Question 3.9: Let T:R² → R² and S:R² → R² be the linear operators given by......

i. T and S are one-one if ker(T)={0} and ker(S)={0} .

\quadFor that, T(X)=0

\begin{array}{l}{{T(x,y)=\overline{{{0}}}}}\\ {{\therefore\ (x+y,x-y)=(0,0).}}\\ {{x+y=0,\;x-y=0}}\end{array}

Solving these equations,

x=0,\ \ y=0.

The solution is trivial; hence, ker(T)={0}.

Hence, T is one-one.

Similarly, S(X)=0

\begin{array}{l}{{S(x,y)=\overline{{{0}}}}}\\ {{\therefore\ (2x+y,x-2y)=(0,0).}}\\ {{2x+y=0,\;x-2y=0}}\end{array}

Solving these equations,

x=0,\ \ y=0.

The solution is trivial; hence, ker(S)={0}.

Hence, S is one-one.

ii. Standard matrix for T=\left[{\begin{array}{c c}{1}&{1}\\ {1}&{-1}\end{array}}\right]

\\[0.5 em]  {\because{T}^{-1}={\frac{\mathrm{adj}({T})}{|{{T}}|}}={\frac{1}{-2}}{\left[\begin{array}{c c}{-1}&{-1}\\ {-1}&{1}\end{array}\right]}={\left[\begin{array}{c c}{{\frac{1}{2}}}&{{\frac{1}{2}}}\\[0.5 em] {{\frac{1}{2}}}&{-{\frac{1}{2}}}\end{array}\right]}}

\\[0.5 em]  \because T^{-1}(x,y)=\!\! \left\lgroup {\frac{x+y}{2}},{\frac{x-y}{2}} \right\rgroup\!\!.

Standard matrix for S=\left[{\begin{array}{c c}{\,2\ }&{\,1}\\ {1}&{\,-2}\end{array}}\right]

\\[0.5 em]  \because S^{-1}={\frac{\mathrm{adj}(S)}{|S|}}={\frac{-1}{5}}{\left[\begin{array}{c c}{-2}&{-1}\\ {-1}&{2}\end{array}\right]}={\left[\begin{array}{c c}{{\frac{2}{5}}}&{{\frac{1}{5}}}\\[0.5 em] {{\frac{1}{5}}}&{{\frac{-2}{5}}}\end{array}\right]}

\\[0.5 em] \because T^{-1}(x,y)=\left\lgroup {\frac{2x+y}{5}},{\frac{x-2y}{5}}\right\rgroup

\\[0.5 em]  T\circ S=T\cdot S=\left[\begin{array}{c c}{{1}}&{{1}}\\ {{1}}&{{-1}} \end{array}\right]\cdot \left[\begin{array}{c c}{{2}}&{{1}}\\ {{1}}&{{-2}}\end{array}\right]

\\[0.5 em] =\left[{\begin{array}{c c}{3}&{-1}\\ {1}&{3}\end{array}}\right]

\\[0.5 em]  (T\cdot S)^{-1}={\frac{T\cdot S}{|T\cdot S|}}={\frac{1}{10}}{\left[\begin{array}{c c}{3}&{1}\\ {-1}&{3}\end{array}\right]}= \left[\begin{array}{c c}{{\frac{3}{10}}}&{{\frac{1}{10}}}\\[0.5 em] {{-\frac{1}{10}}}&{{\frac{3}{10}}}\end{array}\right]

\\[0.5 em] \left(T\circ S\right)^{-1}\left(x,y\right)=\left[\begin{array}{c c}{{\frac{3}{10}}}&{{\frac{1}{10}}}\\[0.5 em] {{-\frac{1}{10}}}&{{\frac{3}{10}}}\end{array}\right]\!\!\!\left[\ \begin{array}{c}{x}\\ {y}\end{array}\ \right]

\\[0.5 em] \left(T\circ S\right)^{-1}(x,y)=\left({\frac{3x+y}{10}},{\frac{-x+3y}{10}}\right).

iii. From (ii) ) LHS of (T\circ S)^{-1}=S^{-1}\circ T^{-1} is

(T\circ S)^{-1}=\left[\begin{array}{c c}{{\frac{3}{10}}}&{{\frac{1}{10}}}\\[0.5 em] {{-\frac{1}{10}}}&{{\frac{3}{10}}}\end{array}\right].

From (ii) T^{-1}=\left[\begin{array}{c c}{{{\frac{1}{2}}}}&{{{\frac{1}{2}}}}\\[0.5 em] {{{\frac{1}{2}}}}&{{-{\frac{1}{2}}}}\end{array}\right] and S^{-1}=\left[\begin{array}{c c}\frac{2}{5}&\frac{1}{5}\\[0.5 em] \frac{1}{5} &\frac{-2}{5}\end{array}\right]

\\[0.5 em] S^{-1}\circ T^{-1}=S^{-1}\cdot T^{-1}=\left[\begin{array}{c c}\frac{2}{5}&\frac{1}{5}\\[0.5 em] \frac{1}{5} &\frac{-2}{5}\end{array}\right]\!\!\!\left[\begin{array}{c c}{{{\frac{1}{2}}}}&{{{\frac{1}{2}}}}\\[0.5 em] {{{\frac{1}{2}}}}&{{-{\frac{1}{2}}}}\end{array}\right]

\\[0.5 em]={\left[\begin{array}{c c}{{\frac{3}{10}}}&{{\frac{1}{10}}}\\[0.5 em] {-{\frac{1}{10}}}&{{\frac{3}{10}}}\end{array}\right]}

\\[0.5 em]  \therefore\left(T\circ S\right)^{-1}=S^{-1}\circ T^{-1}.

Hence, the proof.