Given $D:\mathbb R_{\le n}[x]\to\mathbb R_{\le n}[x], D(p)=p'$ Let $\mathbb R_{\le n}[x]$ be the polynomial space of a degree $\le n$ with coefficients in $\mathbb R$.
Given $D:\mathbb R_{\le n}[x]\to\mathbb R_{\le n}[x]$ defined by $D(p)=p'$ (the derivative transformation).
Let $$J=\begin{bmatrix}1\\
 & 1\\
 &  & ...\\
 &  &  & ...\\
 &  &  &  & 1\\
 &  &  &  &  & 0
\end{bmatrix}\in M_{n+1}(\mathbb R)$$


*

*Find ordered bases $B,C$ of $\mathbb R_{\le n}[x]$ so that $J=[D]_C^B$ (transformation matrix from the base $B$ to $C$)

*Let $E=\{1,x,x^2,\dots,x^n\}$. Find $[D]_E^E$.

*Find an ordered base $B$ of $\mathbb R_{\le n}[x]$ so that $J=[D]_B^B$ or prove that that such ordered base doesn't exist.


my answers: 


*

*as $D(p)=p'$, I tried to build some base as $B=\{x_1,x_2,\dots,x_n\}$ as i is the degree of the every vector $i\leq n$.


then I tried to differentiate $B$. I got confused with the coefficients. what is the right way of finding $B$ and $C$?


*$[D]_E^E=\begin{bmatrix}0\\
 & 1\\
 &  & 2\\
 &  &  & ...\\
 &  &  &  & n\\
 &  &  &  &  & 0\\
 &  &  &  &  &  & 0
\end{bmatrix}$ is that correct?

*I think there's no such base because the only $p$ that will exist $D(p)=p'$ is $p=0$.

 A: For last question : if there was a base $B$ such that $[D]_B^B$ were diagonal, then there would exist $n+1$ polynomials $P_k$, $0\le k\le n$, such that $D(P_k)=P_k$ for $0\le k\le n-1$ and $D(P_n)=0$.
In particular, $P_k$ would be equal to its derivative, which can be true only if $P_k=0$. But this can be a base only if $n=0$.
A: Hints


*

*With $B=(b_0,\ldots,b_n)$ and $C=(c_1,\ldots,c_n)$ as $D(b_n)=\sum_i J_{i,n+1} c_i=0$ you need to have $b_n=1$ (or any constant). Then you can choose the other vectors of $B$ and adjust $C$ accordingly.


For example try $B=(x^n,x^{n-1}, \ldots, x^2,x,1)$ and $C=(nx^{n-1},(n-1) x^{n-2},\ldots,2x,1,x^n)$.


*To compute $[D]_E^E$, for all $e_i=x^i$ you need to write $D(e_i)$ as a linear combination of the $e_j$.


For example $D(e_0)=0=\sum_{j=0}^n 0 \times e_j$ so the first column is filled with $0$.
Then $D(e_1)=1=1 \times e_0+0 \times e_1+\ldots +0 \times e_n$.
Doing so for all the basis leads to:
$$\begin{bmatrix}0&1\\  &0& 2\\  &  & \ddots\\  &  &  & \ddots\\  &  &  &  & &0&n\\  &  &  &  &  & &0 \end{bmatrix}$$


*For the last point your justification is the correct one, you can look at the answer of Nicolas FRANCOIS for a  more precise explanation.

