Question 4.7: Let T:R² → R² be defined by T([x1 x2])=[x1+x2 -2x1+4x2]. Fin......
Let T:R² → R² be defined by T\left\lgroup{\left[\begin{array}{c}{x_{1}}\\ {x_{2}}\end{array}\right]}\right\rgroup = {\left[\begin{array}{c}{x_{1}+x_{2}}\\ {-2x_{1}+4x_{2}}\end{array}\right]}. Find the matrix for T with respect to the basis S_{1}=\{e_{1},e_{2}\} for R², and also find a matrix for T with respect to the basis S_{2}=\left\{u_{1}^{\prime},u_{2}^{\prime}\right\} for R², where u_{1}^{\prime}=\left[{\begin{array}{c}{1}\\ {1}\end{array}}\right]{\mathrm{ard}}\,u_{2}^{\prime}=\left[{\begin{array}{c}{1}\\ {2}\end{array}}\right].
T(e_{1})=T\left\lgroup{\left[\begin{array}{c}{1}\\ {0}\end{array}\right]}\right\rgroup =\left[\begin{array}{c}{{1+0}}\\ {{-2+0}}\end{array}\right]=\left[\begin{array}{c}{{1}}\\ {{-2}}\end{array}\right]
\\[1.5 em] T(e_{2})=T\left\lgroup{\left[\begin{array}{c}{0}\\ {1}\end{array}\right]}\right\rgroup =\left[\begin{array}{c}{{0+1}}\\ {{0+4}}\end{array}\right]=\left[\begin{array}{c}{{1}}\\ {{4}}\end{array}\right].
Let us express e_{1} as a linear combination of e_{1},e_{2}
\begin{array}{c}{T(e_{1})=k_{1}e_{1}+k_{2}e_{2}}\\[1.5 em] \therefore\left[{\begin{array}{c}{1}\\ {-2}\end{array}}\right]=k_{1}\left[{\begin{array}{c}{1}\\ {0}\end{array}}\right]+k_{2}\left[{\begin{array}{c}{0}\\ {1}\end{array}}\right]\\[1.5 em] \therefore\left[{\begin{array}{c}{1}\\ {-2}\end{array}}\right]=\left[{\begin{array}{c}{k_{1}}\\ {k_{2}}\end{array}}\right]. \end{array}
Comparing both sides,
\begin{array}{c}{k_{1}=1,\ k_{2}=-2}\\[1.5 em] {T(e_{1})=u_{1}-2u_{2}}\\[1.5 em] {\pmb{\big[}T(e_{1})\pmb{\big]}_{s_1}={\left[\begin{array}{c}{1}\\ {-2}\end{array}\right].}}\end{array}
Express e_{2} as a linear combination of e_{1},e_{2}
\begin{array}{c}{T(e_{2})=k_{1}e_{1}+k_{2}e_{2}}\\[1.5 em] \therefore\left[{\begin{array}{c}{1}\\ {4}\end{array}}\right]=k_{1}\left[{\begin{array}{c}{1}\\ {0}\end{array}}\right]+k_{2}\left[{\begin{array}{c}{0}\\ {1}\end{array}}\right]\\[1.5 em] \therefore\left[{\begin{array}{c}{1}\\ {4}\end{array}}\right]=\left[{\begin{array}{r}{k_{1}}\\ {k_{2}}\end{array}}\right] \end{array}
\\[1.5 em] \begin{array}{c}{k_{1}=1,\ k_{2}=4}\\[1.5 em] {T(e_{2})=u_{1}+4u_{2}}\\[1.5 em] {\pmb{\big[}T(e_{2})\pmb{\big]}_{s_1}={\left[\begin{array}{c}{1}\\ {4}\end{array}\right].}}\end{array}
The matrix of the transformation T with respect to the basis S_1 is
\pmb{\big[}T\pmb{\big]}_{S_1} =\left[\pmb{\big[}T(e_1)\pmb{\big]}_{s_1}|\pmb{\big[}T(e_{2})\pmb{\big]}_{s_1} \right]=\left[\ {\begin{array}{c c}{1}&{1}\\ {-2}&{4} \end{array}}\ \right].
