
%%metric geometry of Euclidean space
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\centerline{\Head Isometries of figures in Euclidean spaces}
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\centerline{Reinhard Schultz}
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These notes discuss some facts about the metric geometry of Euclidean
spaces that are closely related to the notion of congruence in elementary 
(= high school level) Euclidean geometry. 
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\bigskip
\centerline{\sl Congruence and isometries}
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\medskip
Congruence is a fundamental concept in elementary geometry.  However,
problems arise when one tries to give a definition of this concept
that is both mathematically accurate and comprehensive enough to cover
all the cases one wants to consider ({\it e.g.}, straight line segments,
angles, triangles, parallelograms, circles, ...).  Intuitively, the idea is
that two figures should be congruent if there is a rigid motion that
sends one to the other.  However, this requires a reasonable definition
of rigid motion, which is somewhat beyond the scope of an elementary
course.  One can attempt to circumvent this by defining congruence 
for special types of figures like line segments (distances between the
end points are equal), angles (same measurement), triangles (the vertices
match up in such a way that the sides, or line segments joining 
corresponding vertices, are congruent and the corresponding vertex
angles are congruent), parallelograms (similar with four vertices
rather than three) and circles (congruent radius segments).  However,
at this point one can give a mathematically precise and comprehensive
definition and show that it reduces to the familiar definitions of
elementary geometry.
\medskip
The discussion here indicates one further important fact about 
Euclidean geometry; namely, vector methods are an extremely powerful
tool for analyzing geometric problems.  Calculus and physics books
frequently have sections in which vectors are used to prove classical
theorems in Euclidean geometry; for example, the fact that the diagonals
of a parallelogram bisect each other or the three lines joining the 
vertices of a triangle to the midpoints of their opposite sides all
meet at a single point.  In fact, vector algebra not only allows one 
to write down slick proofs of classical theorems in Euclidean geometry,
but it also allows one to handle problems that are either extremely
difficult or virtually impossible to attack by other methods.
\medskip
\noindent{\bf Definition. }
Let ${\bf E}$ be a Euclidean space (= a finite-dimensional real inner 
product space), and let $A,~B\subset {\bf E}$.  The subsets $A$ and $B$ 
are said to be {\it congruent} if there is a 1--1 correspondence $f:A\to 
B$ that is an isometry with respect to the standard metric.  Specifically,
if $f:A\to B$ is the 1--1 correspondence then we have
$$|f({\bf x})~-~f({\bf y})|~~=~~|{\bf x}~-~{\bf y}|$$
for all ${\bf x},\,{\bf y}\in A$.  If $A$ and 
$B$ are congruent one often writes $A\cong B$ in the classical tradition.
\medskip
Since inverses and composites of isometries are isometries (and the 
identity is an isometry), it follows that congruence is an equivalence
relation.
\medskip
Eventually we shall explain why this definition is equivalent to the
definitions suggested for the special cases.  But first it is useful
to consider another possible definition for congruence.
\medskip
\noindent{\bf Definition. }
Let ${\bf E}$ be a Euclidean space, and let $A,~B\subset {\bf E}$.  The 
subsets $A$ and $B$ are said to be {\it ambiently congruent} if there 
is a 1--1 correspondence $\Phi$ from ${\bf E}$ to itself that is an
isometry and sends $A$ to $B$ ({\it i.e., $\Phi (A) = B$}. 
\medskip
Clearly ambiently congruent subsets are congruent, and we shall prove
a converse to this in Section 4 below.  
\bigskip
\bigskip
\centerline{\bf 1.\qquad Global Euclidean isometries and similarities}
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\medskip
\medskip
We shall begin with the characterization of 
isometries of a finite-dimensional Euclidean space
that is often given in linear algebra textbooks.
\medskip
\noindent{\bf PROPOSITION. }
{\it If ${\bf E}$ is a finte-dimensional Euclidean space and $F$ is 
an isometry from ${\bf E}$ to itself, then $F$ may be expressed
in the form $F({\bf x}) = {\bf b} + A({\bf x})$ where ${\bf b}\in E$ 
is some fixed vector and $A$ is an orthogonal
linear tranformation of ${\bf E}$ ({\it i.e.}, in matrix form we
have that ${}^{{\bf T}}A = A^{=1}$ where ${}^{{\bf T}}A$ denotes the
transpose of $A$).}
\medskip
{\sl Notes. }  It is an elementary exercise to verify that the
composite of two isometries is an isometry (and the inverse of an
isometry is an isometry).  If $A$ is orthogonal, then 
it is elementary to prove that $F({\bf x}) = {\bf b} + A({\bf x})$ 
is an isometry, and in fact this is done in most if not all
undergraduate linear algebra texts.  On the other hand, if $A=I$
then the map above reduces to a {\bf translation} of the form
$F({\bf x}) = {\bf b} + {\bf x}$, and such maps are isometries
because they satisfy the even stronger identity
$$F\bigl({\bf x}\,-\,{\bf y}\bigr)~~=~~
{\bf x}~-~{\bf y}~.$$
Therefore every map of the form 
$F({\bf x}) = {\bf b} + A({\bf x})$, where ${\bf b}\in E$ 
is some fixed vector and $A$ is an orthogonal
linear tranformation of ${\bf E}$, is an isometry of {\bf E}.
Therefore the proposition gives a complete characterization of all
isometries of {\bf E}.
\medskip
\noindent{\bf Sketch of proof. }
This argument is often given in linear algebra texts, and if this
is not done then  hints are frequently given in the exercises, so we 
shall merely indicate the basic steps.
\medskip
First of all, the set of all isometries of {\bf E} is a group
(sometimes called the {\it Galileo group} of {\bf E}).  It contains
both the subgroups of orthogonal matrices and the subgroup of 
translations ($G({\bf x}) = {\bf x} + {\bf c}$ for some fixed 
vector ${\bf c}$), which is 
isomorphic as an additive group to {\bf E} with the vector addition 
operation.  Given $b\in {\bf E}$ let ${\bf S}_{\bf b}$ be translation by 
${\bf b}$, so that $A = {\bf S}_{-F({\bf 0})}\tinycirc F$ is an 
isometry from ${\bf E}$ to itself satisfying $G({\bf 0}) = {\bf 0}$.  
