Express the matrix of $f$ with respect to the basis $\{1,x+1,x^{2}+x\}$. Let $\textbf{P}_{2}(\textbf{R})$ be the real vector space of polynomials of degree less than or equal to $2$.
Let $f:\textbf{P}_{2}(\textbf{R})\rightarrow \textbf{P}_{2}(\textbf{R})$ be the linear map given by differentiation, i.e., $f(p(x)) = p'(x)$.
Compute the matrix of $f$ with respect to the basis $\{1, x + 1, x^2 + x\}$.
MY ATTEMPT
So far I've done   $f(1) = 0$, $f(x+1) = 1$ and $f(x^2 + x) = 2x + 1$.
 A: Let's consider the first equation:
$$
f(1) = 0
$$
Now, we can decompose the input and the output on the polynomial basis  you suggested: 
$$
f(1\times(1) + 0(x+1) + 0(x^2+1))= 0\times(1) + 0(x+1) + 0(x^2+1)
$$
Reading out the coefficients both on the input and the output, we deduce that the sought matrix, call it $f_{mat}$, must be such that
$$
f_{mat}(1,0,0)^T = (0,0,0)^T
$$
This equation is equivalent to state that the first column of $f_{mat}$ is $(0,0,0)^T$.
We can proceed similarly for the other two equations:
$$
f_{mat}(0,1,0)^T = (1,0,0)^T \\
\text{i.e. } f(0\times(1) + 1(x+1) + 0(x^2+1))=1\times(1) + 0(x+1) + 0(x^2+1) \\
f_{mat}(0,0,1)^T = (-1,2,0)^T \\
\text{i.e. } f(0\times(1) + 0(x+1) + 1(x^2+1))=-1\times(1) + 2(x+1) + 0(x^2+1)
$$
 Thus obtaining
$$
 f_{mat}=   \begin{pmatrix}
    0 & 1 & -1 \\
    0 & 0 & 2 \\
    0 & 0 & 0 \\
    \end{pmatrix}
$$
A: Since we are dealing with a linear operator, we can assume the same basis for the domain and the counterdomain.
Given the basis $\mathcal{B} = \{1,x+1,x^{2}+x\}$, we have the following system of equations to solve
\begin{align*}
\begin{cases}
a_{11} + a_{12}(x+1) + a_{13}(x^{2}+x) = f(1) = 0\\\\
a_{21} + a_{22}(x+1) + a_{23}(x^{2} + x) = f(x+1) = 1\\\\
a_{31} + a_{32}(x+1) + a_{33}(x^{2} + x) = f(x^{2}+x) = 2x + 1
\end{cases}
\end{align*}
whence we conclude $f(1) = (0,0,0)_{\mathcal{B}}$, $f(x+1) = (1,0,0)_{\mathcal{B}}$ and $f(x^{2}+x) = (-1,2,0)_{\mathcal{B}}$.
Finally, one has that
\begin{align*}
[f]_{\mathcal{B}} = 
\begin{bmatrix}
0 & 1 & -1\\
0 & 0 & 2\\
0 & 0 & 0
\end{bmatrix}
\end{align*}
and we are done.
Hopefully this helps.
