b) For any slope m, show that PmQm = Pm, where Pm and Qm are transformations defined in Theorem 2.6.5and 2.6.6 (textbook, pages 112 and 113). Reflectionsy00yFigure 2.6.12Let denote the angle between the positive x axis and the line y = mx.The key observation is that the transformation Qm can be accomplished inthree steps: First rotate through -0 (so our line coincides with the x axis), then reflect in the x axis, andfinally rotate back through 0. In other words:QmRe Qo°R_eSince R-0, Qo, and Re are all linear, this (with Theorem 2.6.3) shows that Qm is linear and that its matrixis the product of the matrices of Re, Qo, and R_e. If we write c = cos and s = sin for simplicity, thenthe matrices of Re, R-e, and Qo are√1+m²40Qm (x)ε] [$]·Hence, by Theorem 2.6.3, the matrix of Qm = Reo QoR_e isC636*][19][1ymy = mxFigure 2.6.13mProjectionsPm (x)0Theorem 2.6.5).Solution. The matrix of RXreflection in the y axis is01+m²y = mxSXy = mxwhere c = cos 0 =This gives:XFigure 2.6.14The line through the origin with slope m has equation y = mx, and we letQm: R² → R² denote reflection in the line y = mx.This transformation is described geometrically in Figure 2.6.12. Inwords, Qm(x) is the "mirror image" of x in the line y=mx. If m=0 thenQo is reflection in the x axis, so we already know Qo is linear. While wecould 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 ofQm directly without using Theorem 2.6.2.-1Note that if m= 0, the matrix in Theorem 2.6.5 becomesanalysis 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 matrixX√1+m²so the matrix13The matrix of Recomes from the matrix of Re using the fact that, for all angles 0, cos(-0) = cos 0 andsin(-0) sin(0).14Note that-1 0= lim 10 1m-001+m²S• ] = [ ²2 ₁ ²³22scWe can obtain this matrix in terms of m alone. Figure 2.6.13 showsthat√₁+m² and sin 0.of Qm becomes-플Example 2.6.8Let T: R² → R² be rotation through - followed by reflection in the y axis. Show that I' is areflection in a line through the origin and find the line.is1-m²2m-1 001-1and0 -10-Sc²-5² 2sc2sc s²_c²cos 0 =Theorem 2.6.5Let Qm denote reflection in the line y = mx. Then Qm is a linear1-m²2mtransformation with matrix²2 2mm² - 12mm² -1].and s= sin =matrixFirst observe thatcos(-2)sin(-)Hence the matrix of T is-1- sin(-4)cos(-√1+m²*respectively. 13Se2scs²_c²• [3] = [6] ·-Sm√1+m²01-1 0as before. So, Pm is linear with matrix=[¹ 11 m² 2m2m m² 11+m²as expected. Of course thisHowever it is an easy-114and this is reflection in the line y = -x (take m= −1 inThe method in the proof of Theorem 2.6.5 works more generally. LetPm : R² → R² denote projection on the line y = mx. This transformation isdescribed geometrically in Figure 2.6.14.If m=0, then Po02.6. Linear Transformations 113for all0[B]Hence the argument above for Qm goes through for Pm.Pm RooPooR_0сand the matrix ofC S[3][88][+]=R*]in R², so Po is linear withSCsc s²Theorem 2.6.6Let Pm : R² R² be projection on the line y = mx. Then Pm is a linear transformation with matrix1 mmm²Again, if m= 0, then the matrix in Theorem 2.6.6 reduces to100 0as expected. As the y axis hasno slope, the analysis fails for projection on the y axis, but this transformation is indeed linear with matrix00as is easily verified directly.0 1Note that the formula for the matrix of Qm in Theorem 2.6.5 can be derived from the above formulafor 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.

Question

b) For any slope m, show that PmQm = Pm, where Pm and Qm are transformations defined in Theorem 2.6.5and 2.6.6 (textbook, pages 112 and 113). Reflectionsy00yFigure 2.6.12Let denote the angle between the positive x axis and the line y = mx.The key observation is that the transformation Qm can be accomplished inthree steps: First rotate through -0 (so our line coincides with the x axis), then reflect in the x axis, andfinally rotate back through 0. In other words:QmRe Qo°R_eSince R-0, Qo, and Re are all linear, this (with Theorem 2.6.3) shows that Qm is linear and that its matrixis the product of the matrices of Re, Qo, and R_e. If we write c = cos and s = sin for simplicity, thenthe matrices of Re, R-e, and Qo are√1+m²40Qm (x)ε] [$]·Hence, by Theorem 2.6.3, the matrix of Qm = Reo QoR_e isC636*][19][1ymy = mxFigure 2.6.13mProjectionsPm (x)0Theorem 2.6.5).Solution. The matrix of RXreflection in the y axis is01+m²y = mxSXy = mxwhere c = cos 0 =This gives:XFigure 2.6.14The line through the origin with slope m has equation y = mx, and we letQm: R² → R² denote reflection in the line y = mx.This transformation is described geometrically in Figure 2.6.12. Inwords, Qm(x) is the &#34;mirror image&#34; of x in the line y=mx. If m=0 thenQo is reflection in the x axis, so we already know Qo is linear. While wecould 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 ofQm directly without using Theorem 2.6.2.-1Note that if m= 0, the matrix in Theorem 2.6.5 becomesanalysis 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 matrixX√1+m²so the matrix13The matrix of Recomes from the matrix of Re using the fact that, for all angles 0, cos(-0) = cos 0 andsin(-0) sin(0).14Note that-1 0= lim 10 1m-001+m²S• ] = [ ²2 ₁ ²³22scWe can obtain this matrix in terms of m alone. Figure 2.6.13 showsthat√₁+m² and sin 0.of Qm becomes-플Example 2.6.8Let T: R² → R² be rotation through - followed by reflection in the y axis. Show that I' is areflection in a line through the origin and find the line.is1-m²2m-1 001-1and0 -10-Sc²-5² 2sc2sc s²_c²cos 0 =Theorem 2.6.5Let Qm denote reflection in the line y = mx. Then Qm is a linear1-m²2mtransformation with matrix²2 2mm² - 12mm² -1].and s= sin =matrixFirst observe thatcos(-2)sin(-)Hence the matrix of T is-1- sin(-4)cos(-√1+m²*respectively. 13Se2scs²_c²• [3] = [6] ·-Sm√1+m²01-1 0as before. So, Pm is linear with matrix=[¹ 11 m² 2m2m m² 11+m²as expected. Of course thisHowever it is an easy-114and this is reflection in the line y = -x (take m= −1 inThe method in the proof of Theorem 2.6.5 works more generally. LetPm : R² → R² denote projection on the line y = mx. This transformation isdescribed geometrically in Figure 2.6.14.If m=0, then Po02.6. Linear Transformations 113for all0[B]Hence the argument above for Qm goes through for Pm.Pm RooPooR_0сand the matrix ofC S[3][88][+]=R*]in R², so Po is linear withSCsc s²Theorem 2.6.6Let Pm : R² R² be projection on the line y = mx. Then Pm is a linear transformation with matrix1 mmm²Again, if m= 0, then the matrix in Theorem 2.6.6 reduces to100 0as expected. As the y axis hasno slope, the analysis fails for projection on the y axis, but this transformation is indeed linear with matrix00as is easily verified directly.0 1Note that the formula for the matrix of Qm in Theorem 2.6.5 can be derived from the above formulafor 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.

Accepted Answer

Introduction:
For every linear transformation, there is a related matrix representation. Suppose,…

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).