If we can show that 
$G$ is linear, then it will follow that $G$ is given by an orthogonal
matrix and the proof will be complete.
\medskip
Since $G$ is an isometry it follows that 
$$\bigl|G({\bf x})-G({\bf y})\bigr|^2~~=~~\bigl|{\bf x}-{\bf y}\bigr|^2$$
and since $G(0)= 0$ it also follows that $g$ is length preserving.
If we combine these special cases with the general formula
displayed above we conclude that $\langle G({\bf x}), G({\bf y})\rangle = 
\langle {\bf x},{\bf y}\rangle$ for all ${\bf x},~{\bf y}\in {\bf E}$.  
In particular,
it follows that $G$ sends orthonormal bases to orthonormal bases.
Let $\{{\bf u}_1,~\cdots~,{\bf u}_n\}$ be an orthonormal basis; then we have
$${\bf x} ~~=~~ \sum_{i=1}^n~\langle {\bf x},{\bf u}_i\rangle 
\cdot {\bf u}_i$$
and likewise we have 
$$G({\bf x}) ~~=~~ \sum_{i=1}^n~\langle G({\bf x}),G({\bf u}_i)\rangle 
\cdot G({\bf u}_i)~.$$
Since $G$ preserves inner products we know that 
$$\langle {\bf x},{\bf u}_i\rangle~~=~~
\langle G({\bf x}),G({\bf u}_i)\rangle 
\cdot G({\bf u}_i)$$
for all $i$, and this implies that $G$ is a linear transformation.\bull
\bigskip
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\centerline{\sl Similarities}
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\medskip
In Euclidean geometry the notion of similarity is nearly as important
as congruence.  Informally, it may be viewed as congruence modified
to allow magnification or reduction of distances by a uniform scale
factor.  Formally, one may proceed as follows. 
\medskip
\noindent{\bf Definition. }
Let ${\bf E}$ be a Euclidean space (= a finite-dimensional real inner 
product space), and let $A,~B\subset {\bf E}$.  The subsets $A$ and $B$ 
are said to be {\it similar} if there is a 1--1 correspondence $f:A\to 
B$ and a constant $r>0$ such that 
$$|f({\bf x})~-~f({\bf y})|~~=~~r\cdot |{\bf x}~-~{\bf y}|$$
for all ${\bf x},\,{\bf y}\in A$.  The number $r$ is unique if $A$ and $B$ 
contain more than one point (why?), and it is called the {\it ratio
of similitude} for $f$; the mapping $f$ itself is generally called a 
{\it similarity}.  More generally, the definition is also meaningful 
for metric spaces, but we shall not need this level of abstraction here.  

If $A$ and $B$ are similar one often writes $A\sim B$ in the classical 
tradition, and when it is useful to keep track of the ratio of
similitude we write $A\sim_r B$.  The following elementary observations
show that similarity is an equivalence relation and that congruence
is a stronger relation than similarity.
\medskip
\itemitem{{\bf (1)}} 
For all $A$ and $B$, $A\cong B~\;\Longleftrightarrow~\; 
A\sim_1 B$.\bull
\medskip
\itemitem{{\bf (2)}}
In particular, for all $A$ we have $A\sim_1 A$.\bull
\medskip
\itemitem{{\bf (3)}}
For all $A$ and $B$, $A\sim_r B~\;\Longrightarrow~\; 
B\sim_{1/r} A$ (use the inverse map).\bull
\medskip
\itemitem{{\bf (4)}}
For all $A$, $B$ and $C$, $A\sim_r B$ and 
$B\sim_s C~\;\Longrightarrow~\; A\sim_{rs} C$ (use
the composite of the similarities from $A$ to $B$ and
from $B$ to $C$).\bull
\medskip
To see that similarity is {\it strictly weaker} than congruence, it
is only necessary to consider an arbitrary subset $A$ of a
Euclidean space {\bf E} such that $A$ has more than one point and to
let $B = r\cdot A$ be the set of all points of the form
$r\cdot {\bf a}$ for some ${\bf a}\in A$, where $r$ is an arbitrary 
positive real number not equal to $+1$.  The mapping $f({\bf x}) =
r\cdot {\bf x}$ is then a similarity from $A$ to $B$ with ratio of
similitude $r>1$.  In particular, if $A$ has exactly two points,
then one has this similarity but there is no isometry from 
$A$ to $B$ (why?).  Of course, it is possible for $A$ and 
$B$ to be both isometric and similar with a ratio of similitude
$\neq 1$; for example, if we let $A = {\bf E}$ above, then we
also have $B = {\bf E}$.
\medskip
\noindent{\bf ELEMENTARY OBSERVATION. }
{\it If ${\bf E}$ is a finte-dimensional Euclidean space and $S$ is 
a similarity from ${\bf E}$ to itself with ratio of similitude $r$, 
then $S$ may be expressed in the form $S({\bf x}) = r\cdot F({\bf x})$ 
where $F$ is an isometry of ${\bf E}$.\bull}
\medskip
If we combine this with the description of isometries in terms of
translations and orthogonal linear transformations, we see that
$s$ may be expressed in the form $S({\bf x}) = {\bf b}\, +\, 
r\cdot A({\bf x})$ where ${\bf b}\in E$ and $A$ is an orthogonal
linear tranformation of ${\bf E}$.\bull
\medskip
{\sf SIMILARITIES AND ANGLE MEASUREMENTS. \qquad}
If we are given three vectors ${\bf a},\,{\bf b},\,{\bf c}$ such that
${\bf b}-{\bf a}$ is not a scalar multiple of ${\bf c}-{\bf a}$, then 
the classical geometrical measure of the angle $\angle 
{\bf a}{\bf b}{\bf c}$ is completely
determined by the inner product structure using the
familiar formula
$$\cos \bigl(\angle {\bf a}{\bf b}{\bf c}\bigr)~~=~~
{{({\bf b}~-~{\bf a})\cdot ({\bf c}~-~{\bf a})}\over {\,
|{\bf b}~-~{\bf a}|\cdot|{\bf c}~-~{\bf a}|\,}}~.$$
The following result will be particularly important for our
purposes:
\medskip
\noindent{\bf PROPOSITION. }
{\it Let $T$ be a similarity from the Euclidean space {\bf E}
to itself with ratio of similitude $r$, and let
${\bf a},\,{\bf b}$ and ${\bf c}$ be points of {\bf E} such that 
${\bf b}-{\bf a}$ is not a scalar multiple of ${\bf c}-{\bf a}$.  
