b) For any slope m, show that PmQm = Pm, where Pm and Qm are transformations defined in Theorem 2.6.5 and 2.6.6 (textbook, pages 112 and 113).
b) For any slope m, show that PmQm = Pm, where Pm and Qm are transformations defined in Theorem 2.6.5 and 2.6.6 (textbook, pages 112 and 113).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Transcribed Image Text:b) For any slope m, show that PmQm = Pm, where Pm and Qm are transformations defined in Theorem 2.6.5
and 2.6.6 (textbook, pages 112 and 113).
![Reflections
y
0
0
y
Figure 2.6.12
Let denote the angle between the positive x axis and the line y = mx.
The key observation is that the transformation Qm can be accomplished in
three steps: First rotate through -0 (so our line coincides with the x axis), then reflect in the x axis, and
finally rotate back through 0. In other words:
QmRe Qo°R_e
Since R-0, Qo, and Re are all linear, this (with Theorem 2.6.3) shows that Qm is linear and that its matrix
is the product of the matrices of Re, Qo, and R_e. If we write c = cos and s = sin for simplicity, then
the matrices of Re, R-e, and Qo are
√1+m²
40
Qm (x)
ε] [$]·
Hence, by Theorem 2.6.3, the matrix of Qm = Reo QoR_e is
C
636
*][19][
1
y
m
y = mx
Figure 2.6.13
m
Projections
Pm (x)
0
Theorem 2.6.5).
Solution. The matrix of R
X
reflection in the y axis is
0
1+m²
y = mx
S
X
y = mx
where c = cos 0 =
This gives:
X
Figure 2.6.14
The line through the origin with slope m has equation y = mx, and we let
Qm: R² → R² denote reflection in the line y = mx.
This transformation is described geometrically in Figure 2.6.12. In
words, Qm(x) is the "mirror image" of x in the line y=mx. If m=0 then
Qo is reflection in the x axis, so we already know Qo is linear. While we
could show directly that Qm is linear (with an argument like that for Re),
we prefer to do it another way that is instructive and derives the matrix of
Qm directly without using Theorem 2.6.2.
-1
Note that if m= 0, the matrix in Theorem 2.6.5 becomes
analysis fails for reflection in the y axis because vertical lines have no slope.
exercise to verify directly that reflection in the y axis is indeed linear with matrix
X
√1+m²
so the matrix
13The matrix of Recomes from the matrix of Re using the fact that, for all angles 0, cos(-0) = cos 0 and
sin(-0) sin(0).
14Note that
-1 0
= lim 1
0 1
m-001+m²
S
• ] = [ ²2 ₁ ²³
2
2sc
We can obtain this matrix in terms of m alone. Figure 2.6.13 shows
that
√₁+m² and sin 0.
of Qm becomes
-플
Example 2.6.8
Let T: R² → R² be rotation through - followed by reflection in the y axis. Show that I' is a
reflection in a line through the origin and find the line.
is
1-m²
2m
-1 0
01
-1
and
0 -1
0
-S
c²-5² 2sc
2sc s²_c²
cos 0 =
Theorem 2.6.5
Let Qm denote reflection in the line y = mx. Then Qm is a linear
1-m²
2m
transformation with matrix²2 2m
m² - 1
2m
m² -1
].
and s= sin =
matrix
First observe that
cos(-2)
sin(-)
Hence the matrix of T is
-1
- sin(-4)
cos(-
√1+m²*
respectively. 13
Se
2sc
s²_c²
• [3] = [6] ·
-S
m
√1+m²
01
-1 0
as before. So, Pm is linear with matrix
=[¹ 1
1 m² 2m
2m m² 1
1+m²
as expected. Of course this
However it is an easy
-1
14
and this is reflection in the line y = -x (take m= −1 in
The method in the proof of Theorem 2.6.5 works more generally. Let
Pm : R² → R² denote projection on the line y = mx. This transformation is
described geometrically in Figure 2.6.14.
If m=0, then Po
0
2.6. Linear Transformations 113
for all
0
[B]
Hence the argument above for Qm goes through for Pm.
Pm RooPooR_0
с
and the matrix of
C S
[3][88][+]=R*]
in R², so Po is linear with
SC
sc s²
Theorem 2.6.6
Let Pm : R² R² be projection on the line y = mx. Then Pm is a linear transformation with matrix
1 m
mm²
Again, if m= 0, then the matrix in Theorem 2.6.6 reduces to
10
0 0
as expected. As the y axis has
no slope, the analysis fails for projection on the y axis, but this transformation is indeed linear with matrix
00
as is easily verified directly.
0 1
Note that the formula for the matrix of Qm in Theorem 2.6.5 can be derived from the above formula
for the matrix of Pm. Using Figure 2.6.12, observe that Qm(x) = x+2[Pm (x) -x] so Qm(x) = 2Pm(x) — x.
