Let $M$ be a matrix with entries in $\mathbb C$. The SN (or Jordan-Chevalley) decomposition theorem states that we can find unique matrices $S$ and $N$ such that:

  1. $M=S+N$
  2. $S$ is diagonalizable
  3. $N$ is nilpotent
  4. $SN=NS$.

I would like to prove the uniqueness part of this theorem, since everything else is immediate from the fact that all complex matrices can be put into Jordan normal form for some choice of basis. (If $M=AJA^{-1}$, where $J$ is in Jordan normal form, write $J=J_S+J_N$, where $J_S$ consists of the diagonal part of $J$ with zeroes elsewhere, and $J_N$ consists of the line above the diagonal with zeroes elsewhere. Then let $S=AJ_SA^{-1}$ and $N=AJ_NA^{-1}$ and direct calculation verifies that these meet the criteria above.)

This post attempted to answer the same question, but unfortunately the proof is invalid because it assumes that the difference of two nilpotent matrices is another nilpotent matrix, which is not true (a counterexample is given in this post).

  • $\begingroup$ Do you know that $S$ and $N$ are both polynomials in $M$? $\endgroup$
    – Thorgott
    Oct 3, 2019 at 16:40
  • $\begingroup$ @Thorgott no; maybe they are, but that would have to be proven as well $\endgroup$
    – WillG
    Oct 3, 2019 at 16:44
  • $\begingroup$ Assume that $M=S^{\prime}+N^{\prime}$ is another such decomposition. It's not hard to show that $S^{\prime}=S$ and $N^{\prime}=N$ granted that $S$ and $S^{\prime}$ as well as $N$ and $N^{\prime}$ each commute. This would be an obvious consequence from $S$ and $N$ being polynomials in $M$, but I'm not sure how to prove that without going through an entirely different proof of the Jordan-Chevalley Decomposition altogether. $\endgroup$
    – Thorgott
    Oct 3, 2019 at 16:52
  • $\begingroup$ Actually, a comment in this post suggests that $S$ and $N$ are polynomials in $M$: math.stackexchange.com/questions/2629238/… $\endgroup$
    – WillG
    Oct 4, 2019 at 17:19
  • $\begingroup$ Can you post your proof, assuming they are polynomials in $M$? Then at least it will only be left to prove that this assumption is valid. $\endgroup$
    – WillG
    Oct 4, 2019 at 17:20

2 Answers 2


You don’t need to know that $S$ and $N$ are polynomials in $M$ to prove the uniqueness of the Jordan–Chevalley decomposition.

Let $\lambda_1, \dotsc, \lambda_r$ be the pairwise different eigenvalues of the matrix $S$ and let $V_i$ be the eigenspace of $S$ with respect to the eigenvalue $\lambda_i$. Then $$ \mathbb{C}^n = V_1 \oplus \dotsb \oplus V_r \tag{1} $$ because $S$ is diagonalizable. Each eigenspace $V_i$ is $N$-invariant because the matrices $S$ and $N$ commute. It follows that each eigenspace $V_i$ is $M$-invariant because $M = S + N$.

Let $W_i$ be the generalized eigenspace of $M$ with respect to $\lambda_i$, i.e. $$ W_i = \ker (M - \lambda_i)^m $$ for $m$ sufficiently large. We claim that $W_i = V_i$. Indeed, the matrices $M - \lambda_i I$ and $M - S$ act the same on $V_i$ because $S$ acts on $V_i$ by multiplication with $\lambda_i$. But $M - S = N$ is nilpotent. This means that $M - \lambda_i I$ acts nilpotently on $V_i$, which shows that $V_i \subseteq W_i$. It follows from $(1)$ that already $V_i = W_i$ because the linear subspaces $W_1, \dotsc, W_r$ of $\mathbb{C}^n$ are linearly independent (i.e. their sum is direct).

The generalized eigenspaces $W_i$ are uniquely determined by $M$. We have thus shown that the eigenspaces of $S$ are uniquely determined by $M$. But the matrix $S$ is uniquely determined by its eigenspaces because it is diagonalizable. We have thus shown the uniqueness of $S$. The uniqueness of $N$ now follows from $N = M - S$.

  • $\begingroup$ Can you justify the assertion that each eigenspace $V_i$ is $N$-invariant because $S$ and $N$ commute? I'm not seeing an obvious reason this should be so. $\endgroup$
    – WillG
    Oct 5, 2019 at 17:20
  • 1
    $\begingroup$ Never mind. $v\in V_i\Rightarrow Sv=\lambda_i v\Rightarrow NSv=\lambda_i Nv\Rightarrow SNv=\lambda_i Nv\Rightarrow Nv∈ V_i$ $\endgroup$
    – WillG
    Oct 5, 2019 at 18:21
  • $\begingroup$ Ok great, everything else makes sense to me. And I like the abstract coordinate-independent approach. Thanks! $\endgroup$
    – WillG
    Oct 5, 2019 at 18:39

Granted there are polynomials $S(X),\,N(X)\in\mathbb{C}[X]$ such that $S=S(M)$ and $N=N(M)$.

Assume there is another decomposition $M=S^{\prime}+N^{\prime}$ with $S^{\prime}$ diagonalizable, $N^{\prime}$ nilpotent and $S^{\prime}N^{\prime}=N^{\prime}S^{\prime}$. Since $S^{\prime}$ commutes with $N^{\prime}$, it also commutes with $M=S^{\prime}+N^{\prime}$ and hence with any polynomial in $M$; particularly, $S^{\prime}$ commutes with $S$. Similarly, $N^{\prime}$ commutes with $N$.

A well-known theorem asserts that two commuting diagonalizable matrices are simultaneously diagonalizable (in fact, the converse holds too) and the matrix simultaneously diagonalizing $S$ and $S^{\prime}$ will also diagonalize the linear combination $S-S^{\prime}$ (alternatively, there is a basis of simultaneous Eigenvectors of $S$ and $S^{\prime}$ that will also be Eigenvectors of $S-S^{\prime}$). On the other hand, a binomial expansion shows that any linear combination of two commuting nilpotent matrices is nilpotent, so particularly $N^{\prime}-N$ is nilpotent.

However, $M=S+N=S^{\prime}+N^{\prime}$ implies that $S-S^{\prime}=N^{\prime}-N$, which then is a matrix that is both diagonalizable and nilpotent. The only such matrix is the zero matrix, thus $S=S^{\prime}$ and $N=N^{\prime}$, i.e. the Jordan-Chevalley decomposition is unique.


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