Prove or disprove existence of ordered basis satisfying representing a linear transformation

Question

Let $D: \mathbb {R}_{\le n} [x] \to {R}_{\le n} [x]$ defined as $D\ (p(x)) = p'(x)$.

Prove or disprove: There is an ordered basis $B$ of ${R}_{\le n} [x]$ such that $[D]_{B} = J$ while $J$ being the following matrix:

$$\begin{bmatrix}1&0&\cdots &0\\0&1&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &0\end{bmatrix}$$

I believe the answer is no but I'm not really sure how to show it.

Any help would be appreciated.

Answer 1

Let $b_1$ be the first basis polynomial. Then the first column of the matrix shows that $D(b_1)=b_1$, that is, $$b_1'(x)=b_1(x)\ .$$ There is no polynomial that satisfies this condition except for the zero polynomial, and that can't be part of a basis.