Finding the matrix of a linear transformation with respect to bases

Question

Consider the linear trasnformation $L:\mathbb{P^3}\rightarrow \mathbb{P^2}$ given by the formula $L(p(t))=p'(t)-tp''(t)$ Find the matrix of the linear transformation $L$ with respect to the bases $S=\{1+t^3,t+t^2,t^2−t^3,t^3\}$ and $T=\{1,t,t^2\}$. Here's what I did to attempt the problem. I found the transformation of the standard bases to be :

$L(e_1)=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$

$L(e_2)=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$

$L(e_3)=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$

$L(e_4)=\begin{bmatrix}0\\-3\\0\\0\end{bmatrix}$

So I know that the matrix $A$ representing the linear transformation is :

$A=\begin{bmatrix}0&0&0&0\\0&0&0&-3\\0&0&0&0\\0&0&0&0\end{bmatrix}$

I don't know what to do after this step. How do I find the matrix with respect to both bases $S$ and $T$? I apologize if I did something horrible wrong or if the answer is obvious!

Answer 1

$S=\{1+t^3,t+t^2,t^2−t^3,t^3\}$ is a basis of $\Bbb P^3$, the domain of $L$,

and $T=\{1,t,t^2\}$ is a basis of $\Bbb P^2$, the codomain of $L$.

$L$ maps the elements of $S$ to $-3t^2, 1, 3t^2, -3t^2$, respectively,

so $A=\begin{bmatrix}0&1&0&0\\0&0&0&0\\-3&0&3&-3\end{bmatrix}.$

Note that $A$ has the proper size for mapping a $4$-dimensional space to a $3$-dimensional space.