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Why $J$ is a polynomial of $J^k$, if the diagonal entries of $J$ is $1$?

Here $J=diag(J_{n_1}(1),\cdots, J_{n_s}(1))$, where $J_{n_i}(1)$ is the Jordan block with diagnoal entries being $1$, and $k\geq 1$ is fixed?

How can we show that there exists a polynomial $p$ such that $J=p(J^k)$.

My attempt: It seems right. the $(i,i+1)$ entry of $J^k$ is $C_k^1$, the $(i,i+1)$ entry of $J^{2k}$ is $C_k^2$, $\cdots$. $E,J^k,J^{2k}, \cdots, J^{(n-1)k}$ form a base of $L(E,N,N^2,\cdots,N^{m-1})$, where $N=J-I$. But the representation of $E,J^k,J^{2k}, \cdots, J^{(n-1)k}$ by $E,N,N^2,\cdots,N^{m-1}$ seems hard.

RobPratt
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  • What would be the characteristic polynomial of $J$? – Calvin Lin Jan 05 '25 at 04:25
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    The statement as it stands is false. E.g. when $J=J_2(1)$ over $GF(2)$, we have $J^2=I_2$ but $J$ is not a polynomial of $I_2$. It is true if the characteristic of the underlying field is zero. – user1551 Jan 05 '25 at 06:51

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