol_{k}:\mathcal{M}(V_{0})\rightarrow[0,\infty]$ then implies
$Vol_{k}(B)=Vol_{k}(A)$. Consequently, the definition of
$Vol_{k}:\mathcal{M}(V)\rightarrow[0,\infty]$ is independent of the choice of
$p$.

We continue to extend our terminology concerning Lebesgue measure to affine
subspaces of $\mathbb{E}^{n}$. Again, suppose $V$ is a $k$-dimensional affine
subspace of $\mathbb{E}^{n}$. Say $V=T_{p}(V_{0})$ where $V_{0}$ is a
$k$-dimensional vector subspace of $\mathbb{E}^{n}$ and $p\in\mathbb{E}^{n}$.
Echoing our previous definitions, if u is an ordered orthonormal basis for
$V_{0}$, we call a subset of $V$ a _u -box in $V$_ if it is of the form
$T_{p}(B)$ where $B$ is a u-box in $V_{0}$. Let $\mathbb{B}(V,\textbf{u})$
denote the set of all countable collections of u-boxes in $V$. As above, we
say that a collection $\mathcal{B}\in\mathbb{B}(V,\textbf{u})$ is _almost
disjoint_ if $\\{int_{V}(B):B\in\mathcal{B}\\}$ is pairwise disjoint (where
$int_{V}$ denotes the interior of a subset of $V$ in the subspace topology on
$V$ that it inherits as a subset of $\mathbb{E}^{n}$). We then let
$\mathbb{B}_{0}(V,\textbf{u})=\\{\mathcal{B}\in\mathbb{B}(V,\textbf{u}):\mathcal{B}\text{
is almost disjoint}\\}$. It follows that the properties of Lebesgue measure
asserted for $V_{0}$ by Corollary 3.6, Lemmas 3.8 and 3.9 and Theorem 3.10
effortlessly transfer to properties of $V$. In particular, in one of the
subsequent arguments, we will need:

######  Corollary 3.10′.

Suppose $V$ is a $k$-dimensional affine subspace of $\mathbb{E}^{n}$,
$V=T_{p}(V_{0})$ where $V_{0}$ is a $k$-dimensional vector subspace of
$\mathbb{E}^{n}$ and $p\in\mathbb{E}^{n}$, and u is an ordered orthonormal
basis for $V_{0}$. Then
$\\{\bigcup\mathcal{B}:\mathcal{B}\in\mathbb{B}_{0}(V,\textbf{u})\\}$ is a
dense subset of $\mathcal{M}(V)$ in the topology on $\mathcal{M}(V)$
determined by the symmetric difference pseudometric associated with $Vol_{k}$.
∎

We close this section with a result that will be used later.

######  Proposition 3.11.

Suppose $V$ is a $k$-dimensional affine subspace of $\mathbb{E}^{n}$, $W$ is a
$k$-dimensional inner product space and $L:\mathbb{E}^{n}\rightarrow W$ is a
linear function. Let $V_{0}$ be the $k$-dimensional vector subspace of
$\mathbb{E}^{n}$ such that $V=T_{p}(V_{0})$ where $p\in\mathbb{E}^{n}$, let u
be an ordered orthonormal basis for $V_{0}$, and let
$Vol_{k}:\mathcal{M}(W)\rightarrow[0,\infty]$ denote $k$-dimensional Lebesgue
measure on $W$. Then for every $\mathcal{B}\in\mathbb{B}_{0}(V,\textbf{u})$,
$Vol_{k}(L(\bigcup\mathcal{B})=\sum_{B\in\mathcal{B}}Vol_{k}(L(B))$.

###### Proof.

First consider the special case $V=V_{0}$. Then
$\mathcal{B}\in\mathbb{B}_{0}(V_{0},\textbf{u})$. Let w be an ordered
orthonormal basis for $W$. Then Theorem 3.5 and Lemma 3.7 imply

$Vol_{k}(L(\bigcup\mathcal{B}))=|Det(C_{\textbf{u},\textbf{w}}(L))|\;Vol_{k}(\bigcup\mathcal{B})=|Det(C_{\textbf{u},\textbf{w}}(L))|\;\sum_{B\in\mathcal{B}}Vol_{k}(B)$

$=\sum_{B\in\mathcal{B}}|Det(C_{\textbf{u},\textbf{w}}(L))|\;Vol_{k}(B)=\sum_{B\in\mathcal{B}}Vol_{k}(L(B))$.

Now consider the general case $V=T_{p}(V_{0})$. Given
$\mathcal{B}\in\mathbb{B}_{0}(V,\textbf{u})$, there is a
$\mathcal{B}_{0}\in\mathbb{B}_{0}(V_{0},\textbf{u})$ such that
$\mathcal{B}=\\{T_{p}(B):B\in\mathcal{B}_{0}\\}$. Since
$T_{p}|V_{0}:V_{0}\rightarrow V$ is a homeomorphism, then
$\mathcal{B}_{0}\in\mathbb{B}_{0}(V_{0},\textbf{u})$. Hence, the argument in
the previous paragraph shows that

$Vol_{k}(L(\bigcup\mathcal{B}_{0}))=\sum_{B\in\mathcal{B}_{0}}Vol_{k}(L(B))$.

