Download Math 487 Exam 3 - Practice Problem Solutions and more Schemes and Mind Maps Algebra in PDF only on Docsity!
Math 487 Exam 3 - Practice Problem Solutions
- Consider the following functions:
- f : R → R, f (x) = x^2 + x
- f : R → R, f (x) = 2x
- f : R → R, f (x) = x − 3
- g : R^2 → R^2 , g(x, y) = (x, y^2 )
- g : R^2 → R^2 , g(x, y) = (x^3 , y)
- g : R^2 → R^2 , g(x, y) = (5x, 2 y)
- g : R^2 → R^2 , g(x, y) = (−x, −y)
(a) Which of these mappings are transformations? Justify your answer.
- Consider f (x) = x^2 + x. Since f (0) = 0 and f (−1) = 1 − 1 = 0, f is not 1-1 and hence is not a transformation.
- Consider f (x) = 2x. First suppose f (a) = f (b). Then 2a = 2b, so a = b hence f is 1-1.
Next, let r ∈ R. Let a = r 2. Then f (a) = 2 · r 2 = r. Hence f is also onto, hence f is a transformation.
- Consider f (x) = x − 3. First suppose f (a) = f (b). Then a − 3 = b − 3, so a = b hence f is 1-1.
Next, let r ∈ R. Let a = r + 3. Then f (a) = (r + 3) − 3 = r. Hence f is also onto, hence f is a transformation.
- Consider g(x, y) = (x, y^2 ). Notice that g(0, 1) = (0, 1) and g(0, −1) = (0, 1). Therefore, g is not 1-1 and hence is not a transformation.
- Consider g(x, y) = (x^3 , y). First suppose g((a, b)) = g((c, d)). Then a^3 = c^3 , so a = c and b = d. hence g is 1-1.
Next, let (r, s) ∈ R. Let a = 3
r and b = s. Then g((a, b)) = (r, s). Hence g is also onto, hence g is a transformation.
- Consider g(x, y) = (5x, 2 y). First suppose g((a, b)) = g((c, d)). Then 5a = 5c, so a = c and 2b = 2d, so b = d. hence g is 1-1.
Next, let (r, s) ∈ R. Let a = r 5 and b = s 2. Then g((a, b)) = (r, s). Hence g is also onto, hence g is a transformation.
- Consider g(x, y) = (−x, −y). First suppose g((a, b)) = g((c, d)). Then −a = −c, so a = c and −b = −d, so b = d. hence g is 1-1.
Next, let (r, s) ∈ R. Let a = −r and b = −s. Then g((a, b)) = (r, s). Hence g is also onto, hence g is a transformation.
(b) Which of these mappings are affine transformations? Justify your answer.
From above, f (x) = x^2 +x and g(x, y) = (x, y^2 ) are not transformations, so they are also not affine transformations.
Consider g(x, y) = (x^3 , y). Notice that (0, 0), (1, 1), (2, 2) are collinear since they are all on the line y = x. However, g((0, 0)) = (0, 0), g((1, 1)) = (1, 1), and g((2, 2)) = (8, 2). This first two points are both on the line y = x while the third point is not on y = x. Therefore, this transformation does not preserve lines, hence it cannot be an affine transformation.
It may or may not make sense to call f (x) = 2x and f (x) = x − 3 affine, depending on how we define lines since these are not functions of E. Since they both map R onto R, they preserve the only line in view, so we will consider both of these to be affine.
Consider g(x, y) = (5x, 2 y). We first consider how g acts on vertical lines. Given the line x = a, points on this line all have the form (a, y). Since g((a, y)) = (5a, 2 y), these points are all mapped to the vertical line x = 5a. Moreover, this map is onto , since if we set y = b 2 , we can get g((a, y)) = (5a, b) for any b.
Next, consider a non-vertical line of the form y = mx + b. Then g(a, b) = g(a, ma + b) = (5a, 2 ma + 2b), and g(c, d) = g(c, md + b) = (5c, 2 mc + 2b).
Then the slope of the line between the pair of image points is given by: m′^ = 2 mc+2 5 bc−−^25 maa −^2 b= (^2) 5(mc(c−−aa)) = 25 m.
Since any pair of inputs are mapped to a pair of points that satisfy this slope equation, this function maps the line into another line. The fact that this map is onto is similar to the argument above.
We will show that g(x, y) = (−x, −y) is affine by showing that it is an isometry below.
(c) Which of these mappings are isometries? Justify your answer.
First notice that all maps that are not affine are not isometries since we know that all isometries preserve lines.
Notice that f (x) = 2x is not an isometry since if a = 0 and b = 1, d(0, 1) = 1 while, since f (0) = 0 and f (1) = 2, d(f (0), f (1)) = 2, so distance is not preserved.
f (x) = x − 3 is an isometry since given inputs a and b, d(a, b) = |b − a|, while d(f (a), f (b)) = |(b − 3) − (a − 3)| = |b − a| = d(a, b).
