let $ f\in F[x] $ be a polynomial. and prove that the matrix $ f\left(J_{n}\left(\lambda\right)\right) $ satisfies
$ [f\left(J_{n}\left(\lambda\right)\right)]_{ij}=\begin{cases} \frac{1}{\left(j-i\right)!}f^{(j-i)}\left(\lambda\right) & 1\leq i\leq j\leq n\\ 0 & else \end{cases} $
when $ f^{(j-i)} $ is the (j-i) deriviative of $ f $.
Here's what i tried:
step 1: I proved that
$ [\left(J_{n}\left(0\right)\right)^{k}]=\begin{cases} 1 & j=i+k\\ 0 & else \end{cases} $
step 2: Using the binom formula, I proved that
$ \left(J_{n}\left(\lambda\right)\right)^{k}=\sum_{i=0}^{k}\binom{k}{i}\lambda^{k-i}\left(J_{n}\left(0\right)^{i}\right) $
Now assume $ f\left(x\right)=\sum_{j=0}^{k}a_{j}x^{j} $ then,
$ f\left(J_{n}\left(\lambda\right)\right)=\sum_{j=0}^{k}a_{j}\left(J_{n}\left(\lambda\right)^{j}\right)=\sum_{j=0}^{k}a_{j}\sum_{i=0}^{j}\binom{j}{i}\lambda^{j-i}\left(J_{n}\left(0\right)\right)^{i}=\sum_{j=0}^{k}\sum_{i=0}^{j}a_{j}\lambda^{j-i}\left(J_{n}\left(0\right)\right)^{i} $
Im not sure how to recognize the (j-i) deriviative out of the expression. And I'm not sure how to continue. Any ideas will help. Thanks in advance.
Hint
Your step 1 is OK.
Your step 2 also, but you're not using it completely. During the step 2, you're considering I imagine $f_k(x) = x^k$. Therefore
$$f_k^{(l)}(x)= \begin{cases} \frac{k!}{(k-l)!} x^{k-l} & \text{ for } 0 \le l \le k\\ 0 & \text{ else} \end{cases}$$
Use this and your step 1 to prove the expected formulae for the monomial $f_k(x) = x^k$.
Your last step could be a single sentence noticing that the requested formulae is linear in $f$ as the derivation is.