Then $T({\bf b})-T({\bf a})$ is not a scalar multiple of 
$T({\bf c})-T({\bf a})$, and the cosines of $\angle 
{\bf a}{\bf b}{\bf c}$ and $\angle T({\bf a})T({\bf b})T({\bf c})$ 
are equal.}
\medskip
\noindent{\bf Proof. }  Every similarity can be expressed as
a composite $T = T_1\tinycirc T_2 \tinycirc T_3$  where
$T_3$ is a linear isometry, $T_2$ is multiplication by a 
positive constant and $T_3$ is a translation.  It suffices to
show that each $T_i$ preserves angle cosines and that 
for each $i$ the vector
$T_i({\bf b})-T_i({\bf a})$ is not a scalar multiple of 
$T_i({\bf c})-T_i({\bf a})$.  
\medskip
For linear isometries these are basic facts about orthogonal 
transformations that are established in virtually every linear 
algebra text.  In the
case of multiplication by a positive constant $r\neq 0$ the
condition on  $T({\bf b})-T({\bf a})$ and $T({\bf c})-T({\bf a})$
follows because $T$ is an invertible linear transformation, and
the preservation of angles is a consequence of the following 
chain of identities, in which
$T$ denotes multiplication by $r$:
$$\cos \Bigl(\angle T({\bf a})T({\bf b})T({\bf c})\Bigr)~~=~~
{{r\,({\bf b}~-~{\bf a})\cdot r\,({\bf c}~-~{\bf a})}\over 
{\,|r\,({\bf b}~-~{\bf a})|\cdot|r\,({\bf c}~-~{\bf a})|\,}}~~=$$
$${{r^2\,({\bf b}~-~{\bf a})\cdot ({\bf c}~-~{\bf a})}\over 
{\,r^2\,|{\bf b}~-~{\bf a}|\,|{\bf c}~-~{\bf a}|\,}}~~=~~
{{({\bf b}~-~{\bf a})\cdot ({\bf c}~-~{\bf a})}\over {\,
|{\bf b}~-~{\bf a}|\,|{\bf c}~-~{\bf a}|\,}}~~=~~
\cos \left(\angle {\bf a}{\bf b}{\bf c}\right)~.$$
Therefore it only remains to check the identity in the conclusion
if $T$ is a translation, say $T({\bf x}) = {\bf x}+{\bf p}$ for 
some fixed vector ${\bf p}\in {\bf E}$.  Now if $T$ is a translation
we have the previously mentioned identity
$$T\bigl({\bf x}\,-\,{\bf y}\bigr)~~=~~
{\bf x}~-~{\bf y}$$
which immediately implies that $T({\bf b})-T({\bf a})$ is not a 
scalar multiple of $T({\bf c})-T({\bf a})$.  Furthermore,
we also have
$$\cos \Bigl(\angle T({\bf a})T({\bf b})T({\bf c})\Bigr)~~=~~
{\bigl({T({\bf b})~-~T({\bf a})\,\bigr)\cdot 
\bigl(T({\bf c})~-~T({\bf a})\,\bigr)}\over 
{\,|T({\bf b})~-~T({\bf a})|\cdot|T({\bf c})~-~T({\bf a})|\,}}~~=$$
$${{({\bf b}~-~{\bf a})\cdot ({\bf c}~-~{\bf a})}\over 
{\,|{\bf b}~-~{\bf a}|\,|{\bf c}~-~{\bf a}|\,}}~~=~~
\cos \left(\angle {\bf a}{\bf b}{\bf c}\right)$$
which implies that the translation $T$ preserves angles.  Thus we 
have shown that an arbitrary similarity $T_0$ is a composite of 
similarities that preserve angles, and therefore $T_0$ itself
must also preserve angles.\bull
\medskip
{\sf FIXED POINTS. \qquad}
Given a mapping $T$ from a set $X$ to itself, one says that
$p\in X$ is a {\it fixed point} for $T$ if $T(p) = p$.  For
example, if $X$ is a vector space and $T$ is a linear transformation,
then the zero vector is a fixed point, and a nonzero vector is
a fixed point if and only if $+1$ is an eigenvalue of $T$ and
the vector itself is an eigenvector for this eigenvalue.  Some 
isometries have fixed points while others do not.  Of course
an isometry that comes from an orthogonal linear transformation
will have at least the origin as a fixed point, and in fact if
$+1$ is an eigenvalue then the set of fixed points will be the 
eigenspace for this eigenvalue.  On the other hand, a translation map
of the form $T({\bf v}) = {\bf v} + {\bf c}_0$ for some 
${\bf c}_0\neq {\bf 0}$ will {\bf NEVER}
have a fixed point.  In view of these considerations, the
following results on fixed points of non-isometric similarities
may be somewhat unexpected:
\medskip
\noindent{\bf PROPER SIMILARITY FIXED POINT THEOREM. } 
{\it If $S$ is a similarity from the Euclidean space {\bf E} to
itself with ratio of similitude $r\neq 1$, then $S$ has a 
{\bf unique} fixed point.}
\medskip
The term ``proper similarity'' is meant to suggest that 
$S$ is a similarity but not an isometry.