Finally, g(x, y) = (−x, −y) is an isometry since d(g(a, b), g(c, d)) = d((−a, −b), (−c, −d))
(−c − (−a))^2 + (−d − (−b))^2 =
(−1)^2 (c − a)^2 + (−1)^2 (d − b)^2 =
(c − a)^2 + (d − b)^2 = d((a, b), (c, d)).
- Prove each of the following:
(a) The inverse of a transformation is a transformation.
Let f be a transformation from D to R. By definition, f is a function that is both one-to-one and onto. Then we may define the inverse function f −^1 as follows. For every r ∈ R, since f is 1-1 and onto, there is exactly one d ∈ D such that f (d) = r. Then we define f −^1 (r) = d. Then f −^1 is a function. Since f is well defined, f −^1 is onto (for every d ∈ D, there is an r ∈ R such that f (r) = d). Also, f −^1 is well defined since we for each r ∈ R, there is a d ∈ D so that f −^1 (r) = d and there is exactly one such d since f is 1-1. Finally, f −^1 must be 1-1 since if f −^1 (r 1 ) = f −^1 (r 2 ) = d, f ◦ f −^1 (r 1 ) = f ◦ f −^1 (r 2 ) = f (d), so r 1 = r 2. Hence f −^1 is a transformation.
(b) The inverse of an isometry is an isometry.
We already know from above that the inverse of a transformation is a transformation, so we need only prove that the inverse of an isometry preserves the distance between any pair of points. Let f be an isometry from D to R. Let r 1 , r 2 ∈ R and suppose that d(r 1 , r 2 ) = M. To obtain a contradiction, suppose that d(f −^1 (r 1 ), f −^1 (r 2 )) = M ′^6 = M. Then, since f is an isometry, M ′^ = d
f
f −^1 (r 1 )
, f
f −^1 (r 2 )
= d(r 1 , r 2 ) = M. This is a contradiction. Hence f −^1 is an isometry.
(c) The set of all transformations of a plane forms a group under the operation function composition.
- We showed in HW 3.4 that the composition of two transformations in a transformation. Therefore, the set of transformations of a plane is closed under function composition.
- The identity map is a transformation since it is both 1-1 and onto.
- From part (a) above, the inverse of a transformation is a transformation.
- The fact that function composition is associative is a standard result from the algebra of functions. (i.e. one can show that f ◦ (g ◦ h) = (f ◦ g) ◦ h).
(d) Given f , an isometry of E and a triangle △ABC, if f (A) = A′, f (B) = B′, and f (C) = C′, then △ABC ∼= △A′B′C′.
Suppose that d(A, B) = l 1 , d(A, C) = l 2 and d(B, C) = l 3. Since f is an isometry, d(A′, B′) = d(A, B) = l 1 , d(A′, C′) = d(A, C) = l 2 and d(B′, C′) = d(B, C) = l 3. Then AB ∼= A′B′, AC ∼= A′C′, BC ∼= B′C′. Hence, by the SSS Theorem, △ABC ∼= △A′B′C′.
i. P (2, 3 , 1), Q(5, 3 , 1)
Notice that
P Q = 〈 3 , 0 〉.
Therefore, TP Q =
ii. P (2, 3 , 1), Q(− 1 , 5 , 1)
Notice that
P Q = 〈− 3 , 2 〉.
Therefore, TP Q =
(b) Find the matrix for the transformation RC,θ given:
i. C(0, 0 , 1) and θ = 135◦
Then cos θ = −
√ 2 2 and sin^ θ^ =^
√ 2 2
Hence RC,θ =
√ 2 2 −^
√ 2 √^2 2 2 −^
√ 2 2 0 0 0 1
ii. C(− 1 , 2 , 1) and θ = 30◦
Then cos θ =
√ 3 2 and sin^ θ^ =^
1 2 , and^
OC = 〈− 1 , 2 〉
Notice that TOC =
, TCO =
, and RO,θ =
√ 3 2 −^
1 2 0 1 2
√ 3 2 0 0 0 1
Finally, RC,θ =
√ 3 2 −^
1 2 0 1 2
√ 3 2 0 0 0 1
We leave it to the reader to complete the matrix multiplication.
(c) Find the matrix for Rℓ given: i. ℓ
[
]
Notice that to map h
[
]
onto ℓ, we use rotation about O(0, 0 , 1) with θ = 90◦.
Also notice that cos θ = 0 and sin θ = 1. Then T =
Similarly, rotating about O with θ = 270◦^ gives T −^1 , so T −^1 =
The reflection Rh is given by the matrix A =
Then Rℓ = T AT −^1 =
We leave it to the reader to complete the matrix multiplication.
ii. ℓ
[
]
Notice that ℓ is the line with slope -1 intersecting the x-axis at the point C(− 1 , 0 , 1). Therefore, to map h
[
]
onto ℓ, we use rotation about C(− 1 , 0 , 1) with θ = 135◦.
Also notice that cos θ = −
√ 2 2 and sin^ θ^ =^
√ 2
- Then^ RO,θ^ =
√ 2 2 −^
√ 2 √ 2 0 2 2 −^
√ 2 2 0 0 0 1
,^ TOC =
and TCO =
Hence T = RC, 45 ◦ = TOC RO,θ TCO =
√ 2 2 −^
√ 2 √^2 2 2 −^
√ 2 2 0 0 0 1
Similarly, rotating about C with θ = 225◦^ gives T −^1 , and the reflection Rh is given by the matrix
A =
Then Rℓ = T AT −^1.
