Jordan Block — Definition, Formula & Examples

A Jordan block is a square matrix that has a single eigenvalue repeated along the main diagonal, ones immediately above the diagonal (the superdiagonal), and zeros everywhere else. Jordan blocks are the building blocks of the Jordan normal form of a matrix.

A Jordan block of size $n$ associated with eigenvalue $\lambda$ , denoted $J_n(\lambda)$ , is the $n \times n$ matrix $\lambda I_n + N_n$ , where $I_n$ is the identity matrix and $N_n$ is the $n \times n$ nilpotent matrix with ones on the superdiagonal and zeros elsewhere. Equivalently, $(J_n(\lambda))_{ij} = \lambda$ when $i = j$ , $(J_n(\lambda))_{ij} = 1$ when $j = i + 1$ , and $(J_n(\lambda))_{ij} = 0$ otherwise.

Key Formula

J_n(\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}

Where:

$\lambda$ = The eigenvalue associated with the Jordan block
$n$ = The size (number of rows and columns) of the block

How It Works

Every square matrix over

\mathbb{C}

is similar to a block-diagonal matrix whose diagonal blocks are Jordan blocks — this is the Jordan normal form. The size and number of Jordan blocks for each eigenvalue encode the matrix's structure beyond what eigenvalues alone reveal. A

1 \times 1

Jordan block is simply the scalar

[\lambda]

, corresponding to an eigenvector. Larger Jordan blocks indicate the presence of generalized eigenvectors, meaning the eigenspace is "deficient" in true eigenvectors. A matrix is diagonalizable if and only if all its Jordan blocks are

1 \times 1

Worked Example

Problem: Write the 3×3 Jordan block with eigenvalue 2, and compute its square.

Step 1: Construct the Jordan block

J_3(2)

by placing 2 on the diagonal and 1 on the superdiagonal.

J_3(2) = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}

Step 2: Compute

J_3(2)^2

by matrix multiplication.

J_3(2)^2 = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 4 & 4 & 1 \\ 0 & 4 & 4 \\ 0 & 0 & 4 \end{pmatrix}

Step 3: Notice the pattern: the diagonal has

\lambda^2 = 4

, the superdiagonal has

2\lambda = 4

, and the next diagonal has

1

. Powers of Jordan blocks follow a binomial-like pattern.

Answer:

J_3(2)^2 = \begin{pmatrix} 4 & 4 & 1 \\ 0 & 4 & 4 \\ 0 & 0 & 4 \end{pmatrix}

Why It Matters

Jordan blocks appear in solving systems of linear differential equations, where non-diagonalizable coefficient matrices produce solutions involving polynomial-times-exponential terms. They are also central to understanding matrix functions (exponentials, logarithms) in advanced linear algebra and control theory.

Common Mistakes

Mistake: Placing ones on the subdiagonal (below the main diagonal) instead of the superdiagonal.

Correction: The standard convention puts the ones on the superdiagonal (one position above the diagonal). Some textbooks use the transpose convention, but most courses and references use superdiagonal ones.