Jordan matrix form and polynomial proof.

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.

• Couldn't find the proof yet. A solution can be really helpful. May 31 '20 at 14:14

Hint

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.
• Im sorry I dont understand how it is connected to the rows and columns of the matrix. the i,j term of the matrix should be equal to the (j-i) deriviative multiply by $1/{(j-i)!}$ and I cant see how to proceed with your hint. also I cant see why your hint is correct. I would say $f_{k}^{(l)}\left(x\right)=\begin{cases} \frac{k!}{l!}x^{k-l} & 0\leq l\leq k\\ 0 & else \end{cases}$ rather than what you suggested May 30 '20 at 15:15
• I think that my formulae for the derivation of $f_k$ is correct. Take $l=0$ for example and then proceed by induction. Regarding the hint, use the formulae that you got at step 1 in your step 2 with the derivation formulae that I gave. May 30 '20 at 15:21
• Do you think that for a general polynomial $f_{k}\left(x\right)=\sum_{j=0}^{k}a_{j}x^{j}$ this formulae will work ? : $f_{k}^{(l)}\left(x\right)=\begin{cases} \sum_{j=0}^{k}a_{j}\frac{j!}{(j-l)!}x^{k-l} & 0\leq j\leq l\leq k\\ 0 & else \end{cases}$ Also, the lower bound of the polynomial is 0 while the lower bound in the matrix would be 1 (the first row/column) how do you suggest fixing it? May 30 '20 at 15:32
• $f\left(J_{n}\left(\lambda\right)\right)=\sum_{j=0}^{k}a_{j}\left(J_{n}\left(\lambda\right)\right)^{j}=\sum_{j=0}^{k}\sum_{i=0}^{j}\binom{j}{i}a_{j}\lambda^{j-i}J_{n}^{i}\left(0\right)=\sum_{i=0}^{j}\frac{1}{i!}f^{(i)}\left(\lambda\right)J_{n}^{i}\left(0\right)$ It does look closer to the form i want to get to. But still, I dont know how to proceed. can you elaborate ? May 30 '20 at 15:57