Answer
\begin{array}{c}{T(\mathbf{x}+\mathbf{y})=T(\mathbf{x})+T(\mathbf{y})} \\ {T(c \mathbf{x})=c T(\mathbf{x})}\end{array}
Work Step by Step
Given $$T\left(x_{1}, x_{2}\right)=\left(x_{1}+2 x_{2}, 2 x_{1}-x_{2}\right)$$
\begin{array}{l}{\text { Let } \mathbf{y}=\left(y_{1}, y_{2}\right) \in \mathbb{R}^{2} \text { and } c \in \mathbb{R} \text { scalar. We have to check } 2 \text { properties of } \overline{\mathbf{a}}} \\ {\text { linear transformation. }}\end{array}
So, we have
\begin{aligned} T(\mathbf{x}+\mathbf{y}) &=T\left(\left(x_{1}, x_{2}\right)+\left(y_{1}, y_{2}\right)\right) \\ &=T\left(x_{1}+y_{1}, x_{2}+y_{2}\right) \\
&=\left(\left(x_{1}+y_{1}\right)+2\left(x_{2}+y_{2}\right), 2\left(x_{1}+y_{1}\right)-\left(x_{2}+y_{2}\right)\right) \\ &=\left(x_{1}+y_{1}+2 x_{2}+2 y_{2}, 2 x_{1}+2 y_{1}-x_{2}-y_{2}\right) \\ &=\left(\left(x_{1}+2 x_{2}\right)+\left(y_{1}+2 y_{2}\right),\left(2 x_{1}-x_{2}\right)+\left(2 y_{1}-y_{2}\right)\right) \\ &=\left(x_{1}+2 x_{2}, 2 x_{1}-x_{2}\right)+\left(y_{1}+2 y_{2}, 2 y_{1}-y_{2}\right) \\ &=T(\mathbf{x})+T(\mathbf{y}) \end{aligned}
\begin{aligned} T(c \mathbf{x}) &=T\left(c\left(x_{1}, x_{2}\right)\right) \\ &=t\left(c x_{1}, c x_{2}\right) \\ &=\left(c x_{1}+2 c x_{2}, 2 c x_{1}-c x_{2}\right) \\ &=c\left(x_{1}+2 x_{2}, 2 x_{1}-x_{2}\right) \\ &=c T(\mathbf{x}) \end{aligned}
