Definition 4.1.1. Transformation.
A function is a transformation if and only if it is one-to-one and onto.
Section 4.1 Transformation
Subsection 4.1.1 Planar Transformations
Definition 4.1.2. Planar Transformation.
A transformation is a planar transformation if and only if it is from \(\R^2\) to \(\R^2.\)
For this course all transformations will be transformations of the Euclidean plane.
Checkpoint 4.1.3.
Describe the effect of each of the following transformations by considering the effects on the region (area) with the following vertices. (0,0), (2,0), (3,5), (0,4). Hint: map the vertices, then map the lines connecting the vertices.
(a)
\(T_1:(x,y) \mapsto (y,x)\text{.}\)
Solution.
This translation keeps the same area as the pre-image, it has been reflected on the y—axis and rotated 90 degrees clockwise.
(b)
\(T_2:(x,y) \mapsto (x/2,y/2)\text{.}\)
Solution.
This translation has 1/4th the area of the pre-image.
(c)
\(T_3:(x,y) \mapsto (1-x^3,1-y^3)\text{.}\)
Solution.
The translation has had its origin moved to (1,1) and has been reflected on the y-axis and then reflected again over the x-axis, its area has increased by the third power.
(d)
\(T_4:(x,y) \mapsto (\sqrt{x^2+y^2}x,\sqrt{x^2+y^2}y)\text{.}\)
Solution.
The translation has an increased area.
Subsection 4.1.2 Isometry
Definition 4.1.4. Isometry.
A transformation is an isometry if and only if \(\|P-Q\|=\|T(P)-T(Q)\|.\)
Checkpoint 4.1.5.
Determine which of the following transformations are isometries.
(a)
\(T_1:(x,y) \mapsto (2x,2y)\)
Solution.
In this case, \(\| T_1(P) - T_1(Q) \| = \sqrt{ (2a - 2c)^2 + (2b-2d)^2}\text{.}\) This is plainly not equal to \(\|P-Q\|\text{.}\) This is not an isometry.
(b)
\(T_2:(x,y) \mapsto (-y,-x)\)
Solution.
Calculating, we find that
\begin{align*}
\|T_2(P) - T_2(Q)\| \amp= \sqrt{(-b - (-d))^2 + (-a -(-c))^2} \\
\amp= \sqrt{(d-b)^2 + (c-a)^2} \\
\amp= \sqrt{(-|b-d|)^2 + (-|a-c|)^2}\\
\amp= \sqrt{|a-c|^2 + |b-d|^2} \\
\amp= \|P-Q\|.
\end{align*}
This is an isometry.
(c)
\(T_3:(x,y) \mapsto \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \)
Solution.
This transformation is a rotation around the origin. The transformation could also be written
\begin{equation*}
T_3 : (x,y) \rightarrow (x\cos\theta + y\sin\theta, -x\sin\theta+y\cos\theta).
\end{equation*}
Consequently,
\begin{align*}
\|T_3(P) - T_3(Q)\| \amp =
\left\|
\begin{matrix}
a\cos\theta + b\sin\theta - c\cos\theta - d\sin\theta \\
-a\sin\theta+ b\cos\theta + c\sin\theta - d\cos\theta
\end{matrix}
\right\| \\
\amp=
\left\|
\begin{matrix}
(a-c)\cos\theta + (b-d)\sin\theta \\
(c-a)\sin\theta + (b-d)\cos\theta
\end{matrix}
\right\| \\
\amp=
\sqrt{((a-c)^2 + (b-d)^2)(\cos^2\theta + \sin^2\theta)} \\
\amp = \sqrt{(a-c)^2 + (b-d)^2} \\
\amp= \| P - Q \|.
\end{align*}
This is an isometry.
Lemma 4.1.6.
The composition of two isometries is an isometry.
Theorem 4.1.7. Isometries preserve colinearity.
For any isometry \(T\) if \(A,\) \(B,\) and \(C\) are colinear, then \(T(A),\) \(T(B),\) and \(T(C)\) are colinear.
Proof.
Proof by contraposition. Assume \(T(A)\text{,}\) \(T(B)\text{,}\) and \(T(C)\) are not collinear. Without loss of generality, by triangle inequality \(\|T(A)T(C)\| \lt \|T(A)T(B)\| + \|T(B)T(C)\|\text{.}\) Since \(\|AC\| \lt \|AB\| + \|BC\|\text{,}\) A, B, and C form a triangle. Thus A, B, and C are not collinear.
Corollary 4.1.8. Isometries preserve betweeness.
For any isometry \(T\) if \(A-B-C\) then \(T(A)-T(B)-T(C).\)
Proof.