\medskip
\noindent{\bf Proof. }
We have already noted that $S({\bf x}) = r\,A({\bf x}) + {\bf b}$ where 
$A$ is given by an orthogonal matrix.  By this formula, 
the equation $S({\bf x}) = {\bf x}$ is equivalent to the equation
${\bf x} = r\,A({\bf x}) + {\bf b}$, which in turn is equivalent to 
$$\bigl[I-r\, A\bigr]({\bf x}) = {\bf b}~.$$
The assertion that $S$ has a unique fixed point is equivalent 
to the assertion that the displayed linear equation has a
unique solution.  The latter will happen if $I-r\,A$ is
invertible, or equivalently if $\det(I-r\,A)\neq 0$, and this
is equivalent to saying that $r^{-1}$ is not an eigenvalue of
$A$.  But if $A$ is orthogonal this means that $|A({\bf v})|=
|{\bf v}|$ for
all ${\bf v}$ and hence the only possible eigenvalues are $\pm\,1$;
on the other hand, by construction we have $r>0$ and $r\neq 1$,
and therefore all of the desired conclusions follow.\bull
\bigskip
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\centerline{\bf 2. \qquad Concepts from affine geometry}
\bigskip
\medskip
The following material is generally not discussed very much or
very systematically in undergraduate texts, but it contains the
basic setting that mathematicians frequently use to study certain
geometrical problems using linear algebra.  One of the best references
at the undergraduate textbook level is the chapter of Birkhoff and 
MacLane, {\it Survey of Modern Algebra}; on linear groups (the 
differences among the various editions are relatively minor).  We
shall merely summarize the main points and refer the reader to that
book for more details.
\medskip
Although a broadly based discussion of affine geometry is beyond
the scope of these notes, a few remarks on the terminology might
be informative.  The subject itself have emerged during the
eighteenth century in order to describe an equivalence relation
on geometrical figures that was weaker than similarity but still
indicated that the figures had some properties in common (hence
could be viewed as having some sort of ``affinity'' to each other
that falls short of being a similarity in the mathematical sense).
\medskip








TO BE CONTINUED

\bigskip
\bigskip
\centerline{\bf 3.\qquad Isometries of finite sets}
\bigskip
\medskip
In this section we shall prove a few basic facts about
isometries for finite sets, inducing some special cases
of the Isometry Extension Theorem.  We shall use these
results in the next section to prove the full versions
of the Isometry and Similarity Extension Theorems.
\medskip
It is not difficult to see that a set of three 
collinear points in the Euclidean space {\bf E}
cannot be isometric to a set of three noncollinear
points.  One can see this informally as follows:
If the three points are collinear, then one can
label them as ${\bf p}_1$, ${\bf p}_2$ and 
${\bf p}_3$ such that 
$$|{\bf p}_3 ~-~ {\bf p}_1| ~~=~~  
|{\bf p}_3 ~-~ {\bf p}_2| ~+~ |{\bf p}_2 ~-~ {\bf p}_1|$$
but if they are not collinear then the left hand
side is stricly less than the right for all possible
relabelings of the three points (this is the classical
version of the Triangle Inequality from elementary
geometry).  Likewise, experience with noncoplanar objects
shows that they cannot be flattened into a plane without
stretching or shrinking some distances.  The first result
of this section gives the underlying mathematical reason
for these ``intuitively obvious'' 
facts and their higher dimensional analogs.
\medskip
\noindent{\bf ISOMETRIES AND AFFINE INDEPENDENCE. }
{\it Let $A = \{\,{\bf a}_0,~\cdots~,{\bf a}_m\,\}$
and $B = \{\,{\bf b}_0,~\cdots~,{\bf b}_m\,\}$ be
isometric subsets of some Euclidean space {\bf E}.
Then $A$ is affinely independent if and only if $B$ is.}
\medskip
\noindent{\bf Proof. }
Given two finite sets $X$ and $Y$ such that $Y=X+{\bf w}$
for some vector {\bf w}, the results of the previous
section show that $Y$ is affinely independent if and only
if $Y$ is.  Let $A'$ and $B'$ be obtained from $A$ and $B$
by the translations sending {\bf 0} to $-{\bf a}_0$ and 
$-{\bf b}_0$ respectively.  Then $A$ is affinely independent
if and only if $A'$ is, and $B$ is affinely independent
if and only if $B'$ is.  Therefore it suffices to prove
the result when the first vector in each set is the zero
vector, and by the definition of affine independence this
in turn reduces to showing that 
{\it the set} $A'-\{{\bf 0}\}$ {\it is linearly 
independent if and only if $B'-\{{\bf 0}\}$ is.}  To
simplify notation let us call these sets
$X = \{\,{\bf x}_1,~\cdots~,{\bf x}_m\,\}$
and $Y = \{\,{\bf y}_1,~\cdots~,{\bf y}_m\,\}$. 
\medskip
Let $G_X$ be the $m\times m$ matrix whose 
$$g_{i,j}(X)~~=~~\langle{\bf x}_i,\,{\bf x}_j\rangle$$
({\it i.e.},  the {\it Gram matrix} of the
given set of vectors), 
and define $G_Y$ similarly.  We claim that these two
matrices are equal if $X\cup \{{\bf 0}\}$ and 
$Y\cup \{{\bf 0}\}$ are mapped isometrically by a 
correspondence $T$
sending {\bf 0} to itself and ${\bf x}_i$ to 
${\bf y}_i$ for $i\geq 1$.  If $i=j=k$ this follows
because 
$$g_{k,k}(X)~~=~~|{\bf x}_k|^2~~=~~|{\bf x}_k-{\bf 0}|^2~~=~~
|T({\bf x}_k)-{\bf 0}|^2~~=~~|{\bf y}_k-{\bf 0}|^2~~=~~
|{\bf y}_k|^2~~=~~g_{k,k}(Y)$$
and in general this follows because
$$g_{i,j}(X)~~=~~\langle{\bf x}_i,\,{\bf x}_j\rangle~~=~~-
{1\over 2}\,\Bigl(|{\bf x}_i-{\bf x}_j|^2~-~
|{\bf x}_i|^2~-~|{\bf x}_j|^2\Bigr)~~=$$
$$-{1\over 2}\,\Bigl(|{\bf y}_i-{\bf y}_j|^2~-~
|{\bf y}_i|^2~-~|{\bf y}_j|^2\Bigr)$$
with the latter following from the previously
obtained equation $g_{k,k}(X)=g_{k,k}(Y)$ and the
fact that $T$ is an isometry satisfying 
$T({\bf x}_i)={\bf y}_i$.
