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

MATH152 April 2012

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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) $S\left({\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}\right)={\begin{bmatrix}-v_{1}\\v_{2}v_{3}\\v_{3}+v_{1}\end{bmatrix}}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

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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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (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.

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