Question 4.3: Let T:M22→M22 be the linear operator defined by T(A)=A^T. Fi......
Let T:M_{22}\to M_{22} be the linear operator defined by T(A)=A^{T}. Find the matrix of the operator T with respect to the bases S_{1}=\left\{A_{1},A_{2},A_{3},A_{4}\right\} and S_{2}=\left\{B_{1},B_{2},B_{3},B_{4}\right\}, where A_{1}={\left[\begin{array}{c c}{1}&{0}\\ {0}&{0}\end{array}\right]},\quad A_{2}={\left[\begin{array}{c c}{0}&{1}\\ {0}&{0}\end{array}\right]},\quad A_{3}={\left[\begin{array}{c c}{0}&{0}\\ {1}&{0}\end{array}\right]},\quad A_{4}=\left[{\begin{array}{c c}{0}&{0}\\ {0}&{1}\end{array}}\right],\quad B_{1}=\left[{\begin{array}{c c}{1}&{1}\\ {0}&{0}\end{array}}\right],\quad B_{2}=\left[{\begin{array}{c c}{0}&{1}\\ {0}&{0}\end{array}}\right],\quad B_{3}=\left[{\begin{array}{c c}{0}&{0}\\ {1}&{1}\end{array}}\right],\quad B_{4}=\left[{\begin{array}{c c}{1}&{0}\\ {0}&{1}\end{array}}\right].
Here, T(A_{1})=T\left\lgroup \left[\begin{array}{c c}{{1}}&{{0}}\\ {{0}}&{{0}}\end{array}\right]\right\rgroup=\left[\begin{array}{c c}{{1}}&{{0}}\\ {{0}}&{{0}}\end{array}\right]^{T}=\left[\begin{array}{c c}{{1}}&{{0}}\\ {{0}}&{{0}}\end{array}\right]
T(A_{2})=T\left\lgroup\left[\begin{array}{c c}{{0}}&{{1}}\\ {{0}}&{{0}}\end{array}\right]\right\rgroup=\left[\begin{array}{c c}{{0}}&{{1}}\\ {{0}}&{{0}}\end{array}\right]^{T}=\left[\begin{array}{c c}{{0}}&{{0}}\\ {{1}}&{{0}}\end{array}\right]
T(A_{3})=T\left\lgroup\left[\begin{array}{c c}{{0}}&{{0}}\\ {{1}}&{{0}}\end{array}\right]\right\rgroup=\left[\begin{array}{c c}{{0}}&{{0}}\\ {{1}}&{{0}}\end{array}\right]^{T}=\left[\begin{array}{c c}{{0}}&{{1}}\\ {{0}}&{{0}}\end{array}\right]
T(A_{4})=T\left\lgroup\left[\begin{array}{c c}{{0}}&{{0}}\\ {{0}}&{{1}}\end{array}\right]\right\rgroup=\left[\begin{array}{c c}{{0}}&{{0}}\\ {{0}}&{{1}}\end{array}\right]^{T}=\left[\begin{array}{c c}{{0}}&{{0}}\\ {{0}}&{{1}}\end{array}\right].
Let us express A_{1} as a linear combination of B_{1},B_{2},B_{3},B_{4}
T\left(A_{1}\right)=k_{1}B_{1}+k_{2}B_{2}+k_{3}B_{3}+k_{4}B_{4}
\therefore \left[\begin{array}{c c}{{1}}&{{0}}\\ {{0}}&{{0}}\end{array}\right]=k_{1}\left[\begin{array}{c c}{{1}}&{{1}}\\ {{0}}&{{0}}\end{array}\right]+k_{2}{\left[\begin{array}{c c}{0}&{1}\\ {0}&{0}\end{array}\right]}+k_{3}{\left[\begin{array}{c c}{0}&{0}\\ {1}&{1}\end{array}\right]}+k_{4}{\left[\begin{array}{c c}{1}&{0}\\ {0}&{1}\end{array}\right]}.
\therefore \left[{\begin{array}{c c}{1}&{0}\\ {0}&{0}\end{array}}\right]=\left[~{\begin{array}{c c}{k_{1}+k_{4}}&{k_{1}+k_{2}}\\ {k_{3}}&{k_{3}+k_{4}}\end{array}}~\right].
Comparing both sides,
k_{1}+k_{4}=1,k_{1}+k_{2}=0,k_{3}=0,k_{3}+k_{4}=0.
Using Gauss elimination, we get
\begin{array}{c}k_{1}=1,k_{2}=-1,k_{3}=0,k_{4}=0\\[1 em] \therefore T(A_{1})=B_{1}-B_{2}\\[1 em] {\pmb{\big[}T(A_{1})\pmb{\big]}_{s_1}={\left[~\begin{array}{c}1\\{-1}\\ {0}\\ {0}\end{array}~\right].}}\end{array}
Let us express A_{2} as a linear combination of B_{1},B_{2},B_{3},B_{4}
T\left(A_{2}\right)=k_{1}B_{1}+k_{2}B_{2}+k_{3}B_{3}+k_{4}B_{4}.