Note that $T_{p}(\bigcup\mathcal{B}_{0})=\bigcup\mathcal{B}$. Using
Proposition 2.1 and the fact that
$Vol_{k}:\mathcal{M}(W)\rightarrow[0,\infty]$ is translation invariant, we
have

$Vol_{k}(L(\bigcup\mathcal{B}))=Vol_{k}(L\circ
T_{p}(\bigcup\mathcal{B}_{0}))=Vol_{k}(T_{L(p)}\circ
L(\bigcup\mathcal{B}_{0}))=$

$Vol_{k}(L(\bigcup\mathcal{B}_{0}))=\sum_{B\in\mathcal{B}_{0}}Vol_{k}(L(B))=\sum_{B\in\mathcal{B}_{0}}Vol_{k}(T_{L(p)}\circ
L(B))=$

$\sum_{B\in\mathcal{B}_{0}}Vol_{k}(L\circ
T_{p}(B))=\sum_{B\in\mathcal{B}}Vol_{k}(L(B))$. ∎

##  4\. The Proof of the Product Formula for Volume - for parallelopipeds

Let $V$ be a $k$-dimensional vector subspace of $\mathbb{E}^{n}$ and let
$\textbf{u}=(u_{1},u_{2},\cdots,u_{k})$ be an ordered orthonormal basis for
$V$. If $\textbf{x}=(x_{1},x_{2},\cdots,x_{k})\in V^{k}$, let
$[\textbf{x}]=[x_{1},x_{2},\cdots,x_{k}]$ denote the parallelopiped
$\\{\sum_{1\leq i\leq k}t_{i}x_{i}:0\leq t_{i}\leq 1\text{ for }1\leq i\leq
k\\}$. Then $Vol_{k}([\textbf{u}])=1$ because $Vol_{k}$ coincides with
$k$-dimensional Lebesgue u-measure on $V$ by Corollary 3.4.

Suppose $\textbf{x}=(x_{1},x_{2},\cdots,x_{k})\in V^{k}$ and define the linear
map $L:V\rightarrow V$ by the equations $L(u_{i})=x_{i}$ for $1\leq i\leq k$.
Then
$L(\textbf{u})=(L(u_{1}),L(u_{2}),\cdots,L(u_{k}))=(x_{1},x_{2},\cdots,x_{k})=\textbf{x}$
and $L([\textbf{u}])=[\textbf{x}]$. Proposition 2.3 implies
$C_{\textbf{u},\textbf{u}}(L)=C_{\textbf{u}}(L(\textbf{u}))\\\
=C_{\textbf{u}}(\textbf{x})$. Hence, Theorem 3.5 implies

$(\alpha)\hskip
39.83385ptVol_{k}([\textbf{x}])=|Det(C_{\textbf{u},\textbf{u}}(L))|Vol_{k}([\textbf{u}])=|Det(C_{\textbf{u}}(\textbf{x})|$.

Now suppose $\textbf{x}=(x_{1},x_{2},\cdots,x_{k})$ and
$\textbf{y}=(y_{1},y_{2},\cdots,y_{k})\in V^{k}$. We will prove

$(\beta)\hskip 19.91692pt\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge
x_{k},y_{1}\wedge y_{2}\wedge\cdots\wedge
y_{k}\rangle\rangle=Vol_{k}([\textbf{x}])Vol_{k}([\textbf{y}])$.

For the purpose of proving the Product Formula for Volume for $A=[\textbf{x}]$
and $B=[\textbf{y}]$, the specific orders of the entries of x and y are
immaterial. So let us interchange $y_{1}$ and $y_{2}$ if necessary so that
$Det(C_{\textbf{u}}(\textbf{x}))$ and $Det(C_{\textbf{u}}(\textbf{y}))$ have
the same sign. The definition of $\langle\langle\;,\;\rangle\rangle$ implies

$\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge x_{k},y_{1}\wedge
y_{2}\wedge\cdots\wedge
y_{k}\rangle\rangle=Det(x_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}y_{i})$

Let $(a_{i,j})=C_{\textbf{u}}(\textbf{x})$ and
$(b_{i,j})=C_{\textbf{u}}(\textbf{y})$. Then $x_{i}=\sum_{1\leq r\leq
k}a_{i,r}u_{r}$ and $y_{i}=\sum_{1\leq s\leq k}b_{i,s}u_{s}$. Therefore,

$x_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}y_{j}=(\sum_{r}a_{i,r}u_{r})\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}(\sum_{s}b_{j,s}u_{s})=\sum_{r,s}a_{i,r}b_{j,s}(u_{r}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}u_{s})$

$=\sum_{r}a_{i,r}b_{j,r}=\sum_{r}a_{i,r}(b_{r,j}^{T})$.

