# In the Jordan-Chevalley decomposition $M=D+N$, how obtaining $D$ and $N$ as polynomials in $M$?

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.

• 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. – user120527 Feb 13 at 10:56
• That is not clear to me. Could you expand your comment into an explicit answer? – 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 $$D=\sum_k{\lambda_kP_k(M)}$$.

• 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? – jobe Feb 13 at 20:42
• Yes. That’s Euclid algorithm for polynomials. – Mindlack Feb 13 at 20:43
• Ok. I will spend some additional time to completely understand your answer before accept it. Anyway, thank you very much! – jobe Feb 13 at 20:47