Question 4.5: Let T:P1 → P1 be defined by T(a0+a1x)=a0+a1(x+1), S1={p1,p2}......
Let T\colon P_{1}\rightarrow P_{1} be defined by T(a_{0}+a_{1}x)=a_{0}+a_{1}(x+1),\ \ S_{1}=\{p_{1},p_{2}\} , and S_{2}=\left\{q_{1},q_{2}\right\} where p_{1}=6+3x,\ \ p_{2}=10+2x,\ q_{1}=10+2x\mathrm{,}\ \mathrm{and}\ q_{2}=3{+}2x. Find the matrix of T with respect to the basis S_{1} and matrix of T with respect to the basis S_{2}.
Expressing T(p_{1}) as linear combinations of p_{1} and p_{2}
\begin{array}{c}{{T(p_{1})=k_{1}p_{1}+k_{2}p_{2}}}\\[1.5 em] {{9+3x=k_{1}(6+3x)+k_{2}(10+2x)}}\\[1.5 em] {{\qquad\!\quad\qquad=(6k_{1}+10k_{2})+(3k_{1}+2k_{2})x.}}\end{array}
Equating corresponding coefficients,
6k_{1}+10k_{2}=9,\quad3k_{1}+2k_{2}=3.
Solving these equations,
\begin{array}{c}{{k_{1}={\cfrac{2}{3}},k_{2}={\cfrac{1}{2}}}}\\[1.5 em] {{\therefore{T}{\left(p_{1}\right)}={\cfrac{2}{3}}\,p_{1}+{\cfrac{1}{2}}\,p_{2}}}\\[1.5 em] \therefore\pmb{\big[}T(p_{1})\pmb{\big]}_{S_1}=\left[\begin{array}{c}{{{\cfrac{2}{3}}}}\\[0.5 em] {{\cfrac{1}{2}}}\end{array}\right]. \end{array}
Similarly, T(p_{2})=T(10+2x)=10+2(x+1)=12+2x.
Expressing T(p_{2}) as linear combinations of p_{1} and p_{2}
\begin{array}{c}{{T\bigl(p_{2}\bigr)=k_{1}p_{1}+k_{2}p_{2}}}\\[1.5 em] {{12+2x=k_{1}\bigl(6+3x\bigr)+k_{2}\bigl(10+2x\bigr)}}\\[1.5 em] \qquad\qquad\ \quad{{=\bigl(6k_{1}+10k_{2}\bigr)+\bigl(3k_{1}+2k_{2}\bigr)x.}}\end{array}
Equating corresponding coefficients,
6k_{1}+10k_{2}=12,\quad3k_{1}+2k_{2}=2.
Solving these equations,
\begin{array}{c}{{k_{1}=-{\cfrac{2}{9}},\ k_{2}={\cfrac{4}{3}}}}\\[1.5 em] {{\therefore\,T\left(p_{2}\right)=-{\cfrac{2}{9}}\,p_{1}+{\cfrac{4}{3}}\,p_{2}}} \\[1.5 em] \therefore\pmb{\big[}T(p_{2})\pmb{\big]}_{S_1}=\;\left[\begin{array}{c}{{-\cfrac{2}{9}}}\\[0.5 em] {{\cfrac{4}{3}}}\end{array}\right]\\[1.5 em] \therefore \pmb{\big[}T\pmb{\big]}_{S_1} =\left[\pmb{\big[}T(p_{1})\pmb{\big]}_{S_1}|\pmb{\big[}T(p_{2})\pmb{\big]}_{S_1}\right]\\[1.5 em] \begin{array}{r l}{={\left[\begin{array}{r r}{{\cfrac{2}{3}}}&{-{\cfrac{2}{9}}}\\[0.5 em] {\cfrac{1}{2}}&{{\cfrac{4}{3}}}\end{array}\right]}.}\end{array} \end{array}
Expressing q_{1} as linear combinations of p_{1} and p_{2}
\begin{array}{c}{{q_{1}=k_{1}p_{1}+k_{2}p_{2}}}\\[1.5 em] {{2=k_{1}\left(6+3x\right)+k_{2}\left(10+2x\right)}}\\[1.5 em] \qquad\ {{=\left(6k_{1}+10k_{2}\right)+\left(3k_{1}+2k_{2}\right)x.}}\end{array}
Equating corresponding coefficients,
6k_{1}+10k_{2}=2,\quad3k_{1}+2k_{2}=0.