Thus,
$(x_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}y_{j})=C_{\textbf{u}}(\textbf{x})(C_{\textbf{u}}(\textbf{y})^{T})$.
Hence,

$(\gamma)\hskip 9.95845pt\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge
x_{k},y_{1}\wedge y_{2}\wedge\cdots\wedge
y_{k}\rangle\rangle=Det(C_{\textbf{u}}(\textbf{x}))Det(C_{\textbf{u}}(\textbf{y}))$.

Since $Det(C_{\textbf{u}}(\textbf{x}))$ and $Det(C_{\textbf{u}}(\textbf{y}))$
have the same sign, then we have:

$(\delta)\hskip 3.55658pt\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge
x_{k},y_{1}\wedge y_{2}\wedge\cdots\wedge
y_{k}\rangle\rangle=|Det(C_{\textbf{u}}(\textbf{x}))||Det(C_{\textbf{u}}(\textbf{y}))|$.

$(\delta)$ and $(\alpha)$ clearly imply $(\beta)$.

Again suppose $\textbf{x}=(x_{1},x_{2},\cdots,x_{k})$ and
$\textbf{y}=(y_{1},y_{2},\cdots,y_{k})\in V^{k}$ with entries ordered so that
$Det(C_{\textbf{u}}(\textbf{x}))$ and $Det(C_{\textbf{u}}(\textbf{y}))$ have
the same sign. Let $J\in\mathcal{S}(n,k)$. Recall that
$\mathbb{E}^{J}=\\{(z_{1},z_{2},\cdots,z_{n})\in\mathbb{E}^{n}:z_{i}=0\text{
if }i\notin J\\}$ and $\pi_{J}:\mathbb{E}^{n}\rightarrow\mathbb{E}^{n}$ is
defined by $\pi_{J}(z_{1},z_{2},\cdots,z_{n})=(w_{1},w_{2},\cdots,w_{n})$
where $w_{i}=z_{i}$ if $i\in J$ and $w_{i}=0$ if $i\notin J$. Let
$\textbf{e}=(e_{1},e_{2},\cdots,e_{n})$ be the standard ordered orthonormal
basis for $\mathbb{E}^{n}$. Write $J=\\{j_{1},j_{2},\cdots,j_{k}\\}$ where
$1\leq j_{1}<j_{2}<\cdots<j_{k}\leq n$, let
$\textbf{e}_{J}=(e_{j_{1}},e_{j_{2}},\cdots,e_{j_{k}})$ and let
$\wedge_{J}\textbf{e}=\textit{i}(\textbf{e}_{J})=e_{j_{1}}\wedge
e_{j_{2}}\wedge\cdots\wedge e_{j_{k}}$. We now calculate $\langle\langle
x_{1}\wedge x_{2}\wedge\cdots\wedge x_{k},\wedge_{J}\textbf{e}\rangle\rangle$.
First observe that for $1\leq i\leq k$ and $1\leq t\leq k$,
$x_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}e_{j_{t}}=\pi_{J}(x_{i})\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}e_{j_{t}}$.
Hence, the definition of $\langle\langle\;,\;\rangle\rangle$ implies

$\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge
x_{k},\wedge_{J}\textbf{e}\rangle\rangle=Det(x_{j}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}e_{j_{t}})=Det(\pi_{J}(x_{i})\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}e_{j_{t}})$
$=\langle\langle\pi_{J}(x_{1})\wedge\pi_{J}(x_{2})\wedge\cdots\wedge\pi_{J}(x_{k}),\wedge_{J}\textbf{e}\rangle\rangle.$

Let
$\pi_{J}(\textbf{x})=(\pi_{J}(x_{1}),\pi_{J}(x_{2}),\cdots,\pi_{J}(x_{k}))$.
Since $\mathbb{E}^{J}$ is a $k$-dimensional vector subspace of
$\mathbb{E}^{n}$, and both $\pi_{J}(\textbf{x})$ and
$\textbf{e}_{j}\in(\mathbb{E}^{J})^{k}$, then we can apply $(\gamma)$ to
conclude that

$\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge
x_{k},\wedge_{J}\textbf{e}\rangle\rangle=Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{x})))Det(C_{\textbf{e}_{J}}(\textbf{e}_{J}))=Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{x}))).$

Similarly

$\langle\langle y_{1}\wedge y_{2}\wedge\cdots\wedge
y_{k},\wedge_{J}\textbf{e}\rangle\rangle=Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{y}))).$