We leave it to the reader to complete the computations.
(d) Find the matrix for GP Q given P (2, 3 , 1) and Q(3, 3 , 1).
First notice that
P Q = 〈 1 , 0 〉.
Therefore, TP Q =
Next, notice that since the line ℓ =
P Q is the horizontal line
[
]
, then the translation that takes h
[
]
to ℓ is:
T =
Thus Rℓ = T AT −^1 =
Finally, GP Q = TP QT AT −^1. The details of the matrix multiplication are left to the reader.
- For each statement, determine whether the statement is true or false. Then briefly justify your answer.
(a) Every isometry of E is a translation, a rotation, or a reflection.
This statement is False. Glide reflections are also isometries of E, and these to not fall into any of the categories listed above.
(b) The product of any two distinct rotations is a translation.
This is False. We often get translations by composing two rotations, but in the case where two rotations about same center point C are composed, we do not get a translation. We get another rotation about C.
(c) A nontrivial translation has no invariant points.
True. To see this, suppose that a translation f has an invariant point P. Then f (P ) = P. But then P P ′^ = 〈 0 , 0 〈. Since all points are moved along this same vector, the translation is trivial (i.e. it fixes every point in the plane).
(d) A nontrivial rotation has exactly one invariant point.
True. The only point that is fixed by a non-trivial rotation is the center of the rotation.
(e) A nontrivial translation has no invariant lines.
False. Every line that is parallel to the translation vector in standard position at the origin is invariant under a translation.
Recall that A =
(^) is the matrix for the reflection across the line h
[
]
Hence, the matrix of a reflection that takes P to Q is given by: Rℓ = T ◦ Rh ◦ T −^1 , so the matrix B of Rℓ is given by:
B = T AT −^1 where A is as above, and T is the direct isometry that maps h to ℓ. Notice that T is given by:
RC,θ = TOC RO,θTCO with C( 154 , 0 , 1) and θ = 116. 565 ◦.
The specific computations from here are left to the reader.
(c) Find the matrix of a rotation that maps P to Q.
One way to obtain a rotation that takes P to Q is to consider a rotation about the point C(3, 32 , 1) and to set θ = 180◦.
In this case, TOC =
, TOC =
, and RO,θ =
Hence RC,θ = TOC RO,θTCO =
- Consider the following transformation matrices:
A =
B =
C =
D =
E =
1 2
√ 3 2 0 −
√ 3 2
1 2 0 0 0 1
F =
(a) Which of these are the matrix of an affine transformation of E?
From the form of the matrices, we see that all of these matrices represent affine transformations with the exception of A.
(b) Which of these are the matrix of an isometry of E?
Recall that a transformation is an isometry if and only if its matrix representation has one of the following forms:
cos θ − sin θ a sin θ cos θ b 0 0 1
(^) or
cos θ sin θ a sin θ − cos θ b 0 0 1
From this, we see that C, D, E and F represent isometries while A and B do not.
(c) Which of these are the matrix of an direct isometry of E?
Direct isometries are transformations whose matrix representations have the form:
cos θ − sin θ a sin θ cos θ b 0 0 1
Therefore, we see that E and F represent direct isometries.
(d) Which of these are the matrix of a rotation of E?
Rotations are direct isometries. We will see below that F is a translation, while, looking at the form of E, we see that this matrix represents a rotation (in fact, it is a rotation about O).
(e) Which of these are the matrix of a translation of E?
Translations are also direct isometries. We saw above that E is a rotation. Looking at the form of F , we see that this matrix represents a translation.
- Given the plane figure in E shown below, accurately draw the image of this figure under each of the following isometries:
A
B (^) C
D
E
F (^) G
H
(a) RB, 90
A B^
D C
E F^
H G
A
B (^) C
D
E
F (^) G
H
(b) TF G
A
B (^) C
D
E
F (^) G
A’ H
B’ (^) C’
D’
E’
F’ (^) G’
H’
(c) Rℓ, where ℓ =
HG.
A
C B
D
E
G F
A H
B (^) C
D
E
F (^) G
H
(d) Rℓ, where ℓ =
ED.
A (^) B
D C
E
F H G
A
B (^) C
D
E
F (^) G
H
(e) GCD
A
B C
D
E
F (^) G
H A
B (^) C
D
E
F (^) G
H
A
C B
D
E
G F
H
id R R^2 R^3 V H D 1 D 2 id id R R^2 R^3 V H D 1 D 2 R R R^2 R^3 id D 1 D 2 H V R^2 R^2 R^3 id R H V D 2 D 1 R^3 R^3 id R R^2 D 2 D 1 V H V V D 2 H D 1 id R^2 R^3 R H H D 1 V D 2 R^2 id R R^3 D 1 D 1 V D 2 H R R^3 id R^2 D 2 D 2 H D 1 V R^3 R R^2 id
The multiplication shown in the table above demonstrates that these transformations form a group under compo- sition.
