Let T\colon P_{2}\to P_{2} be the linear operator defined by T(p(x))=p(2x+1). Find the matrix of the operator T with respect to the bases S=\left\{p_{1},p_{2},p_{3}\right\}, where p_{1}=1,p_{2}=x,p_{3}=x^{2}. Also, compute T(2-3x+4x^{2}).
Here, T(p_{1})=T(1)=1
Let us express p_{1} as a linear combination of p_{1},p_{2},p_{3}
\begin{array}{c}{{T(p_{1})=k_{1}p_{1}+k_{2}p_{2}+k_{3}p_{3}}}\\[0.5 em] \therefore 1=k_{1}1+k_{2}x+k_{3}x^{2}. \end{array}Comparing both sides, coefficient of 1,x,x^{2}
\left.\begin{array}{c}{{k_{1}=1,k_{2}=0,k_{3}=0}}\\[0.5 em] {{T(p_{1})=1}}\\[0.5 em] {\pmb{\big[}T(p_{1})\pmb{\big]} _{s}}= {{\left[\ \begin{array}{c}{{1}}\\ {{0}}\\ {{0}}\end{array}\ \right].}}\end{array}\right.Express p_{2} as a linear combination of p_{1},p_{2},p_{3}
\begin{array}{c}{{T(p_{2})=k_{1}p_{1}+k_{2}p_{2}+k_{3}p_{3}}}\\[0.5 em] \therefore 2x+1=k_{1}1+k_{2}x+k_{3}x^{2}. \end{array}Comparing both sides, coefficient of 1,x,x^{2}
\left.\begin{array}{c}{{k_{1}=1,k_{2}=2,k_{3}=0}}\\[0.5 em] T(p_{2})=p_1+2p_2 \\[0.5 em] {\pmb{\big[}T(p_{2})\pmb{\big]} _{s}}= {{\left[\ \begin{array}{c}{{1}}\\ {{2}}\\ {{0}}\end{array}\ \right].}}\end{array}\right.Express p_{3} as a linear combination of p_{1},p_{2},p_{3}
\begin{array}{c}{{T(p_{3})=k_{1}p_{1}+k_{2}p_{2}+k_{3}p_{3}}}\\[0.5 em] \therefore 4x^2+4x+1=k_{1}1+k_{2}x+k_{3}x^{2}. \end{array}Comparing both sides, coefficient of 1,x,x^{2}
\left.\begin{array}{c}{{k_{1}=1,k_{2}=4,k_{3}=4}}\\[0.5 em] T(p_{2})=p_1+4p_2+4p_3 \\[0.5 em] {\pmb{\big[}T(p_{3})\pmb{\big]} _{s}}= {{\left[\ \begin{array}{c}{{1}}\\ {{4}}\\ {{4}}\end{array}\ \right].}}\end{array}\right.The matrix of the operator T with respect to the basis S is
\pmb{\big[}T\pmb{\big]}_s =\left[\pmb{\big[}T(p_{1})\pmb{\big]}_{s}|\pmb{\big[}T(p_{2})\pmb{\big]}_{s}|\pmb{\big[}T(p_{3})\pmb{\big]}_{s}\right]=\left[\ {\begin{array}{c c c}{1}&{1}&1\\ {0}&{2}&4\\ {0}&{0}&4\end{array}}\ \right].