\(A-B-C\) are collinear. Since
\begin{align*}
\|A-B\| = & \|T(A)-T(B)\| = d_1\\
\|B-C\| = & \|T(B)-T(C)\| = d_2\\
\|A-C\| = & \|T(A)-T(C)\| = d_3\\
= & d_1 + d_2
\end{align*}
So \(T(B)\) is between \(T(A)\) and \(T(C)\text{.}\)
Proof.
Let T be an isometry and let \(A, B\text{,}\) and \(C\) exist such that \(A - B - C\text{,}\) where the Ruler postulate tells us the real numbers corresponding to \(A, B\text{,}\) and \(C\text{,}\) say \(a, b\text{,}\) and \(c\text{,}\) exist such that \(a \lt b \lt c\text{.}\) We already know \(T(A), T(B)\text{,}\) and \(T(C)\) are collinear by Theorem 4.1.7. Note \(\|A-B\| + \|B-C\| = \|A-C\|\) implies \(\|T(A)-T(B)\| + \|T(B)-T(C)\| = \|T(A)-T(C)\|\text{,}\) and since \(\|T(A)-T(C)\| > \|T(A)-T(B)\|\) and \(\|T(A)-T(C)\| > \|T(B)-T(C)\|\text{,}\) then \(T(B)\) is in between \(T(A)\) and \(T(C)\) (since \(a \lt b \lt c\)). So, \(T(A) - T(B) - T(C)\text{.}\)
Theorem 4.1.9. Isometries preserve triangles.
For any isometry \(T\) and any three points \(\triangle ABC \cong \triangle T(A)T(B)T(C).\)
Theorem 4.1.10. Isometries preserve angles.
For any isometry \(T\) and any three points \(m\angle ABC = m\angle T(A)T(B)T(C).\)
Proof.
By the definition of triangle congruence, all corresponding sides and angles of the two triangles are congruent. Thus, angle \(ABC\) and angle \(T(A)T(B)T(C)\) are congruent. By the definition of congruent angles, the measures of angle \(ABC\) and angle \(T(A)T(B)T(C)\) are equal.
Theorem 4.1.11. Isometries preserve parallelism.
For any isometry \(T\) \(\ell_1 \parallel \ell_2\) if and only if \(T(\ell_1) \parallel T(\ell_2).\)
Proof.
Suppose that \(l_1\) is parallel to \(l_2\text{.}\) Then by the equidistance postulate, the two line are everywhere equidistant. Since isometries preserve distance, \(T(l_1)\) is everywhere equidistant to \(T(l_2)\text{,}\) so those two lines are parallel. The proof of the converse is identical.
Theorem 4.1.12. Isometries preserve circles.
For any isometry \(T\) circles are mapped to congruent circles.
Subsection 4.1.3 Dilations
Definition 4.1.13. Dilation.
A transformation is a dilation if and only if it can be defined by a point \(Z\) and a ratio \(k\) such that \(T(P)=Q\) where \(Z-P-Q\) and \(\|ZQ\|/\|ZP\|=k.\)
Definition 4.1.14. Similarity.
A transformation is a similarity if and only if it can be expressed as a composition of an isometry and a dilation.
Lemma 4.1.15. Similarity scales segments uniformly.
For any similarity \(T,\) \(\|T(A)T(B)\|/\|AB\| =k.\)
Proof.
\(AB\) is parallel to \(A^\prime B^\prime\) since \(Z-A-A^\prime\) and \(Z-B-B^\prime\) are collinear, and \(\frac{\|ZA^\prime\|}{\|ZA\|} = \frac{\|ZB^\prime\|}{\|ZB\|} = k \text{.}\) So, \(\angle A^\prime B^\prime Z \cong \angle ABZ \) and \(\angle B^\prime A^\prime Z \cong \angle BAZ \text{.}\) Since \(\angle AZB = \angle A^\prime ZB^\prime \text{,}\) by angle-angle-angle similarity, \(\Delta ABZ \) is similar to \(\Delta A^\prime B^\prime Z \text{.}\) So \(\frac{\|ZA^\prime\|}{\|ZA\|} = \frac{\|ZB^\prime\|}{\|ZB\|} = \frac{\|A^\prime B^\prime\|}{\|AB\|} = k \)
Theorem 4.1.16. Similarity preserves colinearity.
For any similarity \(T\) if \(A,\) \(B,\) and \(C\) are colinear, then \(T(A),\) \(T(B),\) and \(T(C)\) are colinear.
Proof.
