Transformations and Matrices in the Cartesian Plane - Lecture Notes




First some definitions and notation. We let MATH and if MATH and MATH are two points in $\QTR{bf}{R}^{2},$ then

Now to define the transformations we will be considering.


Definition: A mapping MATH is called a translation if there is a vector $\overrightarrow{v}$ such that MATH for every point $P$ in $\QTR{bf}{R}^{2}.$ The vector $\overrightarrow{v}$ is called the translation vector. The translation with translation vector $\overrightarrow{v}$ is sometimes denoted MATH

Definition: A mapping $R$ is called a reflection if there is a line $l$ such that, for any point $P$ in $\QTR{bf}{R}^{2}$, MATH where, if $Q$ is the foot of the perpendicular dropped from $P$ to $l,$ MATH The line $l$ is called the axis of reflection and a reflection with axis $l$ is sometimes denoted $R_{l}.\bigskip $

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

First we establish the connection between translations and reflections.


Theorem: Let $l$ and $m$ be two parallel lines. Then the mapping $R_{l}\circ R_{m}$ is a translation.

Proof: Note that we need to find a vector $z$ such that MATH for every $P$ in $\QTR{bf}{R}^{2}. $ Note that $\overrightarrow{z}$ does not depend upon $P.$

Suppose that $P$ is given, that MATH where MATH and MATH Then
MATH
We claim that the vector MATH does not depend upon the initial choice of $P$ and hence is the sought after vector MATH

To see the claim, note that the point MATH is on the line $m$ and the point MATH is on the line $l.$ Note that MATH and MATH are perpendicular to $m$ and $l$ respectively, and hence parallel. It follows that the vector that would take a point from $m$ to $l$ can be computed as follows:
MATH
and thus MATH is independent of the choice of $P.$

Observe that MATH and MATH are the same vector in the figure below (hence does not depend on where $P$ or $Q$ was initially chosen.)


matlct__71.png

Setting MATH now yields that
MATH
for any $P$ in $\QTR{bf}{R}^{2}$ and hence MATH

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 MATH is a translation, then there are parallel lines $l$ and $m$ such that MATH

Proof: Let $S$ be any point in $\QTR{bf}{R}^{2}$ and let $l$ be the line containing $S$ perpendicular to MATH Let $m$ be the line
MATH
Note that the proof of the preceding theorem yields MATH

Rotations: Rotations centered at the origin have a nice representation in terms of matrices. Let us consider a rotation $M$ centered at the origin that has angle of rotation $\phi $. Now let MATH be any point and suppose that MATH [think of polar coordinates.] Note that $M$ will rotate $\left( x,y\right) $ to the point MATH Hence
MATH
or, in matrix form,
MATH



Reflections: As in the case of rotations, if a line $l$ passes through the origin there is a nice matrix representation of $R_{l}$ . Let MATH be any point and suppose that $l$ makes an angle of $\phi $ with the $x$-axis. We would like to get the coordinates of the point MATH

The first difficulty is to get the vector MATH to compute MATH Note that line $l$ has direction MATH and hence a vector that will `move' $P$ to $l$ in a direction perpendicular to $l$ has direction MATH [See Figure 2: There is a delicate point here. Be sure to understand why I chose MATH and not MATH (both are perpendicular to MATH] Now


MATH





matlct__117.png

Let us do the calculations for $x^{\prime }$ and $y^{\prime }$ separately. Before starting, let us recall the trigonometric identities
MATH
First we compute $x^{\prime }.$
MATH
and now we compute $y^{\prime }$
MATH

Summarizing the above in matrix form yields
MATH

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