I have the following operator ($P_n$ is the vector space of all polynomials with degree $\leq n$) $$ D:P_n\rightarrow P_n $$ Such that $D(p)=p'$
I want to know if this operator is diagonalizable.
My attempt: I have that the matrix of representation of $D$ is $$ A:=(a_{ij})_{1\leq i,j\leq (n+1)}=\left\{\begin{array}{ccc} k-1 &if& i=j=k\\ 0&if&i\neq j \end{array} \right. $$ So, the charasteristic polynomial of $A$ is $$ p(\lambda)=\det(A-\lambda I)=\prod_{k=1}^n(k-\lambda) $$ As each eigenvalues has multiplicity $1$ then $D$ is diaginalizable. Am I right?
No. Your matrix representation is wrong.
In fact $D$ is not disgonalizable, because it has no eigenvalues except $0$, for which the only corresponding eigenvectors are the constant polynomials. (This is clear because if $\lambda\ne0$ and $p\ne0$ then $\deg(Dp)<\deg(p)=\deg(\lambda p)$ so $Dp\ne\lambda p$.)