Jordan normal form
Undergraduate (complex matrices)
A Jordan block is a square matrix with a number $\lambda$ on its diagonal, $1$ on the diagonal just above it, and $0$ elsewhere, such as
\[\begin{bmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{bmatrix}.\]Consider a square complex matrix $A$. A Jordan normal form of $A$ is a factorization
\[A = P J P^{-1},\]where $P$ is invertible and $J$ is block diagonal with Jordan blocks on its diagonal.
Examples
Worked examples
- The matrix \(\begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\) : $J = D$, as in the eigendecomposition Undergraduate
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The matrix \(\begin{bmatrix} 3 & 1 \\ -1 & 1 \end{bmatrix}\)
: \(P = \begin{bmatrix} 1 & 1 \\ -1 & 0 \end{bmatrix}\), \(J = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}\)
Undergraduate
The matrix \(A = \begin{bmatrix} 3 & 1 \\ -1 & 1 \end{bmatrix}\) has the eigenvalue $2$ twice, but its eigenvectors are only the multiples of $(1, -1)$, so it is not diagonalizable, and $J$ is a single $2 \times 2$ Jordan block. The columns of $P$ are the eigenvector $v_1 = (1, -1)$ and a vector $v_2 = (1, 0)$ with $(A - 2I) v_2 = v_1$. To check, \(A P = P J = \begin{bmatrix} 2 & 3 \\ -2 & -1 \end{bmatrix}\).
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The matrix \(\begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}\)
: $J = \Lambda$, as in the eigendecomposition
Undergraduate
The matrix \(A = \begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}\) is symmetric positive definite, so it has every decomposition in the table, including a Cholesky decomposition. Its eigendecomposition, Schur decomposition, Jordan normal form, and singular value decomposition all coincide, and its polar and Toeplitz decompositions are trivial.
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The matrix \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\)
: $P = I$, $J = A$
Undergraduate
The matrix \(A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) is already a Jordan block, so it is its own Jordan normal form.
- The matrix \(A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\) : $J = \Lambda$, as in the eigendecomposition Undergraduate
Properties
- Every square complex matrix $A$ has a Jordan normal form, and $J$ is unique up to the order of the blocks. The numbers $\lambda$ in the blocks are the eigenvalues of $A$.
- The matrix $A$ is diagonalizable exactly when every Jordan block in $J$ is $1 \times 1$, and then the Jordan normal form is an eigendecomposition.
- Splitting $J$ into its diagonal and the $1$s above the diagonal gives the Jordan–Chevalley decomposition of $A$.
- The Jordan normal form is not continuous in $A$: an arbitrarily small change in $A$ can change the sizes of the blocks. For this reason, it is not computed numerically; the Schur decomposition is used instead.
Notes
Strictly, the Jordan normal form is the matrix $J$; the factorization $A = P J P^{-1}$ is sometimes called the Jordan decomposition. That name is also used for the Jordan–Chevalley decomposition, so this page avoids it.
Distinctions involving this term
- Types of matrix decompositions: A matrix decomposition writes a matrix in terms of simpler matrices. Most are products: LU, PLU, and Cholesky decompositions use triangular factors, QR and singular value decompositions use orthogonal factors, and eigendecompositions, Schur decompositions, and Jordan normal forms are similarities $A = P M P^{-1}$ to a simpler matrix $M$ (diagonal, triangular, or made of Jordan blocks). In contrast, the Jordan–Chevalley decomposition is a sum of commuting diagonalizable and nilpotent parts, and the Toeplitz decomposition is a sum $H + K$ of a symmetric matrix $H$ and a skew-symmetric matrix $K$.