\medskip
Since $G_X$ and $G_Y$ are equal, one is invertible
if and only if the other is, and therefore it will
suffice to show that {\it the Gram matrix is
invertible if and only if the given set of nonzero
vectors is linearly independent.}
\medskip
Suppose first that $X$ is linearly dependent.  Then
some vector ${\bf x}_\ell$  in the set is a linear
combination of the others, say
$${\bf x}_\ell~~=~~\sum_{i\neq \ell}~c_i{\bf x}_i~.$$
If we let ${\bf g}_i$ denote the $i^{{\rm th}}$
row of the Gram matrix for $X$, then the bilinearity
properties of the inner product imply a similar equation
$${\bf g}_\ell~~=~~\sum_{i\neq \ell}~c_i{\bf g}_i$$
and therefore the rows of $G_X$ are linearly dependent
so that $G_X$ is not invertible.
\medskip
On the other hand, suppose that $X$ is linearly independent,
and let $V$ be the span of $X$; by hypothesis $V$ is 
$m$-dimensional.  Let ${\cal U} = \{\,{\bf u}_1,~\cdots~,
{\bf u}_m\,\}$ be an orthonormal basis for $V$, and 
write the vectors of $X$ as linear combinations of the
vectors in ${\cal U}$:
$${\bf x}_j~~=~~\sum_{i=1}^m~p_{i.j}\,{\bf u_i}$$
Since $X$and $\cal U$ form a basis for the same 
subspace, it follows that the matrix of coefficients
$P=(\,p_{i.j}\,)$ is invertible (of course, its inverse
is the matrix of coefficients for expressing the vectors
in $\cal U$ as linear combinations of the vectors in $X$).
Direct computation shows that the Gram matrix $G_X$ is
equal to ${}^{\bf T}P\cdot P$ where ${}^{\bf T}P$ denotes
the transpose of $P$.  Since $P$ is invertible, its product
with its transpose is also invertible, and therefore
we have shown that $G_X$ is invertible if $X$ is linearly
independent.\bull
\medskip
Here is a useful consequence that reflects the motivation
at the beginning of this section:
\medskip
\noindent{\bf COROLLARY. }
{\it If $A$ and $B$ are isometric subsets of some Euclidean
space {\bf E} and $\alpha$ and $\beta$ are maximal affinely
independent subsets of $A$ and $B$ respectively, then 
$\alpha$ and $\beta$ have the same numbers of elements.}
\medskip
\noindent{\bf Proof. }
Given a subset $S\subset {\bf E}$, let $\aff (S)$ denote the
set of all vectors expressible as (finite) affine combinations
of the vectors in $S$.  It is then an elementary exercise to 
verify that 
$$\aff \bigl( \aff(S) \bigr)~~=~~\aff (S)$$
(an affine combination of affine combinations is an affine
combination).  In particular, $\aff(S)$ is a flat subset
in the sense of Section 2, and therefore $\aff(S)$ has the
form ${\bf x} + W_S$ where ${\bf x}\in S$ is arbitrary and 
$W_S$ depends only upon $S$ and not on {\bf x}.  
\medskip
Suppose now that $M$ is a maximal affinely independent subset
of $S$, and assume that $M$ has $m+1$ elements.  Then 
$\aff (M)$ has the form ${\bf x}' + W_M$ where $\dim W_M = m$.
We claim that $\aff(M) = \aff(S)$; if so, then $W_M = W_S$
and hence $\dim W_S = m$, which implies that the number of
vectors in a maximal affinely independent subset is the same
for all possible choices.
\medskip
It follows immediately that $\aff(M) \subset \aff(S)$.  To prove
the converse, we begin by showing that $S\subset \aff (M)$.
Suppose that ${\bf x}\in S - M$.  Then the set $M\cup\{{\bf x}\}$ 
must be affinely dependent and therefore there is a relation
of the form
$$c_{-1}\,{\bf x}~+~\sum_{i\geq 0}~c_i\,{\bf v}_i~~=~~{\bf 0}$$
where each ${\bf v}_i$ lies in $M$, the coefficients $c_j$ are
not all zero, and $\sum_j~c_j = 0$.  If $c_{-1}$ were zero then
the set $M$ would be affinely indepdendent, so $c_{-1}\neq 0$.
Therefore we may solve the equation above to express
${\bf x}$ as a linear comination of the vectors ${\bf v}_i$,
and for each $i$ the coefficient turns out to be
$${{c_i}\over{\;\sum_{i\geq 0}~c_i\;}}$$
so that the sum of the coefficients is equal to 1.  Therefore
we know that $S \subset \aff (M)$, and by the first paragraph of
the proof we have
$$\aff (S)~~\subset~~\aff \bigl( \aff(M) \bigr)~~=~~\aff (M)$$
which is what we wanted to prove.
\medskip
We now apply this to the situation described in the corollary.
The preceding argument shows that if $p$ and $q$ denote the 
numbers of elements in $\alpha$ and $\beta$ respectively, then 
every maximal affinely independent subset of $A$ contains exactly $p$
elements and likewise every maximal affinely independent subset 
of $B$ contains exactly $q$ elements.  Therefore it suffices to 
show that $B$ contains a maximal affinely independent subset with
exactly $p$ elements.  By the result on isometries and affine
independence, if $T$ is an isometry and $\alpha\subset A$ is
affinely independent, then $T(\alpha)\subset B$ is also 
affinely independent.  To see that it is a {\it maximal} such
subset, suppose that $\beta\subset B$ is a maximal affinely
independent subset that contains $T(\alpha)$.  Since $T^{-1}$
is also an isometry, it follows that $T^{-1}(\beta)\subset A$
is an affinely independent subset of $A$ containing $\alpha$.
By the maximality of the latter it follows that 
$T^{-1}(\beta) = \alpha$ and hence that $\beta = T(\alpha)$.
