Jordan Normal Form — Mathematics Knowledge System

Overview

The Jordan normal form (JNF) is the 'most diagonal' form that any square matrix over an algebraically closed field can be put into by a change of basis. When a matrix is not diagonalizable — because it has eigenvalues with algebraic multiplicity greater than geometric multiplicity — the JNF reveals its structure through Jordan blocks. Every square complex matrix has a unique Jordan normal form (up to ordering of blocks).

Intuition

Diagonalization works perfectly when there are enough independent eigenvectors. When it fails, Jordan form is the next best thing: it groups eigenvalues together in blocks, each block being a scalar on the diagonal plus 1s just above (superdiagonal). A Jordan block of size k for eigenvalue λ represents a 'repeated' eigenvalue that is defective — it only has one eigenvector but needs a chain of generalized eigenvectors to fill out the space.

Formal Definition

Definition

Every matrix A ∈ M_n(C) is similar to a Jordan matrix J = P^{-1} A P where J consists of Jordan blocks:

J_k(\lambda) = \begin{pmatrix}\lambda & 1 & & \\ & \lambda & 1 & \\ & & \ddots & 1 \\ & & & \lambda\end{pmatrix} \in M_k(\mathbb{C})

Jordan block of size k for eigenvalue λ

J = \begin{pmatrix} J_{k_1}(\lambda_1) & & \\ & \ddots & \\ & & J_{k_r}(\lambda_r)\end{pmatrix}

Block diagonal matrix of Jordan blocks; A = P J P^{-1}.

Jordan normal form

\text{Number of Jordan blocks for } \lambda = \dim(\ker(A-\lambda I)) = \text{geometric multiplicity}

Block count

Notation

Notation	Meaning
$J_k(\lambda)$	Jordan block of size k for eigenvalue λ
$J$	Jordan normal form of A
$P$	Change-of-basis matrix of generalized eigenvectors

Properties

Uniqueness

\text{The JNF is unique up to ordering of blocks.}

Algebraic multiplicity

\text{Size of the largest Jordan block for } \lambda = \text{index (nilpotency of } A-\lambda I \text{ restricted to generalized eigenspace).}

Diagonalizable case

A \text{ is diagonalizable iff all Jordan blocks are 1×1, iff algebraic = geometric multiplicity for every eigenvalue.}

Powers of Jordan block

J_k(\lambda)^n = \sum_{j=0}^{k-1} \binom{n}{j}\lambda^{n-j} E_{1,j+1} \text{ (binomial-coefficient structure).}

Worked Examples

1
Find eigenvalues: characteristic polynomial is (λ-4)^2 = 0, so λ=4 with algebraic multiplicity 2.
2
Find geometric multiplicity: dim(ker(A-4I)).
$A - 4I = \begin{pmatrix}0&1\\0&0\end{pmatrix}, \quad \ker(A-4I) = \operatorname{span}\{(1,0)\}, \quad \text{geometric multiplicity} = 1$
3
One Jordan block of size 2 (alg mult=2, geo mult=1).
$J = J_2(4) = \begin{pmatrix}4&1\\0&4\end{pmatrix}$
4
A = J here (A is already in Jordan form). Change of basis P = I.

✓ Answer

J = [[4,1],[0,4]] — a single 2×2 Jordan block for λ=4. A is not diagonalizable.

Practice Problems

Mediumfree response

Without computing P, determine the Jordan normal form of a matrix with characteristic polynomial (λ-1)^2(λ-2) and minimal polynomial (λ-1)^2(λ-2).

Mediumfree response

How does the Jordan normal form relate to the matrix exponential e^{At}? Give the formula for e^{J_k(λ)t}.

Common Mistakes

Common Mistake

Every matrix with repeated eigenvalues is not diagonalizable.

A matrix can have repeated eigenvalues and still be diagonalizable — as long as each repeated eigenvalue has geometric multiplicity equal to its algebraic multiplicity (i.e., enough independent eigenvectors).

Quiz

Summary

Jordan normal form: A = P J P^{-1} where J is block diagonal with Jordan blocks J_k(λ).
A Jordan block J_k(λ) has λ on the diagonal and 1s on the superdiagonal.
Number of blocks for λ = geometric multiplicity; size of largest block = index of λ.
JNF is unique up to ordering of blocks and exists for all complex square matrices.
A is diagonalizable iff all Jordan blocks have size 1 (all algebraic = geometric multiplicities).

References

WebsiteWikipedia — Jordan normal form