In this section we shall prove that isometries which keep the origin fixed are linear in the sense that they preserve sums and scalar products of vectors. Until we have proved this, we shall give a temporary name to isometries that preserve the origin.

First recall that the origin of a Cartesian plane plays a special role in analytic geometry because its coordinates, $O = (0,0)$ are also those of the zero vector, $O = X - X$ for all $X$.

As you have noticed already, we use the same letters for the point and for the vector that displaces the origin to that point. Most of the time this is harmless, but sometimes it leads to mistakes, as it did in the textbook. The context makes clear what we mean. For example, isometries are point transformations, so they take points to points. But we calculate the image point by giving a formula in vector algebra.

Before we continue, let’s look at a few questions.

Question 1.

Does a translation $\tau_D$ have any fixed points?

No. Suppose P is a fixed point of $\tau_D$. Then tau_D(P)=P rArr

 P+D=P D=P-P=O.

Question 2.

How many fixed points does a central dilatation have?

One: the center; suppose P is a fixed point of delta_{Q,r}, r!=1. Then delta_{Q,r}(P)=P rArr

 rP+(1-r)Q=P (1-r)Q=(1-r)P Q=P.

## Definition of an osometry

We can now define the main object of study in this section.

## Properties of osometries

Osometries have four important properties, which we prove next:

Proof of O1. Let X be any point and beta be an osometry. Then

 |X^beta| = |X^beta - O| (zero vector) = |X^beta - O^beta| (by definition of osometry) = |X-O| (by definition of isometry) = |X|. (zero vector)
Question 3.

Do osometries preserve circles centered at the origin?

Yes; by O1, the locus of points equidistant from the origin remain the same distance from the origin.

Proof of O2. First observe that O1 implies (X^beta)^2= |X^beta|^2=|X|^2 = X^2 .

Then for all X!=Y:

 0<$$X^beta-Y^beta$$^2 = (X^beta)^2-2(X^beta Y^beta)+(Y^beta)^2 = X^2 -2(X^beta Y^beta) +Y^2. (by O1)

On the other hand,

 |X^beta-Y^beta|^2 = |X-Y|^2 ( $\beta$ is an isometry ) = X^2-2(X Y)+Y^2.

Equating the right hand sides of equal left hand sides,

 X^2 -2(X^beta Y^beta) +Y^2=X^2-2(X Y)+Y^2 X^beta Y^beta=X Y.

Proof of O3. Again, we use the technique of squaring the difference of two things we wish to prove equal.

 ((rX)^beta -rX^beta)^2 = ((rX)^beta)^2-2(rX)^beta rX^beta + (rX^beta)^2 = (rX)^2 -2(rX)^beta rX^beta +r^2(X^beta)^2 (by O1) = r^2X^2 -2r((rX)^beta X^beta) +r^2X^2 (dot is bilinear) = 2r^2X^2 -2r((rX) X) (by O2) = 2r^2X^2 -2r^2(X X) = 0.

So ((rX)^beta -rX^beta)^2=|(rX)^beta -rX^beta|^2=0. But only the zero vector has length zero, hence (rX)^beta=rX.

Proof of O4. Again we square the difference of items destined to be equal.

 (X^beta+Y^beta-(X+Y)^beta)^2 = (X^beta)^2+(Y^beta)^2+((X+Y)^beta)^2+ 2(X^beta Y^beta)-2(X^beta (X+Y)^beta) -2(Y^beta (X+Y)^beta) = X^2+Y^2+(X+Y)^2+2(X Y)-2(X (X+Y))-2(Y(X+Y)) = (X+Y-(X+Y))^2 = O^2=0.

Hence X^beta+Y^beta=(X+Y)^beta.\ \ square

$\square$

O3 and O4 are called the linearity property of osometries; they show that an osometry is a linear isometry. In general,

Definition.

A point transformation $f$ is linear if for any finite number of $X_n$ and coefficients $r_n$, we have that

$f( \sum_{r=1}^{n} r_n X_n ) = \sum_{r=1}^{n} r_n f(X_n)$.

It is clear that a linear point transformation has to be an osometry. So we have equivalent attributes of a linear isometry. Given an isometry, we use the first, "fixes the origin", to identify it as a linear isometry. Given a linear isometry, we use the linearity properties in its application.

For now, we’ll just note that O3 and O4 imply (X-Y)^beta = (X+(-Y))^beta = X^beta +(-Y)^beta = X^beta-Y^beta, which we have already seen is not true for arbitrary isometries. But it is "nearly" true. Just in what sense this is the case will be treated in the next section.

So every osometry is a linear transformation and in fact, the converse is also true: every linear isometry is also an osometry. Henceforth, we will use these two terms interchangeably. so

Question 4.

Why is every linear isometry an osometry as well?

Let alpha be an isometry such that (sum r_nX_n)^alpha=sum r_nX_n^alpha. To prove alpha is an osometry , we must show that alpha(O)=O. But since O=X-X for any X, we have alpha(O)=alpha(X-X)=alpha(X)-alpha(X)=O.

Note that this seemingly trivial calculation would not be true for a translation, for example.

## The Osometry Factorization Theorem (OFT)

Motivation.

Linear functions tend to be much easier to work with than those that are not linear. The term non-linear unfortunately means more than just "not" linear, in means that the functions are polynomial or transcendental, are therefore much harder to analyze. More generally, any mathematical problem that can be linearized in some way usually becomes much more tractable (this is just one tool mathematicians use to try to simplify problems; another example used in geometry is finding symmetries).

In our case, the distance preserving property of isometries is not quite enough to guarantee linearity. But every isometry is the composition (product) of a linear isometry (osometry) followed by a translation.

Proof. The idea here is simple: determine where alpha takes the origin O and translate back; that will give us a map that fixes the origin. So let alpha(O)=D and define beta:=tau_{-D} \ alpha\  so that for any X,

 X^beta=tau_{-D}(X^alpha)=X^alpha-D.

Then for the origin,

 O^beta=O^alpha-D=D-D=O.

Hence beta fixes the origin. It is also an isometry because the composition of isometries is an isometry. So beta is in fact an osometry. But

 X^beta=X^alpha-D\ \Rightarrow\ X^alpha=X^beta+D.

Which means alpha=tau_D beta.\ \ square

Comment.

Not all linear transformations of the plane are isometries. For example, a shear $(x,y) |-> (x + y, y)$ is not. It is possible to use the above argument for much more general point transformations, which differ from a linear transformation only by a translation of the origin. But this the subject of a course on Linear Algebra.