Hence,

$(\epsilon)\hskip 43.53271pt\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge
x_{k},\wedge_{J}\textbf{e}\rangle\rangle\langle\langle y_{1}\wedge
y_{2}\wedge\cdots\wedge y_{k},\wedge_{J}\textbf{e}\rangle\rangle$

$=Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{x})))Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{y}))).$

We assert that
$Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{x})))Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{y})))\geq
0$. Indeed, since

$C_{\textbf{e}_{J}}(\pi_{J}(\textbf{x}))=C_{\textbf{u}}(\textbf{x})C_{\textbf{u},\textbf{e}_{J}}(\pi_{J})\text{
and
}C_{\textbf{e}_{J}}(\pi_{J}(\textbf{y}))=C_{\textbf{u}}(\textbf{y})C_{\textbf{u},\textbf{e}_{J}}(\pi_{J}),$

then

$Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{x})))Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{y})))=Det(C_{\textbf{u}}(\textbf{x}))Det(C_{\textbf{u}}(\textbf{y}))(Det(C_{\textbf{u},\textbf{e}_{J}}(\pi_{J})))^{2}.$

Since $Det(C_{\textbf{u}}(\textbf{x}))$ and $Det(C_{\textbf{u}}(\textbf{y}))$
have the same sign, our assertion follows. Hence, $(\epsilon)$ implies

$\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge
x_{k},\wedge_{J}\textbf{e}\rangle\rangle\langle\langle y_{1}\wedge
y_{2}\wedge\cdots\wedge y_{k},\wedge_{J}\textbf{e}\rangle\rangle$
$=|Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{x})))|\;|Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{y})))|.$

Since $[\pi_{J}(\textbf{x})]=\pi_{J}([\textbf{x}])$ and
$[\pi_{J}(\textbf{y})]=\pi_{J}([\textbf{y}])$, then $(\alpha)$ implies

$|Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{x})))|=Vol_{k}(\pi_{J}([\textbf{x}]))\text{
and
}|Det(C_{\textbf{e}_{J}}(\pi_{J}(\textbf{y})))|=Vol_{k}(\pi_{J}([\textbf{y}])).$

It follows that

$(\zeta)\hskip 43.53271pt\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge
x_{k},\wedge_{J}\textbf{e}\rangle\rangle\langle\langle y_{1}\wedge
y_{2}\wedge\cdots\wedge y_{k},\wedge_{J}\textbf{e}\rangle\rangle$

$=Vol_{k}(\pi_{J}([\textbf{x}]))Vol_{k}(\pi_{J}([\textbf{y}])).$

Because $\\{\wedge_{J}\textbf{e}:J\in\mathcal{S}(n,k)\\}$ is an orthonormal
basis for $\Lambda_{k}(\mathbb{E}^{n})$, the Product Formula for Inner
Products when applied to $\Lambda_{k}(\mathbb{E}^{n})$ with its inner product
$\langle\langle\;,\;\rangle\rangle$ tells us:

$\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge x_{k},y_{1}\wedge
y_{2}\wedge\cdots\wedge y_{k}\rangle\rangle=$

$\sum_{J\in\mathcal{S}(n,k)}\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge
x_{k},\wedge_{J}\textbf{e}\rangle\rangle\langle\langle y_{1}\wedge
y_{2}\wedge\cdots\wedge y_{k},\wedge_{J}\textbf{e}\rangle\rangle$.

Combining this equation with $(\beta)$ and $(\zeta)$ yields the Product
Formula for

$k$-dimensional parallelopipeds that lie in a $k$-dimensional vector subspace
of $\mathbb{E}^{n}$:

$(\eta)\hskip
24.18483ptVol_{k}([\textbf{x}])Vol_{k}([\textbf{y}])=\sum_{J\in\mathcal{S}(n,k)}Vol_{k}(\pi_{J}([\textbf{x}]))Vol_{k}(\pi_{j}([\textbf{y}])).$
∎

##  5\. The Proof of the Product Formula for Volume - for Lebesgue measurable
subsets of parallel $k$-dimensional affine subspaces of $\mathbb{E}^{n}$

Let $V$ and $W$ be identical or parallel $k$-dimensional affine subspaces of
$\mathbb{E}^{n}$. We first prove the Product Formula for Volume for boxes in V
and W. There is a $k$-dimensional vector subspace $U$ of $\mathbb{E}^{n}$ and
points $p,q\in\mathbb{E}^{n}$ such that $V=T_{p}(U)$ and $W=T_{q}(U)$. Let
$\textbf{u}=(u_{1},u_{2},\cdots,u_{k})$ be an ordered orthonormal basis for
$U$. Suppose $B$ is a u-box in $V$ and $C$ is a u-box in $W$. Then
$B=T_{p}(B_{0})$ and $C=T_{q}(C_{0})$ where $B_{0}$ and $C_{0}$ are u-boxes in
$U$. Therefore, $Vol_{k}(B)=Vol_{k}(B_{0})$ and $Vol_{k}(C)=Vol_{k}(C_{0})$.
Suppose

