Jordan Decomposition Theorem

The Jordan decomposition expresses any matrix (over an algebraically closed field) as the sum of a diagonalizable part and a nilpotent part that commute with each other. This additive decomposition is the algebraic counterpart of the Jordan normal form and is fundamental to the structure theory of linear operators.

Statement

Theorem7.4Jordan (Additive) Decomposition

Let $A \in M_{n \times n}(F)$ where $F$ is algebraically closed. Then there exist unique matrices $D$ and $N$ such that:

$A = D + N$ ,
$D$ is diagonalizable,
$N$ is nilpotent,
$DN = ND$ (they commute).

Moreover, $D$ and $N$ are polynomials in $A$ : there exist $p, q \in F[\lambda]$ with $D = p(A)$ and $N = q(A)$ .

RemarkJordan decomposition from Jordan normal form

If $A = PJP^{-1}$ with $J = \operatorname{diag}(J_{k_1}(\lambda_1), \ldots)$ , then decompose each block: $J_k(\lambda) = \lambda I_k + N_k$ . So $J = \Lambda + \tilde{N}$ where $\Lambda = \operatorname{diag}(\lambda_1 I_{k_1}, \ldots)$ is diagonal and $\tilde{N}$ is the nilpotent part. Then $D = P\Lambda P^{-1}$ and $N = P\tilde{N}P^{-1}$ .

Examples

ExampleJordan decomposition of a 2x2 matrix

$A = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = D + N$ .

$D = 3I$ is diagonalizable, $N^2 = 0$ is nilpotent, $DN = ND = \begin{pmatrix} 0 & 3 \\ 0 & 0 \end{pmatrix}$ ✓.

ExampleJordan decomposition of a 3x3 matrix

$A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{pmatrix}$ .

Jordan form: $\operatorname{diag}(J_2(2), J_1(3))$ , so in the Jordan basis:

$D' = \operatorname{diag}(2, 2, 3)$ , $N' = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ .

Since $A$ is already upper triangular with Jordan structure, $D = \operatorname{diag}(2, 2, 3)$ and $N = A - D = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ .

Wait -- check commutativity: $DN = \begin{pmatrix} 0 & 2 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{pmatrix}$ and $ND = \begin{pmatrix} 0 & 2 & 0 \\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{pmatrix}$ . These are not equal, so this naive decomposition is wrong.

The correct $D$ and $N$ must commute. Using the Jordan form basis: if $P^{-1}AP = \operatorname{diag}(J_2(2), 3)$ , then $D = P\operatorname{diag}(2, 2, 3)P^{-1}$ and $N = P\operatorname{diag}(N_2, 0)P^{-1}$ . These are the correct $D, N$ which commute.

ExampleJordan decomposition of a diagonalizable matrix

If $A$ is already diagonalizable: $D = A$ , $N = 0$ .

$A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ : $D = A$ , $N = 0$ .

ExampleJordan decomposition of a nilpotent matrix

If $A$ is nilpotent: $D = 0$ , $N = A$ .

$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ : $D = 0$ , $N = A$ .

Uniqueness

TheoremUniqueness of the Jordan decomposition

The decomposition $A = D + N$ with $D$ diagonalizable, $N$ nilpotent, and $DN = ND$ is unique.

ProofProof of uniqueness

Suppose $A = D_1 + N_1 = D_2 + N_2$ are two such decompositions. Then $D_1 - D_2 = N_2 - N_1$ .

Since $D_1, N_1$ commute and $D_2, N_2$ commute, and both pairs are polynomials in $A$ , all four matrices commute pairwise.

$D_1 - D_2$ is diagonalizable (difference of commuting diagonalizable matrices). $N_2 - N_1$ is nilpotent (sum of commuting nilpotent matrices). A matrix that is both diagonalizable and nilpotent must be zero (its only eigenvalue is $0$ , but it equals its diagonal form, which is $0$ ). So $D_1 = D_2$ and $N_1 = N_2$ .

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ExampleVerifying uniqueness fails without commutativity

$A = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$ . Jordan decomposition: $D = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ , $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ .

$DN = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = ND$ ? $ND = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \neq DN$ .

So this obvious splitting does NOT satisfy $DN = ND$ . The correct decomposition requires passing to the Jordan basis.

$A$ has eigenvalues $1, 0$ , so it is actually diagonalizable (distinct eigenvalues). The correct decomposition is $D = A$ , $N = 0$ .

The polynomial property

ExampleD and N as polynomials in A

$A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ : eigenvalues $2, 3$ , diagonalizable (distinct eigenvalues).