Therefore $T(\alpha)$ is a maximal affinely independent subset
of $B$ with exactly $p$ elements, and accordingly we must have
$p = q$ as required.\bull
\medskip
The next result is an important special case of the main 
result on extending isometries.
\medskip
\noindent{\bf PROPOSITION. }
{\it Let {\bf E} be an $m$-dimensional Euclidean space,
and let $A = \{\,{\bf a}_0,~\cdots~,{\bf a}_m\,\}$
and $B = \{\,{\bf b}_0,~\cdots~,{\bf b}_m\,\}$ be
isometric affinely independent subsets of {\bf E}.  Then
there is a unique isometry $T$ from {\bf E} to itself
such that $T({\bf a}_i) = {\bf b}_i$ for all $i$.}
\medskip
\noindent{\bf Proof. }
We start with essentially the same arguments used at the
beginning of the proof of the previous result.  Using
translations we can reduce the proof to the special case
where ${\bf a}_0={\bf b}_0={\bf 0}$, and since a linear
isometry sends the zero vector to itself it suffices to 
prove that if $A'$ and $B'$ are the translations of 
$A$ and $B$ sending {\bf 0} to $-{\bf a}_0$ and 
$-{\bf b}_0$ respectively, then there is a unique isometry
sending the ordered list of vectors in $A'$ to the corresponding
ordered list of vectors in $B'$.  Since isometries sending the
zero vector to itself are orthogonal linear transformations
and the nonzero vectors in $A'$ and $B'$ are bases for {\bf E},
it suffices to prove that {\it there is a (necessarily unique)
{\bf orthogonal} linear transformation from {\bf E} to itself
sending the ordered basis $A'-\{{\bf 0}\}$ to
$B'-\{{\bf 0}\}$}.  As in the previous proof, to
simplify notation let us call these bases
$X = \{\,{\bf x}_1,~\cdots~,{\bf x}_m\,\}$
and $Y = \{\,{\bf y}_1,~\cdots~,{\bf y}_m\,\}$. 
\medskip
General considerations from linear algebra imply that there 
is a unique invertible linear transformation $T$ such that
$T({\bf x}_i) = {\bf y}_i$ for all $i$.  We need to prove that
$T$ is an isometry under the hypotheses on $A$ and $B$ (and
their reformulations to conditions on $X$ and $Y$).
\medskip
As in the previous proof, if $G_X$ and $G_Y$ are the so-called
Gram matrices defined by the formulas
$$g_{i,j}(X)~~=~~\langle{\bf x}_i,\,{\bf x}_j\rangle~,\qquad\qquad
g_{i,j}(Y)~~=~~\langle{\bf y}_i,\,{\bf y}_j\rangle$$
then one can check that $G_X = G_Y$ using the fact that $T$ maps 
$X\cup \{{\bf 0}\}$ isometrically to $Y\cup \{{\bf 0}\}$.  Therefore
the proof reduces to verifyint the following assertion:
\medskip
\noindent{\bf LEMMA. }
{\it Let {\bf E} be a Euclidean space, let
$X = \{\,{\bf x}_1,~\cdots~,{\bf x}_m\,\}$
and $Y = \{\,{\bf y}_1,~\cdots~,{\bf y}_m\,\}$
be ordered bases for {\bf E} whose Gram matrices
are equal, and suppose that
$T$ is a linear transformation from $T$ to itself 
such that  $T({\bf x}_i) = {\bf y}_i$ for all $i$ 
(recall that the latter implies invertibility).  Then
$T$ is orthogonal.}
\medskip
\noindent{\bf Proof. }
This is an elementary but inelegant computation.  Let
{\bf p} and {\bf q} be arbitrary vectors in {\bf E},
and write them as linear combinations
$${\bf p}~~=~~\sum_{i=1}^m~s_i\,{\bf x}_i~,\qquad\qquad\qquad
{\bf q}~~=~~\sum_{j=1}^m~t_j\,{\bf x}_j~.$$
Therefore the inner product $\langle{\bf p},\,{\bf q}\rangle$
is equal to 
$$\sum_{i,j}~s_i\,t_j\,\langle{\bf x_i},\,{\bf x}_j\rangle$$
and the inner product $\langle T({\bf p}),\,T({\bf q})\rangle$
is equal to the corresponding expression in which each term
of the form ${\bf x}_k$ is replaced by the corresponding
vector ${\bf y}_k$.  Since the condition on $T$ implies that 
$\langle{\bf x}_i,\,{\bf x}_j\rangle$ is equal to 
$\langle{\bf y}_i,\,{\bf y}_j\rangle$ for all $i$ and $j$,
this means that $\langle{\bf p},\,{\bf q}\rangle$ and
$\langle T({\bf p}),\,T({\bf q})\rangle$ are equal, which in
turn implies that $T$ is an orthogonal linear transformation.\bull
\medskip
\noindent{\bf COMPLEMENT. }
{\it If in the preceding result we alter the hypothesis on
the dimension of {\bf E} to $\dim {\bf E} \geq m$, the 
existence portion of the conclusion remains valid.}
\medskip
\noindent{\bf Sketch of proof. }
As before, it suffices to establish the case where the isometry
sends {\bf 0} to itself, so we shall limit our attention to
this case and leave the general case to the reader as an exercise.