Let \(T\) be a similarity so \(A, B,\) and \(C\) are colinear. Remember that a similarity is expressed as a composition of an isometry and a dilation. \(\frac{\|ZT(A)\|}{\|ZA\|} = \frac{\|ZT(B)\|}{\|ZB\|} = \frac{\|ZT(C)\|}{\|ZC\|} = k\) from the definition of dilation. From the definition of isometry: \(\|A - B\| = \|T(A) - T(B)\|\text{,}\) \(\|B - C\| = \|T(B) - T(C)\|\text{,}\) and \(\|A - C\| = \|T(B) - T(C)\|\text{.}\) Without loss of generality, suppose \(A - B - C,\) so \(\|A - B\| + \|B - C\| = \|A - C\|.\) Now suppose that \(T(A), T(B),\) and \(T(C)\) are non-colinear. We can construct segments between these three points and so \(\|T(A) - T(B)\| + \|T(B) - T(C)\| \neq \|T(A) - T(C)\|\text{.}\) This means that \(k \cdot (\|A - B\| + \|B - C\|) \neq k \cdot (\|A - C\|).\) This simplifies to \(\|A - B\| + \|B - C\| \neq \|A - C\|\) which is a contradiction. Thus, \(T(A), T(B), and T(C)\) are colinear.
Corollary 4.1.17. Similarity preserves betweeness.
For any similarity \(T\) if \(A-B-C\) then \(T(A)-T(B)-T(C).\)
Theorem 4.1.18. Similarity preserves triangle similarity.
For any similarity \(T\) and any three points \(\triangle ABC \sim \triangle T(A)T(B)T(C).\)
Theorem 4.1.19. Similarity preserves angles.
For any similarity \(T\) and any three points \(m\angle ABC = m\angle T(A)T(B)T(C).\)
Proof.
From Theorem 4.1.18, \(\triangle ABC \sim \triangle T(A)T(B)T(C)\text{.}\)
The corresponding angles of the two triangles are congruent by definition of triangle similarity. So, \(\angle ABC \cong \angle T(A)T(B)T(C)\text{.}\)
The measures of the angles are equal by definition of angle congruence. Thus, \(m\angle ABC = m\angle T(A)T(B)T(C)\text{.}\)
Theorem 4.1.20. Similarity preserves parallelism.
For any similarity \(T\) \(\ell_1 \parallel \ell_2\) if and only if \(T(\ell_1) \parallel T(\ell_2).\)
Proof.
Let \(T\) be a similarity and let \(l_1\) and \(l_2\) be parallel lines. Since \(l_1\) and \(l_2\) are parallel, they are everywhere equidistant by the equidistance postulate. Since \(T\) is a similarity, it scales distances uniformly, if \(A\) is any point on \(l_1\text{,}\) then \(\mathrm{d}(T(A),T(l_2) = k\mathrm{d}(A,l_2)\) for some \(k\text{.}\) But then \(T(l_1)\) and \(T(l_2)\) are everywhere equidistant, and thus parallel. The converse is identical. (Take a similarity \(T^\prime = T^{-1}\text{.}\))
Theorem 4.1.21.
For any similarity \(T\) circles are mapped to circles.
Lemma 4.1.22. Similarities are closed.
The composition of two similarities is a similarity.
Proof.
We first show that isometries and dilations commute. Let \(M\) be an isometry and \(D\) a dilation with center \(Z\) and ratio \(k\text{.}\) Let \(A\) be a point in \(\R^2\) and suppose that \(M(D(A)) = P\text{.}\) Since \(D\) is a dilation, \(Z - A - D(A)\) and \(\|Z D(A)\| = k\|Z A\|\text{.}\) Since \(M\) is an isometry \(\|M(Z) P\| = \|Z D(A)\|\text{.}\)
Consider \(D(M(A))\text{.}\) By the definition of a dilation, \(D(M(A))\) must be colinear with \(M(Z)\) and \(M(A)\) and \(\|M(Z) D(M(A))\| = k\|M(Z) M(A)\|\text{.}\) But this last distance is just \(k \|Z A\|\text{,}\) since \(M\) is an isometry. By the ruler postulate, there is a unique point on the line \(M(Z)M(A)\) this distance away from \(M(Z)\text{.}\) Therefore \(D(M(A)) = P = M(D(A))\text{.}\)
The theorem follows immediately.
Suppose \(T_1\) and \(T_2\) are similarities. Then there exist isometries \(M_1\) and \(M_2\) and dilations \(D_1\) and \(D_2\) so that \(T_1 = M_1 \circ D_1\) and \(T_2 = M_2 \circ D_2\text{.}\) We show that \(T_1 \circ T_2\) is a similarity. Then, since dilations and isometries commute,
\begin{align*}
T_1 \circ T_2 \amp = M_1 \circ D_1 \circ M_2 \circ D_2 \\
\amp = M_1 \circ M_2 \circ D_1 \circ D_2.
\end{align*}
Since isometries and dilations are closed, this is the composition of an isometry and a dilation, so \(T_1 \circ T_2\) is a similarity. (Really, this requires an additional lemma to the effect that dilations are closed; one can easily convince oneself this is true, but the margin is too small to contain the proof.)
