Jordan Normal Form

The Jordan normal form is the "next best thing" after diagonalization. When a matrix cannot be diagonalized, it can still be brought into a nearly diagonal form consisting of Jordan blocks. This form is unique (up to ordering of blocks) and completely classifies matrices up to similarity over an algebraically closed field.

Jordan blocks

Definition7.1Jordan block

A Jordan block of size $k$ with eigenvalue $\lambda$ is the $k \times k$ matrix:

$J_k(\lambda) = \lambda I_k + N_k,$

where $N_k$ is the $k \times k$ matrix with $1$ s on the superdiagonal and $0$ s elsewhere. Explicitly:

$J_k(\lambda) = \begin{pmatrix} \lambda & 1 & 0 & \cdots & 0 \\ 0 & \lambda & 1 & \cdots & 0 \\ \vdots & & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda & 1 \\ 0 & 0 & \cdots & 0 & \lambda \end{pmatrix}.$

ExampleSmall Jordan blocks

$J_1(\lambda) = (\lambda)$ -- a $1 \times 1$ block (just a scalar).
$J_2(3) = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}$ .
$J_3(0) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ (a nilpotent Jordan block).
$J_4(2) = \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{pmatrix}$ .

ExamplePowers of Jordan blocks

$(J_k(\lambda))^n = \sum_{j=0}^{k-1} \binom{n}{j} \lambda^{n-j} N_k^j$ , so the $(i, i+j)$ entry of $J_k(\lambda)^n$ is $\binom{n}{j}\lambda^{n-j}$ for $j = 0, 1, \ldots, k-1$ .

For $J_2(3)^n = \begin{pmatrix} 3^n & n \cdot 3^{n-1} \\ 0 & 3^n \end{pmatrix}$ .

For $J_3(1)^n = \begin{pmatrix} 1 & n & \binom{n}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{pmatrix}$ .

Jordan normal form

Definition7.2Jordan normal form

A matrix $A \in M_{n \times n}(F)$ (where $F$ is algebraically closed) is in Jordan normal form if it is a block diagonal matrix of Jordan blocks:

$J = \operatorname{diag}(J_{k_1}(\lambda_1), J_{k_2}(\lambda_2), \ldots, J_{k_r}(\lambda_r)),$

where $k_1 + k_2 + \cdots + k_r = n$ .

Theorem7.1Jordan Normal Form Theorem

Let $A \in M_{n \times n}(F)$ where $F$ is algebraically closed (e.g., $F = \mathbb{C}$ ). Then $A$ is similar to a unique (up to permutation of blocks) Jordan normal form $J$ . That is, there exists an invertible $P$ such that $P^{-1}AP = J$ .

ExampleJordan forms for 2x2 matrices

Every $2 \times 2$ matrix over $\mathbb{C}$ has one of two Jordan forms:

Two distinct eigenvalues $\lambda_1 \neq \lambda_2$ : $J = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$ (diagonal).
One repeated eigenvalue $\lambda$ with $m_g = 2$ : $J = \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} = \lambda I$ (already diagonal).
One repeated eigenvalue $\lambda$ with $m_g = 1$ : $J = \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}$ (a single Jordan block).

ExampleAll Jordan forms for 3x3 matrices

The possible Jordan forms for $3 \times 3$ matrices (grouped by eigenvalue pattern):

Three distinct eigenvalues $\lambda_1, \lambda_2, \lambda_3$ : $\operatorname{diag}(\lambda_1, \lambda_2, \lambda_3)$ .

One repeated eigenvalue (say $\lambda$ with $m_a = 2$ , plus $\mu \neq \lambda$ ):

$\operatorname{diag}(\lambda, \lambda, \mu)$ if $m_g(\lambda) = 2$ .
$\operatorname{diag}(J_2(\lambda), \mu)$ if $m_g(\lambda) = 1$ .

Triple eigenvalue $\lambda$ (with $m_a = 3$ ):

$\lambda I_3$ if $m_g = 3$ .
$\operatorname{diag}(J_2(\lambda), \lambda)$ if $m_g = 2$ .
$J_3(\lambda)$ if $m_g = 1$ .

Computing Jordan form

ExampleFinding the Jordan form

$A = \begin{pmatrix} 5 & 4 & 2 & 1 \\ 0 & 1 & -1 & -1 \\ -1 & -1 & 3 & 0 \\ 1 & 1 & -1 & 2 \end{pmatrix}$ .

Step 1: Characteristic polynomial. $p_A(\lambda) = (\lambda - 1)(\lambda - 2)(\lambda - 4)^2$ ... (assume after computation).

Eigenvalues: $1, 2, 4$ (with $4$ having $m_a = 2$ ).

Step 2: For $\lambda = 4$ , compute $\dim\ker(A - 4I)$ . If $m_g(4) = 2$ : diagonal block $\operatorname{diag}(4, 4)$ . If $m_g(4) = 1$ : Jordan block $J_2(4)$ .

Step 3: The Jordan form is either $\operatorname{diag}(1, 2, 4, 4)$ or $\operatorname{diag}(1, 2, J_2(4))$ .

ExampleNilpotent 3x3 matrix

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

Eigenvalue: $0$ with $m_a = 3$ . $\operatorname{rank}(A) = 1$ , so $m_g(0) = \dim\ker A = 2$ .

Since $m_g = 2$ , there are two Jordan blocks for $\lambda = 0$ : one $J_2(0)$ and one $J_1(0)$ .