\therefore \left[\begin{array}{c c}{{0}}&{{0}}\\ {{1}}&{{0}}\end{array}\right]=k_{1}\left[\begin{array}{c c}{{1}}&{{1}}\\ {{0}}&{{0}}\end{array}\right]+k_{2}{\left[\begin{array}{c c}{0}&{1}\\ {0}&{0}\end{array}\right]}+k_{3}{\left[\begin{array}{c c}{0}&{0}\\ {1}&{1}\end{array}\right]}+k_{4}{\left[\begin{array}{c c}{1}&{0}\\ {0}&{1}\end{array}\right]}
\therefore \left[{\begin{array}{c c}{0}&{0}\\ {1}&{0}\end{array}}\right]=\left[~{\begin{array}{c c}{k_{1}+k_{4}}&{k_{1}+k_{2}}\\ {k_{3}}&{k_{3}+k_{4}}\end{array}}~\right].
Comparing both sides,
k_{1}+k_{4}=1,\,k_{1}+k_{2}=0,\,k_{3}=0,\,k_{3}+k_{4}=0.
Using Gauss elimination, we get
\begin{array}{c}k_{1}=1,k_{2}=-1,k_{3}=1,k_{4}=-1 \\[1 em] \therefore T(A_{2})=B_{1}-B_{2} +B_{3}-B_{4} \\[1 em] {\pmb{\big[}T(A_{2})\pmb{\big]}_{s_1}={\left[~\begin{array}{c}1\\{-1}\\1\\{-1}\end{array}~\right].}}\end{array}
Let us express A_{3} as a linear combination of B_{1},B_{2},B_{3},B_{4}
T\left(A_{3}\right)=k_{1}B_{1}+k_{2}B_{2}+k_{3}B_{3}+k_{4}B_{4}.
\therefore \left[\begin{array}{c c}{{0}}&{{1}}\\ {{0}}&{{0}}\end{array}\right]=k_{1}\left[\begin{array}{c c}{{1}}&{{1}}\\ {{0}}&{{0}}\end{array}\right]+k_{2}{\left[\begin{array}{c c}{0}&{1}\\ {0}&{0}\end{array}\right]}+k_{3}{\left[\begin{array}{c c}{0}&{0}\\ {1}&{1}\end{array}\right]}+k_{4}{\left[\begin{array}{c c}{1}&{0}\\ {0}&{1}\end{array}\right]}
\therefore \left[{\begin{array}{c c}{0}&{1}\\ {0}&{0}\end{array}}\right]=\left[~{\begin{array}{c c}{k_{1}+k_{4}}&{k_{1}+k_{2}}\\ {k_{3}}&{k_{3}+k_{4}}\end{array}}~\right].
Comparing both sides,
k_{1}+k_{4}=0,\,k_{1}+k_{2}=1,\,k_{3}=0,\,k_{3}+k_{4}=0.
Using Gauss elimination, we get
\begin{array}{c}k_{1}=0,k_{2}=1,k_{3}=0,k_{4}=0 \\[1 em] \therefore T(A_{3})=B_{2} \\[1 em] {\pmb{\big[}T(A_{3})\pmb{\big]}_{s_1}={\left[~\begin{array}{c}{0}\\{1}\\0\\{0}\end{array}~\right].}}\end{array}
Let us express A_{4} as a linear combination of B_{1},B_{2},B_{3},B_{4}
T\left(A_{4}\right)=k_{1}B_{1}+k_{2}B_{2}+k_{3}B_{3}+k_{4}B_{4}.
\therefore \left[\begin{array}{c c}{{0}}&{{0}}\\ {{0}}&{{1}}\end{array}\right]=k_{1}\left[\begin{array}{c c}{{1}}&{{1}}\\ {{0}}&{{0}}\end{array}\right]+k_{2}{\left[\begin{array}{c c}{0}&{1}\\ {0}&{0}\end{array}\right]}+k_{3}{\left[\begin{array}{c c}{0}&{0}\\ {1}&{1}\end{array}\right]}+k_{4}{\left[\begin{array}{c c}{1}&{0}\\ {0}&{1}\end{array}\right]}
\therefore \left[{\begin{array}{c c}{{0}}&{{0}}\\ {{0}}&{{1}}\end{array}}\right]=\left[~{\begin{array}{c c}{k_{1}+k_{4}}&{k_{1}+k_{2}}\\ {k_{3}}&{k_{3}+k_{4}}\end{array}}~\right].
Comparing both sides,
k_{1}+k_{4}=0,\,k_{1}+k_{2}=0,\,k_{3}=0,\,k_{3}+k_{4}=1.
Using Gauss elimination, we get
\begin{array}{c}k_{1}=-1,k_{2}=1,k_{3}=0,k_{4}=1 \\[1 em] \therefore T(A_{4})=-B_{1}+B_{2}+B_{4} \\[1 em] {\pmb{\big[}T(A_{2})\pmb{\big]}_{s_1}={\left[~\begin{array}{c}{-1}\\{1}\\0\\{1}\end{array}~\right].}}\end{array}
The matrix of the transformation T with respect to the bases S_1 and S_2 is
\pmb{\big[}T\pmb{\big]}_{s_2,s_1} =\left[\pmb{\big[}T(A_{1})\pmb{\big]}_{s_2}|\pmb{\big[}T(A_{2})\pmb{\big]}_{s_2}|\pmb{\big[}T(A_{3})\pmb{\big]}_{s_2}|\pmb{\big[}T(A_{4})\pmb{\big]}_{s_2} \right]=\left[\ {\begin{array}{c c c c}{1}&{1}&0&-1\\ -1&-1&1&1\\ {0}&{1}&0&0\\ {0}&{-1}&0&1 \end{array}}\ \right].