The Jordan-Chevalley expresses a linear operator $M$ as $$ M = D + N, $$ where $D$ is semisimple (diagonalizable), $N$ is nilpotent and $DN=ND$. Although it is stated in many sources that $D$ and $N$ can be written as polynomials in $M$, I never see any method for obtaining such polynomials. I want to emphasize that I am not interested in a proof about the existence of these polynomials, but in a procedure for explicitly obtaining them given the Jordan-Chevalley decomposition.

  • $\begingroup$ If you already know $D$, $N$, expressing them as linear combination of $I,M,M^2,...,M^{n-1}$ is a simple Gaussian elimination problem. $\endgroup$ – user120527 Feb 13 at 10:56
  • $\begingroup$ That is not clear to me. Could you expand your comment into an explicit answer? $\endgroup$ – jobe Feb 13 at 11:04

Write $\chi_M(X)=\prod_{k=1}^s{(X-\lambda_k)^{\alpha_k}}$ (characteristic polynomial).

From among others Cayley-Hamilton, we know that $\bigoplus_{k=1}^s{\ker((M-\lambda_kI)^{\alpha_k})}=\mathbb{C}^n$ (for instance).

Let, for each $k$, $A_k$, $B_k$ be polynomials such that $A_k(X)(X-\lambda_k)^{\alpha_k}+B_k(X)\prod_{l \neq k}{(X-\lambda_l)^{\alpha_l}}=1$.

Let $P_k(X)=B_k(X)\prod_{l \neq k}{(X-\lambda_l)^{\alpha_l}}$.

Then, if $(M-\lambda_kI)^{\alpha_k}v=0$, then $P_k(M)v=v$. On the other hand, if $l \neq k$ and $(M-\lambda_lI)^{\alpha_l}v=0$, then $P_k(M)v=0$.

Then $D=\sum_k{\lambda_kP_k(M)}$.

  • $\begingroup$ I am having a hard time to fully understand your answer. One point: how can one compute $A_k(X)$ and $B_k(X)$? I have tried for some cases, but I could not. Is the relation between $A_k$ and $B_k$ a version of the Bézout's identity? $\endgroup$ – jobe Feb 13 at 20:42
  • $\begingroup$ Yes. That’s Euclid algorithm for polynomials. $\endgroup$ – Mindlack Feb 13 at 20:43
  • $\begingroup$ Ok. I will spend some additional time to completely understand your answer before accept it. Anyway, thank you very much! $\endgroup$ – jobe Feb 13 at 20:47

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