Jordan form: $\operatorname{diag}(J_2(0), 0) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ .

The matrix is already in Jordan form.

ExampleJordan form with complex eigenvalues

$A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ over $\mathbb{C}$ .

Eigenvalues: $i, -i$ (distinct), so the Jordan form is $\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$ .

Over $\mathbb{R}$ : $A$ is already in its real canonical form (a rotation block).

Generalized eigenvectors

Definition7.3Generalized eigenvector

A vector $v$ is a generalized eigenvector of $A$ for eigenvalue $\lambda$ of order $k$ if:

$(A - \lambda I)^k v = 0 \quad \text{but} \quad (A - \lambda I)^{k-1} v \neq 0.$

The generalized eigenspace is $G_\lambda = \ker(A - \lambda I)^{m_a(\lambda)}$ , and $\dim G_\lambda = m_a(\lambda)$ .

ExampleFinding generalized eigenvectors

$A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ , $\lambda = 2$ , $m_a = 2$ , $m_g = 1$ .

Eigenvector: $v_1 = (1, 0)$ , with $(A - 2I)v_1 = 0$ .

Generalized eigenvector: solve $(A - 2I)v_2 = v_1$ , i.e., $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} v_2 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ . Solution: $v_2 = (0, 1)$ .

$(A - 2I)^2 v_2 = (A-2I)v_1 = 0$ ✓, $(A-2I)v_2 = v_1 \neq 0$ ✓.

Jordan chain: $v_1 \leftarrow v_2$ . Transition matrix: $P = (v_1, v_2) = I$ , $P^{-1}AP = J_2(2)$ .

ExampleGeneralized eigenvectors for a 3x3 matrix

$A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ , $\lambda = 1$ , $m_a = 3$ , $m_g = 1$ (since $\operatorname{rank}(A - I) = 2$ ).

$(A - I) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ , $(A-I)^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ , $(A-I)^3 = 0$ .

Jordan chain: start with $v_3 = (0, 0, 1)$ (in $\ker(A-I)^3 \setminus \ker(A-I)^2$ ).

$v_2 = (A-I)v_3 = (0, 1, 0)$ , $v_1 = (A-I)v_2 = (1, 0, 0)$ .

$P = (v_1, v_2, v_3) = I$ , confirming $A$ is already in Jordan form $J_3(1)$ .

Properties of Jordan form

ExampleInvariants from Jordan form

$J = \operatorname{diag}(J_2(3), J_1(5))$ :

$\det(J) = 3^2 \cdot 5 = 45$ .
$\operatorname{tr}(J) = 3 + 3 + 5 = 11$ .
Characteristic polynomial: $(\lambda - 3)^2(\lambda - 5)$ .
Minimal polynomial: $(\lambda - 3)^2(\lambda - 5)$ (the largest Jordan block for $\lambda = 3$ has size $2$ ).

ExampleNilpotent index from Jordan form

If $A$ has Jordan form $\operatorname{diag}(J_3(0), J_2(0), J_1(0))$ :

$A^2 \neq 0$ (the $J_3(0)$ block needs $3$ applications to vanish).
$A^3 = 0$ (all blocks are killed by the third power).
The nilpotent index is the size of the largest Jordan block.

ExampleJordan form classifies similarity

Two matrices are similar iff they have the same Jordan form (up to ordering of blocks).

$A = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 2 \\ 0 & 3 \end{pmatrix}$ : both have Jordan form $J_2(3)$ . They are similar (any upper triangular matrix with eigenvalue $3$ and $m_g = 1$ in $2 \times 2$ is similar to $J_2(3)$ ). Indeed $P = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ gives $P^{-1}BP = J_2(3)$ .

$A = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}$ and $C = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$ : different Jordan forms ( $J_2(3)$ vs $\operatorname{diag}(3, 3)$ ), so not similar.

Applications of Jordan form

ExampleMatrix exponential via Jordan form

$e^{J_k(\lambda)t} = e^{\lambda t} \begin{pmatrix} 1 & t & t^2/2 & \cdots \\ 0 & 1 & t & \cdots \\ \vdots & & \ddots & \\ 0 & 0 & \cdots & 1 \end{pmatrix}$ .

For $J_2(3)$ : $e^{J_2(3)t} = e^{3t}\begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}$ .

The solution to $x' = J_2(3)x$ is $x(t) = e^{3t}(c_1 + c_2 t, c_2)$ -- note the polynomial growth factor $t$ due to the Jordan block. Diagonal matrices give pure exponentials; Jordan blocks add polynomial factors.

ExampleLong-term behavior from Jordan form

So $A^n \to 0$ iff all eigenvalues satisfy $|\lambda| < 1$ (the spectral radius $\rho(A) < 1$ ).

Summary

RemarkJordan form as the complete similarity invariant

The Jordan normal form provides:

A canonical representative for each similarity class of matrices.
A classification: two matrices are similar iff they have the same Jordan form.
Explicit formulas for $A^n$ and $e^{At}$ in terms of eigenvalues and block sizes.
The minimal polynomial as $\prod_i (\lambda - \lambda_i)^{k_i}$ where $k_i$ is the largest block size for $\lambda_i$ .
A decomposition of $\mathbb{C}^n$ into generalized eigenspaces: $\mathbb{C}^n = G_{\lambda_1} \oplus \cdots \oplus G_{\lambda_m}$ .