$D = A$ , $N = 0$ . $D = p(A)$ with $p(\lambda) = \lambda$ , $N = q(A)$ with $q(\lambda) = 0$ .

$A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ : $D = 2I$ , $N = A - 2I$ .

$D = p(A)$ where $p(\lambda) = 2$ (constant polynomial). $N = q(A)$ where $q(\lambda) = \lambda - 2$ .

Since $D$ and $N$ are polynomials in $A$ , they commute with every matrix that commutes with $A$ (not just with each other). This is a powerful structural property.

ExampleConstructing D and N via interpolation

For $A$ with distinct eigenvalues $\lambda_1, \ldots, \lambda_k$ (over the algebraic closure), construct $p(\lambda)$ satisfying:

$p(\lambda) \equiv \lambda_i \pmod{(\lambda - \lambda_i)^{m_i}}$ for each $i$ ,

where $m_i$ is the algebraic multiplicity. This is a Chinese Remainder Theorem problem. Then $D = p(A)$ and $N = A - p(A)$ .

For $A$ with $p_A(\lambda) = (\lambda - 2)^2(\lambda - 5)$ : find $p$ with $p(\lambda) \equiv 2 \pmod{(\lambda-2)^2}$ and $p(\lambda) \equiv 5 \pmod{(\lambda-5)}$ . By CRT: $p(\lambda) = 2 + \frac{3(\lambda - 2)^2}{(5 - 2)^2} \cdot 3 = \ldots$ (a polynomial of degree $\leq 2$ ).

Multiplicative Jordan decomposition

TheoremMultiplicative Jordan decomposition

If $A$ is invertible, then $A = D_s \cdot U$ where $D_s$ is diagonalizable (semisimple), $U$ is unipotent ( $U - I$ is nilpotent), and $D_s U = U D_s$ .

This follows from the additive decomposition: $A = D + N = D(I + D^{-1}N)$ , so $D_s = D$ and $U = I + D^{-1}N$ .

ExampleMultiplicative decomposition example

$A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ : $D = 2I$ , $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ .

$D_s = 2I$ , $U = I + \frac{1}{2}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1/2 \\ 0 & 1 \end{pmatrix}$ .

$A = D_s U = 2I \cdot \begin{pmatrix} 1 & 1/2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ ✓.

$U - I = \begin{pmatrix} 0 & 1/2 \\ 0 & 0 \end{pmatrix}$ is nilpotent ✓.

Applications

ExampleMatrix exponential via Jordan decomposition

$e^{A} = e^{D+N} = e^D \cdot e^N$ (since $DN = ND$ , the exponential splits).

$e^D$ is easy (diagonalizable: $e^D = Pe^{\Lambda}P^{-1}$ ). $e^N$ is a finite sum (nilpotent: $e^N = I + N + N^2/2! + \cdots + N^{k-1}/(k-1)!$ ).

For $A = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}$ : $e^A = e^3 \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ .

ExampleMatrix powers via Jordan decomposition

$A^n = (D + N)^n = \sum_{k=0}^{n} \binom{n}{k} D^{n-k} N^k$ (using $DN = ND$ , the binomial theorem applies).

Since $N$ is nilpotent with $N^m = 0$ , the sum is finite (at most $m$ terms).

For $A = J_2(\lambda)$ : $A^n = \lambda^n I + n\lambda^{n-1} N = \begin{pmatrix} \lambda^n & n\lambda^{n-1} \\ 0 & \lambda^n \end{pmatrix}$ .

ExampleSolving differential equations

$x' = Ax$ has solution $x(t) = e^{At}x_0 = e^{Dt}e^{Nt}x_0$ .

The diagonalizable part $e^{Dt}$ gives exponential modes. The nilpotent part $e^{Nt}$ gives polynomial corrections (polynomial times exponential solutions).

For $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ : $x(t) = e^t\begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}x_0$ . The solution is $x_1(t) = e^t(c_1 + c_2 t)$ , $x_2(t) = c_2 e^t$ .

Summary

RemarkThe Jordan decomposition as structure theorem

The Jordan decomposition $A = D + N$ is one of the most fundamental results in linear algebra:

It exists and is unique for matrices over algebraically closed fields.
$D$ captures the eigenvalue behavior (the diagonal part), $N$ captures the deficiency from diagonalizability.
Both $D$ and $N$ are polynomials in $A$ , so they commute with everything that commutes with $A$ .
The decomposition enables computation of $e^{At}$ , $A^n$ , and solutions to differential equations.
It generalizes to the Jordan--Chevalley decomposition in the theory of Lie algebras and algebraic groups.