Question 4.4: Let T:R² → R² be the linear operator defined by T([x1 x2])=[......
Let T:R² → R² be the linear operator 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 of the operator T with respect to the basis S=\{u_{1},u_{2}\}, where u_{1}=\left[\ {\begin{array}{c}{1}\\ {1}\end{array}}\ \right] and u_{2}={\left[\ \begin{array}{c}{1}\\ {2}\end{array}\ \right]}. Also, verify [T]_{S}[X]_{S}=[T(X)]_{S}.
Here, T(u_{1})=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 em] T(u_{2})=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].
Let us express u_{1} as a linear combination of u_{1},u_{2}
\begin{array}{c} T(u_{1})=k_1u_{1}+k_2u_{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,
k_{1}+k_{2}=2,k_{1}+2k_{2}=2.
Using Gauss elimination method,
\begin{array}{c}{{k_{1}=2,k_{2}=0}}\\[1 em] {{T(u_{1})=2u_{1}}} \\[1 em] \pmb{\big[}T(u_{1})\pmb{\big]}_{s}=\left[\ \begin{array}{c}{{2}}\\ {{0}}\end{array}\ \right].\end{array}
Express u_{2} as a linear combination of u_{1},u_{2}
\begin{array}{c} T(u_{2})=k_1u_{1}+k_2u_{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]}} \\[1.5 em] k_{1}+k_{2}=3,k_{1}+2k_{2}=6. \end{array}
Using Gauss elimination method,
\ \begin{array}{c}{{k_{1}=0,k_{2}=3}}\\[1 em] {{T(u_{2})=3u_{2}}} \\[1 em] \pmb{\big[}T(u_{2})\pmb{\big]}_{s}=\left[\ \begin{array}{c}{{0}}\\ {{3}}\end{array}\ \right].\end{array}
The matrix of the transformation T with respect to the basis S is
\pmb{\big[}T\pmb{\big]}_S =\Big[\big[T(u_{1})\big]_{s}| \big[ T(u_{2})\big]\Big]_{s}=\left[\ {\begin{array}{c c}{2}&{0} \\ {0}&{3}\end{array}}\ \right].
If X\! =\!\left[\begin{array}{c}{x_{1}}\\ {x_{2}}\end{array}\right] is any vector in R², then the formula for T(X)=\left[\!\!\!\begin{array}{c c}{{\begin{array}{c}{{x_{1}+x_{2}}}\\ {{-2x_{1}+4x_{2}}}\end{array}}}\end{array}\!\!\!\right].
To find \pmb{\big[}T(X)\pmb{\big]}_{S} and [X]_{S}, we must express \pmb{\big[}T(X)\pmb{\big]}_{S} and [X]_{S} as a linear combination of u_1 and u_2.
Therefore, X =k_{1}u_{1}+k_{2}u_{2}
\\[1 em] \begin{array}{c}\therefore\left[\ {\begin{array}{c}{x_{1}}\\ {x_{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 k_{1}+k_{2}=x_{1},\ k_{1}+2k_{2}=x_{2}\\[1.5 em] \therefore k_{1}=2x_{1}-x_{2},\ k_{2}=-x_{1}+x_{2} \\[1.5 em] \therefore \pmb{[} X \pmb]_{S}={\left[\begin{array}{c}{\;\;2x_{1}-x_{2}}\\ {-x_{1}+x_{2}}\end{array}\right]} \\[1.5 em] \therefore \pmb{[} T \pmb]_{S}\pmb{[} X \pmb]_{S}={\left[\begin{array}{l l}{2}&{0}\\ {0}&{3}\end{array}\right]}\!\!{\left[\begin{array}{c}{\,\,2x_{1}-x_{2}}\\ {\,\,-x_{1}+x_{2}}\end{array}\right]}={\left[\begin{array}{c}{\,\,4x_{1}-2x_{2}}\\ {-3x_{1}+3x_{2}}\end{array}\right]} \qquad\qquad (4.2) \end{array}\\[1.5 em]
For \pmb{[}T( X) \pmb]_{S^{\prime}}
\begin{array}{c}T(X)=c_{1}u_{1}+c_{2}u_{2} \\[1.5 em] \therefore \!\left[{\begin{array}{c}{x_{1}+x_{2}}\\ {-2x_{1}+4x_{2}}\end{array}}\right]=c_1\biggl[\ \begin{array}{c}{{1}}\\ {{1}}\end{array}\ \biggr]+c_{2}\biggl[\ \begin{array}{c}{{1}}\\ {{2}}\end{array}\ \biggr]\\[1.5 em] \therefore \!\left[{\begin{array}{c}{x_{1}+x_{2}}\\ {-2x_{1}+4x_{2}}\end{array}}\right]=\left[\begin{array}{c}{c_{1}+c_{2}}\\ {c_{1}+2c_{2}}\end{array}\right] \\[1.5 em] c_{1}+c_{2}=x_{1}+x_{2},\quad c_{1}+2c_{2}=-2x_{1}+4x_{2}.\end{array}\\[1.5 em]
To find c_1,c_2, use Gauss elimination.
So, c_{1}=4x_{1}-2x_{2} and c_{2}=-3x_{1}+3x_{2}.
Thus,\qquad\qquad\qquad\qquad\qquad\pmb{\big[}T(X)\pmb{\big]} _{S} = \left[{\begin{array}{r}{4x_{1}-2x_{2}}\\ {-3x_{1}+3x_{2}}\end{array}}\right].\qquad\qquad\qquad\qquad\qquad\qquad(4.3)
From equations (4.2) and (4.3),
\therefore \pmb{\big[}T\pmb{\big]} _{S}\pmb{\big[}X\pmb{\big]} _{S} = \pmb{\big[}T(X)\pmb{\big]} _{S}\ .
Hence, it is verified.