To find \pmb{\big[}T\pmb{\big]}_{S_2}
\begin{array}{c} u_{1}^{\prime}={\left[\begin{array}{c}{1}\\ {1}\end{array}\right]},\ u_{2}^{\prime}={\left[\begin{array}{c}{1}\\ {2}\end{array}\right]} \\[1.5 em] T\Big(u_{1}^{\prime}\Big)=T\left\lgroup{\left[\begin{array}{c}{1}\\ {1}\end{array}\right]}\right\rgroup=\left[\begin{array}{c}{{1+1}}\\ {{-2+4}}\end{array}\right]=\left[\begin{array}{c}{{2}}\\ {{2}}\end{array}\right] \\[1.5 em] T\Big(u_{2}^{\prime}\Big)=T\left\lgroup{\left[\begin{array}{c}{1}\\ {2}\end{array}\right]}\right\rgroup =\left[\begin{array}{c}{{1+2}}\\ {{-2+8}}\end{array}\right]=\left[\begin{array}{c}{{3}}\\ {{6}}\end{array}\right].\end{array}
Let us express u^{\prime}_{1} as a linear combination of u^{\prime}_{1},u^{\prime}_{2}
\begin{array}{c}{T(u^{\prime}_{1})=k_{1}u^{\prime}_{1}+k_{2}u^{\prime}_{2}}\\[1.5 em] \therefore\left[{\begin{array}{c}{2}\\ {2}\end{array}}\right]=k_{1}\left[{\begin{array}{c}{1}\\ {1}\end{array}}\right]+k_{2}\left[{\begin{array}{c}{1}\\ {2}\end{array}}\right]\\[1.5 em] \therefore\left[{\begin{array}{c}{2}\\ {2}\end{array}}\right]=\left[{\begin{array}{c}{k_{1}+k_{2}}\\ {k_{1}+2k_{2}}\end{array}}\right]. \end{array}
Comparing both sides,
\begin{array}{c}{k_{1}+k_{2}=2,\ k_{1}+2k_{2}=2}\\[1.5 em] {k_{1}=2,\ \ k_{2}=0}\\[1.5 em] T\Big(u_{2}^{\prime}\Big)=2u_{1} \\[1.5 em] \Big[T\Big(u_{2}^{\prime}\Big)\Big]_{s_{2}}=\left[\begin{array}{c}{{2}}\\ {{0}}\end{array}\right]. \end{array}
Let us express u^{\prime}_{2} as a linear combination of u^{\prime}_{1},u^{\prime}_{2}
\begin{array}{c}{T(u^{\prime}_{2})=k_{1}u^{\prime}_{1}+k_{2}u^{\prime}_{2}}\\[1.5 em] \therefore\left[{\begin{array}{c}{3}\\ {6}\end{array}}\right]=k_{1}\left[{\begin{array}{c}{1}\\ {1}\end{array}}\right]+k_{2}\left[{\begin{array}{c}{1}\\ {2}\end{array}}\right]\\[1.5 em] \therefore\left[{\begin{array}{c}{3}\\ {6}\end{array}}\right]=\left[{\begin{array}{c}{k_{1}+k_{2}}\\ {k_{1}+2k_{2}}\end{array}}\right]. \end{array}
Comparing both sides,
\begin{array}{c}{k_{1}+k_{2}=3,k_{1}+2k_{2}=6}\\[1.5 em] {k_{1}=0,\ \ k_{2}=3}\\[1.5 em] T\Big(u_{2}^{\prime}\Big)=3u_{2} \\[1.5 em] \Big[T\Big(u_{2}^{\prime}\Big)\Big]_{s_{2}}=\left[\begin{array}{c}{{0}}\\ {{3}}\end{array}\right]. \end{array}
The matrix of the transformation T with respect to the basis S_{2} is
\pmb{\big[}T\pmb{\big]}_{S_2} =\left[\Big[T\Big(u_{1}^{\prime}\Big)\Big]_{s_{2}}|\Big[T\Big(u_{2}^{\prime}\Big)\Big]_{s_{2}} \right]=\left[\ {\begin{array}{c c}{2}&{0}\\ {0}&{3} \end{array}}\ \right].