Answer
See below
Work Step by Step
Given: $T:P_4(R)\rightarrow P_3(R)\\
p(x)=3-4x+6x^2+6x^3-2x^4$
We can see that $T(E_{11})=T(E_{22})=T(E_{33})=tr(E_{11})=tr(E_{22})=tr(E_{33})=1$
a) Computing $[T]^C_B$
$T(E_{11})=1.1\\
T(E_{12})=0.1\\
T(E_{13})=0.1\\
T(E_{21})=0.1\\
T(E_{22})=1.1\\
T(E_{23})=0.1\\
T(E_{31})=0.1\\
T(E_{32})=0.1\\
T(E_{33})=1.1$
then we have $[T(E_{11})]_0=1\\
[T(E_{12})]_C=0\\
[T(E_{13})]_C=0\\
[T(E_{21})]_C=0\\
[T(E_{22})]_C=1\\
[T(E_{23})]_C=0\\
[T(E_{31})]_C=0\\
[T(E_{32})]_C=0\\
[T(E_{33})]_C=1$
Hence, $[T]_B^C=\begin{bmatrix}
1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1\end{bmatrix}$
We can notice that $[A]_B=\begin{bmatrix}
2\\ -6\\ 0 \\ 1 \\ 4 \\ -4\\0 \\ 0 \\ 3 \end{bmatrix}$
Consequently, $[T(v)]_C=[T]^C_B[A]_B=\begin{bmatrix}
1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}
2\\ -6\\ 0 \\ 1 \\ 4 \\ -4\\0 \\ 0 \\ 3 \end{bmatrix}\begin{bmatrix}
-4 \\
12 \\ 18 \\ -8
\end{bmatrix}\\
\rightarrow T(A)=3$
b) From part (a), we obtain: $T(p(x))=tr(A)=tr(\begin{bmatrix}
2 & -6 & 0\\
1 & 4 & -4\\
0 & 0 & -3
\end{bmatrix})=2+4-3=3$