$B_{0}=\\{\sum_{1\leq i\leq k}t_{i}u_{i}:b_{i}\leq t_{i}\leq
b_{i}^{\prime}\text{ for }1\leq i\leq k\\}$

and

$C_{0}=\\{\sum_{1\leq i\leq k}t_{i}u_{i}:c_{i}\leq t_{i}\leq
c_{i}^{\prime}\text{ for }1\leq i\leq k\\}$

where $b_{i}<b_{i}^{\prime}$ and $c_{i}<c_{i}^{\prime}$ are real numbers for
$1\leq i\leq k$. Let $\textbf{x}=(x_{1},x_{2},\cdots,x_{k})$ and let
$\textbf{y}=(y_{1},y_{2},\cdots,y_{k})$ where $x_{i}=b_{i}^{\prime}-b_{i}$ and
$y_{i}=c_{i}^{\prime}-c_{i}$ for $1\leq i\leq k$. Also let
$p^{\prime}=\sum_{1\leq i\leq k}b_{i}u_{i}$ and $q^{\prime}=\sum_{1\leq i\leq
k}c_{i}u_{i}$. Then $B_{0}=T_{p^{\prime}}([\textbf{x}])$ and
$C_{0}=T_{q^{\prime}}([\textbf{y}])$. Since $p^{\prime}$ and $q^{\prime}\in
U$, then $T_{p^{\prime}}$ and $T_{q^{\prime}}$ map $U$ onto itself. Since
$Vol_{k}$ is translation invariant on $U$, it follows that
$Vol_{k}(B_{0})=Vol_{k}([\textbf{x}])$ and
$Vol_{k}(C_{0})=Vol_{k}([\textbf{y}])$. We conclude that

$(\theta)\hskip
70.56284ptVol_{k}(B)Vol_{k}(C)=Vol_{k}([\textbf{x}])Vol_{k}([\textbf{y}]).$

Let $J\in\mathcal{S}(n,k)$. For $z\in\mathbb{E}^{n}$, since
$\pi_{J}(z)\in\mathbb{E}^{J}$, then $T_{\pi_{J}(z)}$ maps $\mathbb{E}^{J}$ to
itself. Since $Vol_{k}$ is translation invariant on $\mathbb{E}^{J}$, then we
have $Vol_{k}(T_{\pi_{J}(z)}(A))=Vol_{k}(A)$ for every Lebesgue measurable
subset $A$ of $\mathbb{E}^{J}$. Recall that for $z,w\in\mathbb{E}^{n}$,
$T_{z}\circ T_{w}=T_{z+w}$ and, by Proposition 2.1 applied to the linear map
$\pi_{J}:\mathbb{E}^{n}\rightarrow\mathbb{E}^{J}$, $\pi_{J}\circ
T_{z}=T_{\pi_{J}(z)}\circ\pi_{J}$. Hence,

$Vol_{k}(\pi_{J}(B))=Vol_{k}(\pi_{J}(T_{p}(B_{0})))=Vol_{k}(\pi_{J}(T_{p}(T_{p^{\prime}}([\textbf{x}]))))=$
$Vol_{k}(\pi_{J}\circ
T_{p+p^{\prime}}([\textbf{x}]))=Vol_{k}(T_{\pi_{J}(p+p^{\prime})}(\pi_{J}([\textbf{x}]))=Vol_{k}(\pi_{J}([\textbf{x}])).$

Similarly, $Vol_{k}(\pi_{J}(C))=Vol_{k}(\pi_{J}([\textbf{y}]))$. Therefore,

$(\kappa)\hskip
33.2897ptVol_{k}(\pi_{J}(B))Vol_{k}(\pi_{J}(C))=Vol_{k}(\pi_{J}([\textbf{x}]))Vol_{k}(\pi_{J}([\textbf{y}]))$

for every $J\in\mathcal{S}(n,k)$.

Combining $(\eta)$, $(\theta)$ and $(\kappa)$ yields a Product Formula for
Volume that holds for u-boxes $B$ and $C$ in parallel $k$-dimensional affine
subspaces $V$ and $W$ of $\mathbb{E}^{n}$:

$Vol_{k}(B)Vol_{k}(C)=\sum_{J\in\mathcal{S}(n,k)}Vol_{k}(\pi_{J}(B))Vol_{k}(\pi_{J}(C))$.

