Transformations and Matrices in the Cartesian Plane - Lecture Notes
First some definitions and notation. We let

and if

and

are two points in

then
the (Euclidean) distance from

to

is

The vector

is

with initial point

and terminal point

Note that we use angle brackets to denote a vector and round parentheses to
denote a point.
If

and

is a vector, then

is
the point

Now to define the transformations we will be considering.
Definition: A mapping

is
called a translation if there is a vector

such
that

for
every point

in

The
vector

is
called the translation vector. The translation with translation vector

is sometimes denoted

Definition: A mapping

is
called a reflection if there is a line

such
that, for any point

in

,

where,
if

is
the foot of the perpendicular dropped from

to


The
line

is
called the axis of reflection and a reflection with axis

is
sometimes denoted

Note that the translation vector does not depend upon the point being
translated. In the definition of reflection, the vector

does depend upon the point being reflected.
First we establish the connection between translations and
reflections.
Theorem: Let

and

be
two parallel lines. Then the mapping

is
a translation.
Proof: Note that we need to find a vector

such that

for every

in

Note that

does not depend upon

Suppose that

is given, that

where

and

Then
We
claim that the vector

does not depend upon the initial choice of

and hence is the sought after vector

To see the claim, note that the point

is on the line

and the point

is on the line

Note that

and

are perpendicular to

and

respectively, and hence parallel. It follows that the vector that would take a
point from

to

can be computed as follows:

and
thus

is independent of the choice of

Observe that

and

are the same vector in the figure below (hence does not depend on where

or

was initially chosen.)

Setting

now yields that

for
any

in

and hence

We can use the above calculations to establish the converse of the theorem:
any translation can be described as the composition of two
reflections.
Corollary: It

is
a translation, then there are parallel lines

and

such
that

Proof: Let

be any point in

and let

be the line containing

perpendicular to

Let

be the line

Note
that the proof of the preceding theorem yields

Rotations: Rotations centered at the origin have a nice
representation in terms of matrices. Let us consider a rotation

centered at the origin that has angle of rotation

.
Now let

be any point and suppose that

[think of polar coordinates.] Note that

will rotate

to the point

Hence

or,
in matrix form,

Reflections: As in the case of rotations, if a line

passes through the origin there is a nice matrix representation of

. Let

be any point and suppose that

makes an angle of

with the

-axis.
We would like to get the coordinates of the point

The first difficulty is to get the vector

to compute

Note that line

has direction

and hence a vector that will `move'

to

in a direction perpendicular to

has direction

[See Figure 2: There is a delicate point here. Be sure to understand why I
chose

and not

(both are perpendicular to

]
Now


Let us do the calculations for

and

separately. Before starting, let us recall the trigonometric identities

First
we compute



and
now we compute



Summarizing the above in matrix form
yields