\medskip
Let $V$ be the span of the linearly independent set $X$ and let 
$W$ be the span of the linearly independent set $Y$ such that
$X\cup\{ {\bf 0} \}$ is congruent to $Y\cup\{ {\bf 0} \}$ by
an isometry sending {\bf 0} to itself.  Then
the dimensions of $V$ and $W$ are equal, and similarly the 
dimensions of their orthogonal complements $V^\perp$ and $W^\perp$
are equal.  The methods of the preceding result yield an invertible
orthogonal linear transformation from $V$ to $W$ sending $X$ to $Y$,
and we may extend this to an orthogonal linear transformation 
from {\bf E} to itself by choosing ordered orthonormal bases for
$V^\perp$ and $W^\perp$ and stipulating that the linear transformation
send the ordered basis for the former to the ordered basis for
the latter.  This yields an invertible linear transformation from
{\bf E} to itself, and it is orthogonal because it preserves inner
products on both $V$ and $V^\perp$.\bull
\medskip
{\sl Note. }
If $\dim V < \dim {\bf E}$ in the preceding result, then the choices 
of orthonormal bases for $V^\perp$ and $W^\perp$ are not unique, and 
therefore the extension to an isometry will not be unique.\bull
\medskip
Finally, the following sharpening of the first result is also
important in the proof of the main result in Section 4:
\medskip
\noindent{\bf ISOMETRIES AND BARYCENTRIC COORDINATES. 	}
{\it Let {\bf E} be an $m$-dimensional Euclidean space,
and let $A = \{\,{\bf a}_0,~\cdots~,{\bf a}_m\,\}$
and $B = \{\,{\bf b}_0,~\cdots~,{\bf b}_m\,\}$ be ordered
affinely independent subsets of {\bf E}, and let
{\bf p} and {\bf q} be points of {\bf E} such that
$A\;\cup\;\{\,{\bf p}\,\}$ and 
$B\;\cup\;\{\,{\bf q}\,\}$ are
isometric by an isometry $T$ sending ${\bf a}_i$ to 
${\bf b}_i$ for all $i$ and sending {\bf p} to
{\bf q}.  Then the barycentric coordinates of
{\bf p} with respect to $A$ are equal to the
barycentric coordinates of {\bf q} with respect to 
$B$.}
\medskip
\noindent{\bf Proof. }
Once again the argument begins with a reduction to
the case where ${\bf a}_0={\bf b}_0={\bf 0}$; this is
true because if $S_A$ and $S_B$ denote translation by
$-{\bf a}_0$ and $-{\bf b}_0$ respectively, then 
the barycentric coordinates of $S_A({\bf p})$ with respect 
to $S_A(A)$ are equal to to those of {\bf p} with respect
to $A$, and the barycentric coordinates of $S_B({\bf q})$ 
with respect to $S_B(B)$ are equal to to those of {\bf q} 
with respect to $B$.  
\medskip
Once again, to simplify notation we denote the 
ordered bases $A'-\{{\bf 0}\}$ and $B'-\{{\bf 0}\}$ by
$X = \{\,{\bf x}_1,~\cdots~,{\bf x}_m\,\}$
and $Y = \{\,{\bf y}_1,~\cdots~,{\bf y}_m\,\}$
respectively. If $U$ is the unique linear transformation
sending ${\bf x}_i$ to ${\bf y}_i$ for each $i$, then the
previous results of this section imply that $U$ is
an isometry, and the proof of the proposition reduces to
verifying that ${\bf q} = U({\bf p})$.
\medskip
Since $|{\bf p}|^2=|{\bf q}|^2$, 
$|{\bf x}_i|^2=|{\bf y}_i|^2$ for all $i$
and $|{\bf p} - {\bf x}_i|^2 = 
|{\bf q} - {\bf y}_i|^2$ for all $i$, it follows from the 
bilinearity properties of inner products that
$$\langle {\bf p},\,{\bf x}_i\rangle ~~=~~ 
\langle {\bf q},\,{\bf y}_i\rangle$$
for all $i$.  On the other hand, since $U$ is an orthogonal
transformation it also follows that 
$$\langle {\bf p},\,{\bf x}_i\rangle ~~=~~ 
\langle U({\bf p}),\,{\bf y}_i\rangle$$
for all $i$.  Therefore it suffices to show that {\it two vectors
in {\bf E} are equal if their inner products with each basis
vector ${\bf y}_i$ are equal\/}.
\medskip
Let ${\cal E}=\{\,{\bf e}_1,~\cdots~,{\bf e}_m\,\}$ be an 
orthonormal basis for {\bf E}, so that
$${\bf v}~~=~~\sum_{i=0}^m~\langle {\bf v},\,{\bf e}_i\rangle\,
{\bf e}_i$$
for all ${\bf V}\in {\bf E}$.
Since $Y$ and $\cal E$ are both orthonormal bases, it follows
that each vector in $\cal E$ can be expressed as a linear combination
of the vectors in $Y$, and therefore one may write each function
$\langle {\bf v},\,{\bf e}_i\rangle$ explicitly as a linear 
combination of the inner product functions $\langle {\bf v},
\,{\bf x}_j\rangle$.  Therefore, if we have two vectors
{\bf q} and {\bf r} such that 
$$\langle {\bf r},\,{\bf y}_j\rangle ~~=~~ 
\langle {\bf q},\,{\bf y}_j\rangle$$
for all $j$, the we also have 
$$\langle {\bf r},\,{\bf e}_i\rangle ~~=~~ 
\langle {\bf q},\,{\bf e}_i\rangle$$
for all $i$, and by the formula at the beginning of this paragraph
it follows that ${\bf q} = {\bf r}$.  By the remarks in the 
previous paragraph, the condition on inner products holds if 
${\bf r} = U({\bf p})$.  Therefore we have shown that ${\bf q} = 
U({\bf p})$ and accordingly that the barycentric
coordinates of {\bf p} with respect to $A$ are equal to the 
barycentric coordinates of {\bf q} to $B$.\bull
\bigskip
\bigskip
\centerline{\it Extensions to similarities}
\bigskip
\medskip
All of the results above extend to similarities.  Specifically,
suppose that the hypothesis involving an isometry in 
each result above is replaced by a hypothesis involving a 
similarity with ratio of similitude $r>0$.  If $M$ denotes the
invertible linear transformation of {\bf E} to itself corresponding
to multiplication by $r^{-1}$, then two subsets $A$ and $B$ in
{\bf E} are similar with ratio of similitude $r$ if and only if
$A$ and $M(B)$ are isometric, and a set of vectors $X$ 
is affinely or linearly independent if and only if either of the
sets $M^{\pm\,1}(X)$ is affinely or linearly independent respectively.
Furthermore, the mappings $M^{\pm\,1}$ preserve barycentric coordinates.