Next we prove the Product Formula for Volume for two unions of almost disjoint
collections of u-boxes in $V$ and $W$. Let
$\mathcal{B}\in\mathbb{B}_{0}(V,\textbf{u})$ and
$\mathcal{C}\in\mathbb{B}_{0}(W,\textbf{u})$. For each $J\in\mathcal{S}(n,k)$,
since $\pi_{J}:\mathbb{E}^{n}\rightarrow\mathbb{E}^{J}$ is a linear map, then
Corollary 3.10 implies that
$Vol_{k}(\pi_{J}(\bigcup\mathcal{B}))=\sum_{B\in\mathcal{B}}Vol_{k}(\pi_{J}(B))$
and
$Vol_{k}(\pi_{J}(\bigcup\mathcal{C}))=\sum_{C\in\mathcal{C}}Vol_{k}(\pi_{J}(C))$.
Hence, we have:

$Vol_{k}(\bigcup\mathcal{B})Vol_{k}(\bigcup\mathcal{C})=(\sum_{B\in\mathcal{B}}Vol_{k}(B))(\sum_{C\in\mathcal{C}}Vol_{k}(C))=$

$\sum_{B\in\mathcal{B}}\sum_{C\in\mathcal{C}}Vol_{k}(B)Vol_{k}(C)=$

$\sum_{B\in\mathcal{B}}\sum_{C\in\mathcal{C}}(\sum_{J\in\mathcal{S}(n,k)}Vol_{k}(\pi_{J}(B))Vol_{k}(\pi_{J}(C)))=$

$\sum_{J\in\mathcal{S}(n,k)}(\sum_{B\in\mathcal{B}}\sum_{C\in\mathcal{C}}Vol_{k}(\pi_{J}(B))Vol_{k}(\pi_{J}(C)))=$

$\sum_{J\in\mathcal{S}(n,k)}(\sum_{B\in\mathcal{B}}Vol_{k}(\pi_{J}(B)))(\sum_{C\in\mathcal{C}}Vol_{k}(\pi_{J}(C)))=$

$\sum_{J\in\mathcal{S}(n,k)}Vol_{k}(\pi_{J}(\bigcup\mathcal{B}))Vol_{k}(\pi_{J}(\bigcup\mathcal{C}))$.

To complete the proof of the Product Formula for Volume, it suffices to prove
that the function
$\Psi:\mathcal{M}(V)\times\mathcal{M}(W)\rightarrow\mathbb{R}$ defined by

$\Psi(A,B)=Vol_{k}(A)Vol_{k}(B)-\sum_{J\in\mathcal{S}(n,k)}Vol_{k}(\pi_{J}(A))Vol_{k}(\pi_{J}(B))$

equals $0$ at each point $(A,B)\in\mathcal{M}(V)\times\mathcal{M}(W)$. We
argue that $\Psi$ is continuous. A remark following the definition of the
_symmetric difference pseudometric_ in section 3 established that measures are
continuous. In particular, the following functions are continuous:

$Vol_{k}:\mathcal{M}(V)\rightarrow\mathbb{R}\text{,
}Vol_{k}:\mathcal{M}(W)\rightarrow\mathbb{R}\text{ and }$
$Vol_{k}:\mathcal{M}(\mathbb{E}^{J})\rightarrow\mathbb{R}\text{ for each
}J\in\mathcal{S}(n,k).$

Since the functions $\pi_{J}|V:V\rightarrow\mathbb{E}^{J}$ and
$\pi_{J}|W:W\rightarrow\mathbb{E}^{J}$ are linear for each
$J\in\mathcal{S}(n,k)$, then Corollary 3.6 implies that the following
functions are dilations and, hence, are continuous:

$(\pi_{J}|V)_{*}:\mathcal{M}(V)\rightarrow\mathcal{M}(\mathbb{E}^{J})\text{
and
}(\pi_{J}|V)_{*}:\mathcal{M}(V)\rightarrow\mathcal{M}(\mathbb{E}^{J})\text{
for each }J\in\mathcal{S}(n,k).$

Consequently, the compositions

$Vol_{k}\circ(\pi_{J}|V)_{*}:\mathcal{M}(V)\rightarrow\mathbb{R}\text{ and
}Vol_{k}\circ(\pi_{J}|W)_{*}:\mathcal{M}(W)\rightarrow\mathbb{R}$

are continuous for each $J\in\mathcal{S}(n,k)$. Thus, the function

$(A,B)\mapsto(Vol_{k}(A),Vol_{k}(B),(Vol_{k}\circ\pi_{J}(A),Vol_{k}\circ\pi_{J}(B))_{J\in\mathcal{S}(n,k)})$
$:\mathcal{M}(V)\times\mathcal{M}(W)\rightarrow\mathbb{R}^{2}\times(\mathbb{R}^{2})^{|\mathcal{S}(n,k)|}$

is continuous, where $|\mathcal{S}(n,k)|$ denotes the number of elements in
the finite set $\mathcal{S}(n,k)$. Also the function

$(s_{V},s_{W},(t_{V,J},t_{W,J})_{J\in\mathcal{S}(n,k)})\mapsto
s_{V}s_{W}-\sum_{J\in\mathcal{S}(n,k)}t_{V,J}t_{W,J}$