Substituting the matrices for P(x) and 12 (x) gives the desired formula.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F293ec77a-bb0f-4498-a317-90c98547c742%2F1bfc6408-8c13-47a3-afd8-4562aace9252%2Fjizewr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Reflections
y
0
0
y
Figure 2.6.12
Let denote the angle between the positive x axis and the line y = mx.
The key observation is that the transformation Qm can be accomplished in
three steps: First rotate through -0 (so our line coincides with the x axis), then reflect in the x axis, and
finally rotate back through 0. In other words:
QmRe Qo°R_e
Since R-0, Qo, and Re are all linear, this (with Theorem 2.6.3) shows that Qm is linear and that its matrix
is the product of the matrices of Re, Qo, and R_e. If we write c = cos and s = sin for simplicity, then
the matrices of Re, R-e, and Qo are
√1+m²
40
Qm (x)
ε] [$]·
Hence, by Theorem 2.6.3, the matrix of Qm = Reo QoR_e is
C
636
*][19][
1
y
m
y = mx
Figure 2.6.13
m
Projections
Pm (x)
0
Theorem 2.6.5).
Solution. The matrix of R
X
reflection in the y axis is
0
1+m²
y = mx
S
X
y = mx
where c = cos 0 =
This gives:
X
Figure 2.6.14
The line through the origin with slope m has equation y = mx, and we let
Qm: R² → R² denote reflection in the line y = mx.
This transformation is described geometrically in Figure 2.6.12. In
words, Qm(x) is the "mirror image" of x in the line y=mx. If m=0 then
Qo is reflection in the x axis, so we already know Qo is linear. While we
could show directly that Qm is linear (with an argument like that for Re),
we prefer to do it another way that is instructive and derives the matrix of
Qm directly without using Theorem 2.6.2.
-1
Note that if m= 0, the matrix in Theorem 2.6.5 becomes
analysis fails for reflection in the y axis because vertical lines have no slope.
exercise to verify directly that reflection in the y axis is indeed linear with matrix
X
√1+m²
so the matrix
13The matrix of Recomes from the matrix of Re using the fact that, for all angles 0, cos(-0) = cos 0 and
sin(-0) sin(0).
14Note that
-1 0
= lim 1
0 1
m-001+m²
S
• ] = [ ²2 ₁ ²³
2
2sc
We can obtain this matrix in terms of m alone. Figure 2.6.13 shows
that
√₁+m² and sin 0.
of Qm becomes
-플
Example 2.6.8
Let T: R² → R² be rotation through - followed by reflection in the y axis. Show that I' is a
reflection in a line through the origin and find the line.
is
1-m²
2m
-1 0
01
-1
and
0 -1
0
-S
c²-5² 2sc
2sc s²_c²
cos 0 =
Theorem 2.6.5
Let Qm denote reflection in the line y = mx. Then Qm is a linear
1-m²
2m
transformation with matrix²2 2m
m² - 1
2m
m² -1
].
and s= sin =
matrix
First observe that
cos(-2)
sin(-)
Hence the matrix of T is
-1
- sin(-4)
cos(-
√1+m²*
respectively. 13
Se
2sc
s²_c²
• [3] = [6] ·
-S
m
√1+m²
01
-1 0
as before. So, Pm is linear with matrix
=[¹ 1
1 m² 2m
2m m² 1
1+m²
as expected. Of course this
However it is an easy
-1
14
and this is reflection in the line y = -x (take m= −1 in
The method in the proof of Theorem 2.6.5 works more generally. Let
Pm : R² → R² denote projection on the line y = mx. This transformation is
described geometrically in Figure 2.6.14.
If m=0, then Po
0
2.6. Linear Transformations 113
for all
0
[B]
Hence the argument above for Qm goes through for Pm.
Pm RooPooR_0
с
and the matrix of
C S
[3][88][+]=R*]
in R², so Po is linear with
SC
sc s²
Theorem 2.6.6
Let Pm : R² R² be projection on the line y = mx. Then Pm is a linear transformation with matrix
1 m
mm²
Again, if m= 0, then the matrix in Theorem 2.6.6 reduces to
10
0 0
as expected. As the y axis has
no slope, the analysis fails for projection on the y axis, but this transformation is indeed linear with matrix
00
as is easily verified directly.
0 1
Note that the formula for the matrix of Qm in Theorem 2.6.5 can be derived from the above formula
for the matrix of Pm. Using Figure 2.6.12, observe that Qm(x) = x+2[Pm (x) -x] so Qm(x) = 2Pm(x) — x.
Substituting the matrices for P(x) and 12 (x) gives the desired formula.
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