Given are the following bases of the vector space $V = \{p \in \mathbb{R}[t] \mid \operatorname{deg}(p) ≤ 3\}$ of all polynomials of degree less than or equal to $3$:
$\mathcal{A} = (t^3, t^2, t, 1)$
$\mathcal{B} = (1, t + 1, t^2 + t + 1, t^3 + t^2 + t + 1)$a) Display the polynomial $p(t)=3t^3+t^2-1$ in coordinate representation with respect to base $\mathcal{A}$ and $\mathcal{B}$.
b) Let $f:V\to V$ be the linear mapping that maps each polynomial from $V$ to its derivative. Determine $M^\mathcal{A}_\mathcal{A}(f)$ and $M_\mathcal{B}^\mathcal{A}(f)$.
a)
$\lambda_1 t^3+\lambda_2 t^2+\lambda_3 t^1+\lambda 4 t^0 = 3t^3+t^2-1t^0\implies \lambda_1=3,\; \lambda_2=1,\;\lambda_3=0,\;\lambda_4=-1$
coordinate representation:$\begin{pmatrix}3\\1\\0\\-1\end{pmatrix}$
$\lambda_1 +\lambda_2 (t+1)+\lambda_3 (t^2 + t + 1) +\lambda 4 (t^3 + t^2 + t + 1) = 3t^3+t^2-1t^0$$\implies \lambda_1=-2,\; \lambda_2=-1,\;\lambda_3=-2,\;\lambda_4=3$
coordinate representation:$\begin{pmatrix}-2\\-1\\-2\\3\end{pmatrix}$
b)
I really don't know how I can calculate $M^\mathcal{A}_\mathcal{A}(f)$ and $M^\mathcal{A}_\mathcal{B}(f)$, can you give me any hints, please?
Part (b): Abbreviate $M:=M_{\mathcal{A}}^{\mathcal{A}}(f)$ and using the convention to write $f(v)=M\hat{v}$, where $\hat{v}$ denotes the coordindate represntation of $v$ in terms of basis $\mathcal{A}$, you can determine $M$ by evaluation $f$ on the basis vectors in $\mathcal{A}$, i. e. $f(t^3)=3t^2$, thus the first column vector of $M$ is $(0,3,0,0)^t$, which is the coordinate representation of the image in terms of $\mathcal{A}$.
If you now want to compute $M_{\mathcal{B}}^{\mathcal{A}}(f)$ you do the same thing, starting with basis elements in $\mathcal{A}$ but then the columns of $M_{\mathcal{B}}^{\mathcal{A}}(f)$ are coordinate representations of the images in terms of basis $\mathcal{B}$. As an example: we have $f(t^3)=3t^2=-3(t+1)+3(t^2 +t+1)$, hence the first column vector of $M_{\mathcal{B}}^{\mathcal{A}}(f)$ is $(0,-3,3,0)^t$.