$:\mathbb{R}^{2}\times(\mathbb{R}^{2})^{|\mathcal{S}(n,k)|}\rightarrow\mathbb{R}$

is continuous. Since $\Psi$ is the composition of the latter two functions, it
is continuous. Let
$\mathcal{D}_{V}=\\{\bigcup\mathcal{B}:\mathcal{B}\in\mathbb{B}_{0}(V,\textbf{u})\\}$
and
$\mathcal{D}_{W}=\\{\bigcup\mathcal{C}:\mathcal{C}\in\mathbb{B}_{0}(W,\textbf{u})\\}$.
The preceding proof establishes that $\Psi$ equals $0$ at each point
$(\bigcup\mathcal{D},\bigcup\mathcal{E})$ of the subset
$\mathcal{D}_{V}\times\mathcal{D}_{W}$ of
$\mathcal{M}(V)\times\mathcal{M}(W)$. Since, by Proposition 3.10′,
$\mathcal{D}_{V}$ is a dense subset of $\mathcal{M}(V)$ and $\mathcal{D}_{W}$
is a dense subset of $\mathcal{M}(W)$, then
$\mathcal{D}_{V}\times\mathcal{D}_{W}$ is a dense subset of
$\mathcal{M}(V)\times\mathcal{M}(W)$. It follows that $\Psi$ equals $0$ at
every point of $\mathcal{M}(V)\times\mathcal{M}(W)$. ∎

As noted in the Abstract, the Pythagorean Theorem for Volume follows
immediately from the Product Formula for Volume by setting $A=B$.

##  6\. The Binet-Cauchy Formula and the inner product on
$\Lambda_{k}(\mathbb{E}^{n})$

For $1\leq k\leq n$, if $A=(a_{i,j})$ is a $k\times n$ matrix and
$J=\\{j_{1},j_{2},\cdots,j_{k}\\}\in\mathcal{S}(n,k)$ where $1\leq
j_{1}<j_{2}<\cdots<j_{k}\leq n$, let $A^{[J]}=(a^{[J]}_{i,t})$ denote the
$k\times k$ matrix determined by $a^{[J]}_{i,t}=a_{i,j_{t}}$. (In other words,
for $1\leq t\leq k$, the $t^{th}$ column of $A^{[J]}$ is the $j_{t}^{th}$
column of $A$.) For $1\leq k\leq n$, any two $k\times n$ matrices $A$ and $B$
satisfy the equation:

_The Binet-Cauchy (or Cauchy-Binet) Formula:_

$Det(AB^{T})=\sum_{J\in\mathcal{S}(n,k)}Det(A^{[J]})Det(B^{[J]})$.

We will show that the Binet-Cauchy Formula is equivalent to the equation we
have called the Product Formula for Inner Products specialized to the inner
product $\langle\langle\;,\;\rangle\rangle$ on $\Lambda_{k}(\mathbb{E}^{n})$.

Let $1\leq k\leq n$ and let $A$ and $B$ be $k\times n$ matrices. For $1\leq
i\leq k$, let $x_{i}=(x_{i,1},x_{i,2},\cdots,x_{i,n})\in\mathbb{E}^{n}$ be the
$i^{th}$ row of $A$ and let
$y_{i}=(y_{i,1},y_{i,2},\cdots,y_{i,n})\in\mathbb{E}^{n}$ be the $i^{th}$ row
of $B$. Recall that $\textbf{e}=(e_{1},e_{2},\cdots,e_{n})$ denotes the
standard ordered orthonormal basis for $\mathbb{E}^{n}$; and for
$J=\\{j_{1},j_{2},\cdots,j_{k}\\}\in\mathcal{S}(n,k)$ where $1\leq
j_{1}<j_{2}<\cdots<j_{k}\leq n$,
$\textbf{e}_{J}=(e_{j_{1}},e_{j_{2}},\cdots,e_{j_{k}})$ and
$\wedge_{J}\textbf{e}=e_{j_{1}}\wedge e_{j_{2}}\wedge\cdots\wedge e_{j_{k}}$.
The Product Formula for the inner product $\langle\langle\;,\;\rangle\rangle$
on $\Lambda_{k}(\mathbb{E}^{n})$ implies:

$\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge x_{k},y_{1}\wedge
y_{2}\wedge\cdots\wedge y_{k}\rangle\rangle=$

$\sum_{J\in\mathcal{S}(n,k)}\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge
x_{k},\wedge_{J}\textbf{e}\rangle\rangle\langle\langle y_{1}\wedge
y_{2}\wedge\cdots\wedge y_{k},\wedge_{J}\textbf{e}\rangle\rangle$.

