Answer
a. $2-t+3t^2-t^3+t^4$
b. See solution
c. The matrix for T relative to the given bases is
$M=\begin{bmatrix}
1&0&0\\0&1&0\\1&0&1\\0&1&0\\0&0&1
\end{bmatrix}$
Work Step by Step
a. $T(2-t+t^2)=(2-t+t^2)+t^2(2-t+t^2)=$$2-t+3t^2-t^3+t^4$
b. Let c and d be scalars and $\vec{u}$ and $\vec{v}$ be vectors
$T(c\vec{u}+d\vec{v})=(c\vec{u}+d\vec{v})+t^2(c\vec{u}+d\vec{v})=$$
c\vec{u}+d\vec{v}+t^2c\vec{u}+t^2d\vec{v}=
c(\vec{u}+t^2\vec{u})+d(\vec{v}+t^2\vec{v})=$
$cT(\vec{u})+dT(\vec{v})$
c. Let basis C be $\{1,t,t^2,t^3,t^4\}$
$T(1)=1+t^2$
$[T(1)]_C=\begin{bmatrix}
1\\0\\1\\0\\0
\end{bmatrix}$
$T(t)=t+t^3$
$[T(t)]_C=\begin{bmatrix}
0\\1\\0\\1\\0
\end{bmatrix}$
$T(t^2)=t^2+t^4$
$[T(t^2)]_C=\begin{bmatrix}
0\\0\\1\\0\\1
\end{bmatrix}$
The matrix for T relative to the given bases is
$M=\begin{bmatrix}
1&0&0\\0&1&0\\1&0&1\\0&1&0\\0&0&1
\end{bmatrix}$