Details of verifying the extensions of each result to similarities
are left to the reader as an exercise.\bull
\bigskip 
\bigskip
\centerline{\bf 4.\qquad Extensions of isometries and similarities}
\bigskip
\medskip
We are now ready to prove the main result:
\medskip
\noindent{\bf ISOMETRY EXTENSION THEOREM. }
{\it Let ${\bf E}$ be a Euclidean space, let $A,~B\subset {\bf E}$,
and let $T:A\to B$ be an isometry.  Then there is a global
isometry  $\Phi$ from ${\bf E}$ to itself such that $T$ is the
restriction of $\Phi$ to $A$.  Furthermore, if $A$ contains a
maximal affinely independent subset of $\dim {\bf E} + 1$ 
vectors, then $\Phi$ is the unique extension of $T$.}
\medskip
There is a corresponding result for similarities; formulation of
the statement and proof of this result will be left to the
reader as an exercise.
\medskip
\noindent{\bf Proof. }
We shall follow the strategy of Section 3, considering the special
case where $T({\bf 0}) = {\bf 0}$ and showing that the general case
follows from the latter.  However, it will be more convenient to 
reverse the order of these steps and to begin with the special case.
\medskip
Let $A_0$ be a maximal affinely independent subset of $A$
containing {\bf 0}, and let $V$ be the span of $A_0$.  By maximality
it follows that every point of $A$ is a linear combination of vectors
in $A_0$ and therefore $A\subset V$.  Furthermore, the affine
independence of $A_0$ implies that $X_0 = A_0 - \{ {\bf 0} \}$
is linearly independent and hence forms a basis for $V$.
\medskip
By the results of Section 3 we also know that $B_0 = T(A_0)$ is 
a maximal affinely independent subset of $B$ containing {\bf 0}.
If $W$ is the span of $B_0$, it follows that $B\subset W$ and
$Y_0 = B_0 - \{ {\bf 0} \}$ forms a basis for $W$.  Therefore
the results of Section 3 also imply the existence of an orthogonal
linear transformation $\Phi$ on {\bf E} such that $\Phi = T$
on $X_0$. The final steps in the proof of the special case are to
prove that $\Phi = t$ on all of $A$ and that the extension is
unique if $A$ is a maximal affinely independent subset of {\bf E}
(so that $A$ spans {\bf E}).  Uniqueness follows quickly because
two linear transformations that agree on $A$ must agree everywhere
since $A$ is a spanning set.  To verify that $\Phi = T$ on $A$, we
shall use the result on isometries and barycentric coordinates from
Section 3.  According to the latter, if ${\bf a}\in A$ and we write
$${\bf a}~~=~~c_0\,{\bf 0}~+~\sum_{i=1}^m~c_i\,{\bf x}_i$$
where $\sum_{i\geq 0}~s_i = 1$, then 
$$T({\bf a})~~=~~c_0\,{\bf 0}~+~\sum_{i=1}^m~c_i\,{\bf y}_i~~=~~
\sum_{i=1}^m~c_i\,T({\bf x}_i)~~=~~
\sum_{i=1}^m~c_i\,\Phi({\bf x}_i)~~=~~\Phi({\bf a})$$
(since $\Phi$ is linear), so that $\Phi = T$ on $A$.
\medskip
We turn now to the general case.  Let ${\bf v}\in A$ and ${\bf w}\in 
B$ be such that $T({\bf v}) = {\bf w}$, and for an arbitrary vector
${\bf z}\in {\bf E}$ let $S_{\bf z}$ denote translation by {\bf z}.
Then the composite ${\widehat T} = S_{-{\bf w}}\tinycirc T\tinycirc
S_{{\bf v}}$ is an isometry from $S_{-{\bf v}}(A)$ to $S_{-{\bf w}}
(B)$ sending {\bf 0} to itself, and therefore ${\widehat T}$ 
extendes to an orthogonal linear transformation ${\widehat \Phi}$
on {\bf E}.  Furthermore this extension is uniqu if the linear
span of $S_{-{\bf v}}(A)$ is equal to {\bf E}, which is equivalent
to saying that $A$ contains an affinely independent subset with
$\dim {\bf E} + 1$ vectors.  If we set $\Phi$ equal to 
$S_{{\bf w}}\tinycirc {\widehat \Phi}\tinycirc S_{-{\bf v}}$, it 
follows that $\Phi$ is an isometry of {\bf E} such that $\Phi = T$
on $A$.  Finally, suppose that $\Psi$ is an arbitrary isometry of 
{\bf E} such that $\Psi = T$ on $A$.  Then both ${\widehat \Phi}$ and
$S_{-{\bf w}}\tinycirc {\widehat \Psi}\tinycirc S_{{\bf v}}$ are
orthogonal transformations with the same values on a spanning
subset of {\bf E} and hence they are equal.  Since the composite
$S_{\bf z}\tinycirc S_{-{\bf z}}$ is the identity for all {\bf z}, 
we therefore have
$$\Psi~~=~~{\rm Id}\tinycirc \Psi\tinycirc {\rm Id}~~=~~
\bigl(S_{\bf w}\tinycirc S_{-{\bf w}}\bigr)\tinycirc 
{\widehat \Psi}\tinycirc \bigl(S_{{\bf v}}\tinycirc S_{-{\bf v}}
\bigr)~~=$$
$$S_{\bf w}\tinycirc \bigl(S_{-{\bf w}} \tinycirc 
{\widehat \Psi}\tinycirc S_{{\bf v}}\bigr)\tinycirc 
S_{-{\bf v}}~~=~~
S_{{\bf w}}\tinycirc {\widehat \Phi}\tinycirc S_{-{\bf v}}~~=~~\Phi$$
and it follows that $\Phi$ must be unique.\bull
\bigskip
\bigskip
\centerline{\bf 5.\qquad Isometries, similarities and classical geometry}
\bigskip
\medskip
In this section we shall indicate how the standard notions of 
congruence and similarity in elementary geometry reduce to special 
cases of isometry and similarity in the sense of these notes.
\medskip

TO BE COMPLETED
\bye