Thus, in order to convert this equation into the Binet-Cauchy Formula, it
suffices to prove that

$\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge x_{k},y_{1}\wedge
y_{2}\wedge\cdots\wedge y_{k}\rangle\rangle=Det(AB^{T}),$

and

$\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge
x_{k},\wedge_{J}\textbf{e}\rangle\rangle=Det(A^{[J]})\text{ and
}\langle\langle y_{1}\wedge y_{2}\wedge\cdots\wedge
y_{k},\wedge_{J}\textbf{e}\rangle\rangle=Det(B^{[J]})$

for every $J\in\mathcal{S}(n,k)$. Observe that

$\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge x_{k},y_{1}\wedge
y_{2}\wedge\cdots\wedge
y_{k}\rangle\rangle=Det(x_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}y_{j}),$
$\langle\langle x_{1}\wedge x_{2}\wedge\cdots\wedge
x_{k},\wedge_{J}\textbf{e}\rangle\rangle=\langle\langle x_{1}\wedge
x_{2}\wedge\cdots\wedge x_{k},e_{j_{1}}\wedge e_{j_{2}}\wedge\cdots\wedge
e_{j_{k}}\rangle\rangle=Det(x_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}e_{j_{t}})$

and

$\langle\langle y_{1}\wedge y_{2}\wedge\cdots\wedge
y_{k},\wedge_{J}\textbf{e}\rangle\rangle=\langle\langle y_{1}\wedge
y_{2}\wedge\cdots\wedge y_{k},e_{j_{1}}\wedge e_{j_{2}}\wedge\cdots\wedge
e_{j_{k}}\rangle\rangle=Det(y_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}e_{j_{t}})$

for $J=\\{j_{1},j_{2},\cdots,j_{k}\\}\in\mathcal{S}(n,k)$ where $1\leq
j_{1}<j_{2}<\cdots<j_{k}\leq n$. Hence, converting the Product Formula for
$\langle\langle\;,\;\rangle\rangle$ to the Binet-Cauchy Formula is simply a
matter of comparing the entries of the matrices
$(x_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}y_{j})$,
$(x_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}e_{j_{t}})$ and
$(y_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}e_{j_{t}})$ to the
corresponding entries of the matrices $AB^{T}$, $A^{[J]}$ and $B^{[J]}$.
Clearly, the $(i,j)^{th}$ entries of
$(x_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}y_{j})$ and $AB^{T}$
are both equal to
$x_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}y_{j}$, the
$(i,t)^{th}$ entries of
$(x_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}e_{j_{t}})$ and
$A^{[J]}$ are both equal to $x_{i,j_{t}}$, and the $(i,t)^{th}$ entries of
$(y_{i}\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}e_{j_{t}})$ and
$B^{[J]}$ are both equal to $y_{i,j_{t}}$. We conclude that the Binet-Cauchy
Formula is simply a reformulation of the Product Formula for
$\langle\langle\;,\;\rangle\rangle$. ∎

We note that one consequence of the Binet-Cauchy Formula is the following
occasionally useful fact.

######  Corollary 6.1.

For $1\leq k\leq n$, a $k\times n$ matrix $A$ has rank $k$ if and only if
$Det(A^{[J]})\neq 0$ for some $J\in\mathcal{S}(n,k)$.

###### Proof.

We set $B=A$ in the Binet-Cauchy Formula to obtain the equation

$Det(AA^{T})=\sum_{J\in\mathcal{S}(n,k)}(Det(A^{[J]}))^{2}$.

Then with the help of Lemma 6.2 below, we obtain the equivalence of the
following three statements. i) $rank(A)=k$. ii) $Det(AA^{T})\neq 0$. iii)
$Det(A^{[J]})\neq 0$ for some $J\in\mathcal{S}(n,k)$. ∎

######  Lemma 6.2.

For $1\leq k\leq n$, a $k\times k$ matrix $A$ has rank $k$ if and only if
$Det(AA^{T})\neq 0$.

###### Proof.

Consider the following statements. i) $rank(A)<k$. ii) The rows of A are
linearly dependent. iii) There is an $a\in\mathbb{E}^{k}-\\{0_{k}\\}$ such
that $aA=0_{n}$ (where $0_{k}$ and $0_{n}$ are the additive identities of
$\mathbb{E}^{k}$ and $\mathbb{E}^{n}$, respectively). iv) There is an
$a\in\mathbb{E}^{k}-\\{0_{k}\\}$ such that
$a(AA^{T})a^{T}=(aA)\mathbin{\vbox{\hbox{\scalebox{0.6}{$\bullet$}}}}(aA)=0$.
v) There is an $a\in\mathbb{E}^{k}-\\{0_{k}\\}$ such that $a(AA^{T})=0_{k}$.
vi) The rows of $AA^{T}$ are linearly dependent. vii) $Det(AA^{T})=0$. Observe
that i), ii) and iii) are equivalent, iii) $\implies$ v) $\implies$ iv)
$\implies$ iii), and v), vi) and vii) are equivalent. ∎

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