Science:Math Exam Resources/Courses/MATH221/December 2008/Question 08 (b)

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MATH221 December 2008

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Question 08 (b)
True or false (explain your answer): Let $T$ be the transformation of $\mathbb {R} ^{2}\rightarrow \mathbb {R} ^{2}$ given by $T{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}x+y\\x-y\end{pmatrix}}$ . Then $T$ is a linear transformation.

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Hint
Apply the definition of linearity to check if the given function is indeed linear.

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Solution
The statement is true. In order to prove T is a linear transformation, we need to show that for any $c\in \mathbb {R}$ and vectors $v_{1}={\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}},v_{2}={\begin{pmatrix}x_{2}\\y_{2}\end{pmatrix}}\in \mathbb {R} ^{2}$ , the properties $T(cv_{1})=cT(v_{1})$ , and $T(v_{1}+v_{2})=T(v_{1})+T(v_{2})$ are satisfied . Let us check these properties: $T(c{\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}})=T{\begin{pmatrix}cx_{1}\\cy_{1}\end{pmatrix}}={\begin{pmatrix}cx_{1}+cy_{1}\\cx_{1}-cy_{1}\end{pmatrix}}={\begin{pmatrix}c(x_{1}+y_{1})\\c(x_{1}-y_{1})\end{pmatrix}}=c{\begin{pmatrix}x_{1}+y_{1}\\x_{1}-y_{1}\end{pmatrix}}=cT{\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}$ , ${\begin{aligned}T({\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\end{pmatrix}})&=T{\begin{pmatrix}x_{1}+x_{2}\\y_{1}+y_{2}\end{pmatrix}}\\&={\begin{pmatrix}x_{1}+x_{2}+y_{1}+y_{2}\\x_{1}+x_{2}-y_{1}-y_{2}\end{pmatrix}}\\&={\begin{pmatrix}x_{1}+y_{1}\\x_{1}-y_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}+y_{2}\\x_{2}-y_{2}\end{pmatrix}}\\&=T{\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}+T{\begin{pmatrix}x_{2}\\y_{2}\end{pmatrix}}.\end{aligned}}$ Therefore, $T$ is a linear transformation.