Science:Math Exam Resources/Courses/MATH152/April 2012/Question 06 (e)

MATH152 April 2012

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[hide] Question 06 (e)
This question requires little or no calculations. Is each of the following transformations linear? Answer Yes or No (no justification required). (1) The reflection at the line x=0 in the plane. (2) $T\left({\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}\right)={\begin{bmatrix}v_{1}+2v_{2}\\-3v_{1}\end{bmatrix}}$ . (3) $R\left({\begin{bmatrix}v_{1}\\v_{2}\end{bmatrix}}\right)={\begin{bmatrix}2v_{1}-5v_{2}\\\cos ^{2}(v_{1})\end{bmatrix}}$ . (4) $S\left({\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}\right)={\begin{bmatrix}-v_{1}\\v_{2}v_{3}\\v_{3}+v_{1}\end{bmatrix}}$ .

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[show] Hint
Remember: A transformation T is linear if the following hold: i) T(u + v) = T(u) + T(v) ii) T(cu) = cT(u) for any vectors u,v and any constant c.

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[show] Solution
(1) YES. (2) YES. (3) NO. (4) NO. Justification for each answer is given below. Remember: A transformation T is linear if the following hold: i) T(u + v) = T(u) + T(v) ii) T(cu) = cT(u) for any vectors u,v and any constant c. (1) YES. The reflection at the x = 0 plane results in the component of the vector parallel to x being mapped to its negative while all other components are unchanged. For a formal proof, let $\mathbf {u} ,\mathbf {v} \in \mathbb {R} ^{3}$ . If we let M be the transformation in question, then $M(\mathbf {u} )={\begin{bmatrix}-u_{1}\\u_{2}\\u_{3}\end{bmatrix}}.$ If we check that the above conditions i), ii) hold, we see that they are satisfied: ${\begin{aligned}{\mbox{i)}}\quad M(\mathbf {u} +\mathbf {v} )&={\begin{bmatrix}-(u_{1}+v_{1})\\u_{2}+v_{2}\\u_{3}+v_{3}\end{bmatrix}}\\&={\begin{bmatrix}-u_{1}\\u_{2}\\u_{3}\end{bmatrix}}+{\begin{bmatrix}-v_{1}\\v_{2}\\v_{3}\end{bmatrix}}\\&=M(\mathbf {u} )+M(\mathbf {v} )\end{aligned}}$ ${\begin{aligned}{\mbox{ii)}}\quad M(c\mathbf {u} )={\begin{bmatrix}-cu_{1}\\cu_{2}\\uc_{3}\end{bmatrix}}=c{\begin{bmatrix}-u_{1}\\u_{2}\\u_{3}\end{bmatrix}}=cM(\mathbf {u} )\end{aligned}}$ Therefore, the reflection in the x = 0 line/plane is a linear transformation. (2) YES. We need only check the conditions i), ii) above to confirm T is linear ${\begin{aligned}{\mbox{i)}}\quad T(\mathbf {u} +\mathbf {v} )&={\begin{bmatrix}(u_{1}+v_{1})+2(u_{2}+v_{2})\\-3(u_{1}+v_{1})\end{bmatrix}}\\&={\begin{bmatrix}u_{1}+2u_{1}\\-3u_{1}\end{bmatrix}}+{\begin{bmatrix}v_{1}+2v_{1}\\-3v_{1}\end{bmatrix}}\\&=T(\mathbf {u} )+T(\mathbf {v} )\end{aligned}}$ ${\begin{aligned}{\mbox{ii)}}\quad T(c\mathbf {u} )={\begin{bmatrix}cu_{1}+2cu_{1}\\-3cu_{1}\end{bmatrix}}=c{\begin{bmatrix}u_{1}+2u_{1}\\-3u_{1}\end{bmatrix}}=cT(\mathbf {u} )\end{aligned}}$ Therefore, T is linear. (3) NO. In this case, we can see right way that R fails condition i) ${\begin{aligned}{\mbox{i)}}\quad R(\mathbf {u} +\mathbf {v} )={\begin{bmatrix}2(u_{1}+v_{1})-5(u_{2}+v_{2})\\\cos ^{2}(u_{1}+v_{1})\end{bmatrix}}\end{aligned}}$ If we let $u_{1}=\pi ,\,v_{1}=-\pi$ , then we see ${\begin{aligned}\cos ^{2}(u_{1}+v_{1})&=\cos ^{2}(\pi -\pi )=(1)^{2}=1\\\cos ^{2}(u_{1})+\cos ^{2}(v_{1})&=\cos ^{2}(\pi )+\cos ^{2}(-\pi )=(-1)^{2}+(-1)^{2}=2\end{aligned}}$ Hence ${\begin{aligned}{\mbox{i)}}\quad R(\mathbf {u} +\mathbf {v} )&={\begin{bmatrix}2(u_{1}+v_{1})-5(u_{2}+v_{2})\\\cos ^{2}(u_{1}+v_{1})\end{bmatrix}}\\&\neq {\begin{bmatrix}2u_{1}-5u_{2}\\\cos ^{2}(u_{1})\end{bmatrix}}+{\begin{bmatrix}2v_{1}-5v_{2}\\\cos ^{2}(v_{1})\end{bmatrix}}\\&=R(\mathbf {u} )+R(\mathbf {v} )\end{aligned}}$ Therefore, R is NOT linear. (4) NO. In this case, we can see right way that S fails condition ii) ${\begin{aligned}{\mbox{ii)}}&\quad S(c\mathbf {u} )={\begin{bmatrix}-cu_{1}\\(cu_{2})(cu_{3})\\cu_{3}+cu_{1}\end{bmatrix}}={\begin{bmatrix}-cu_{1}\\c^{2}u_{2}u_{3}\\c(u_{3}+u_{1})\end{bmatrix}}\neq cS(\mathbf {u} )\end{aligned}}$ Therefore, S is NOT linear.