Solving these equations,
\begin{array}{c}{{k_{1}=-{\cfrac{2}{9}},\ k_{2}={\cfrac{1}{3}}}}\\[1.5 em] \therefore{{ q_{1}=-{\cfrac{2}{9}}\ p_{1}+{\cfrac{1}{3}}\ p_{2}}}\\[1.5 em] \therefore\pmb{\big[}q_{1}\pmb{\big]}_{S_1}=\left[\begin{array}{c}{{{-\cfrac{2}{9}}}}\\[0.5 em] {{\cfrac{1}{3}}}\end{array}\right]. \end{array}
Expressing q_{2} as linear combinations of p_{1} and p_{2}
\begin{array}{c}{{q_{2}=k_{1}p_{1}+k_{2}p_{2}}}\\[1.5 em] {{3+2x=k_{1}(6+3x)+k_{2}(10+2x)}}\\[1.5 em] \qquad\qquad\quad {{=(6k_{1}+10k_{2})+(3k_{1}+2k_{2})x.}}\end{array}
Equating corresponding coefficients,
6k_{1}+10k_{2}=3,\quad3k_{1}+2k_{2}=2.
Solving these equations,
\begin{array}{l}{{k_{1}={\cfrac{7}{9}},\,k_{2}=-{\cfrac{1}{6}}}}\\[1.5 em] {{\therefore{q_{2}}={\cfrac{7}{9}}\ p_{1}-{\cfrac{1}{6}}\ p_{2}}}\\[1.5 em] \therefore\pmb{\big[}q_{2}\pmb{\big]}_{S_1}=\left[\begin{array}{c}{{{\cfrac{7}{9}}}}\\[0.5 em] {{\! -\cfrac{1}{6}}}\end{array}\right]. \end{array}
Hence, the transition matrix from S_{2} to S_{1} is
\begin{array}{c c}P= \Big[\big[q_{1}\big]_{s_1}|\big[ q_{2}\big]_{s_1}\Big] \\[1.5 em ] P=\left[\ {\begin{array}{c c}{-{\cfrac{2}{9}}}& \ \ \ \cfrac{7}{9} \\[0.6 em] {\quad\cfrac{1}{3}}&{-{\cfrac{1}{6}}}\end{array}}\ \right] . \end{array}\\[1.5 em] Thus, P^{-1}=\left[\begin{array}{c c}{{\cfrac{3}{4}}}&{{\cfrac{7}{2}}}\\[0.6 em] {{\cfrac{3}{2}}}&{{1}}\end{array}\right].
\left[T\right]_{S_{2}}=P^{-1}\left[T\right]_{S_{1}}P= \left[\begin{array}{c c}{{\cfrac{3}{4}}}&{{\cfrac{7}{2}}}\\[0.6 em] {{\cfrac{3}{2}}}&{{1}}\end{array}\right]\!\! \left[\begin{array}{c c}{{\cfrac{2}{3}}}&{-{\cfrac{2}{9}}}\\[0.6 em] {\cfrac{1}{2}}&{\ \ \ {\cfrac{4}{3}}}\end{array}\right]\!\! \left[\begin{array}{r r}{{-\cfrac{2}{9}}}&{{\cfrac{7}{9}}}\\[0.6 em] {{\cfrac{1}{3}}}&{{-\cfrac{1}{6}}}\end{array}\right]\\[1.5 em] \qquad\qquad\qquad\qquad =\left[{\begin{array}{r r}{1}&{1}\\ {0}&{1}\end{array}}\right]In the next section, we will discuss the similarity of the linear matrix transformation and its examples.