pg 538 - R.C20: Compute the matrix representation of T relative to the bases B and C.
\(T:P3\) –> \(C^3\)
\(T(a+bx+cx^2+dx)=\left[\begin{array}{rrr}2a-3b+4c-2d \\a+b-c+d \\3a+0b+2c-3d-d\end{array}\right]\)
\(B=[1,x,x^2,x^3]\), C=[\({\left[\begin{array}{rrr}1 \\0 \\0\end{array}\right]}\), \({\left[\begin{array}{rrr}1 \\1 \\0\end{array}\right]}\), \({\left[\begin{array}{rrr}1 \\1 \\1\end{array}\right]}\)]
We are going to take the \(\rho_C\) for each T() value and find a linear combination of the vectors.
For \(T(1)\): \(\rho_C({\left[\begin{array}{rrr}2 \\1 \\3\end{array}\right]})\)= \(\rho_C(1{\left[\begin{array}{rrr}1 \\0 \\0\end{array}\right]}\) \({-2\left[\begin{array}{rrr}1 \\1 \\0\end{array}\right]}\)+ \({3\left[\begin{array}{rrr}1 \\1 \\1\end{array}\right]}\))= \({\left[\begin{array}{rrr}1 \\-2 \\3\end{array}\right]}\)
For \(T(x)\): \(\rho_C({\left[\begin{array}{rrr}-3 \\1 \\0\end{array}\right]})\)= \(\rho_C(-4{\left[\begin{array}{rrr}1 \\0 \\0\end{array}\right]}\)+ \({1\left[\begin{array}{rrr}1 \\1 \\0\end{array}\right]}\)+ \({0\left[\begin{array}{rrr}1 \\1 \\1\end{array}\right]}\))= \({\left[\begin{array}{rrr}-4 \\1 \\0\end{array}\right]}\)
For \(T(x^2)\): \(\rho_C({\left[\begin{array}{rrr}4 \\-1 \\2\end{array}\right]})\)= \(\rho_C(5{\left[\begin{array}{rrr}1 \\0 \\0\end{array}\right]}\) \({-3\left[\begin{array}{rrr}1 \\1 \\0\end{array}\right]}\)+ \({2\left[\begin{array}{rrr}1 \\1 \\1\end{array}\right]}\))= \({\left[\begin{array}{rrr}5 \\-3 \\2\end{array}\right]}\)
For \(T(x^3)\): \(\rho_C({\left[\begin{array}{rrr}-2 \\1 \\-3\end{array}\right]})\)= \(\rho_C(-3{\left[\begin{array}{rrr}1 \\0 \\0\end{array}\right]}\)+ \({4\left[\begin{array}{rrr}1 \\1 \\0\end{array}\right]}\) \({-3\left[\begin{array}{rrr}1 \\1 \\1\end{array}\right]}\))= \({\left[\begin{array}{rrr}-3 \\4 \\-3\end{array}\right]}\)
The four vectors that we find will become the columns for the matrix representation.
\(M^T_B={4\left[\begin{array}{rrr}1&-4&5&3 \\-2&1&-3&4\\3&0&2&-3\end{array}\right]}\)