Jordan Blocks and Chains¶

Definition¶

A Jordan block $\boldsymbol{J}(\lambda)$ is a square matrix with eigenvalue $\lambda$ on the main diagonal, $1$s immediately above the diagonal, and $0$s elsewhere.

A Jordan chain for eigenvalue $\lambda$ of a square matrix $\boldsymbol{A}$ is a sequence of generalized eigenvectors connected by

ending with an ordinary eigenvector and then $\boldsymbol{0}$.

Interpretation¶

Jordan blocks are the building pieces of Jordan normal form. A Jordan block of size at least $2$ is the simplest kind of matrix that is not diagonalizable: it has a single eigenvalue but only one independent eigenvector. Jordan chains explain where those blocks come from and how the generalized eigenvectors are ordered.

The figure shows $\boldsymbol{J}(\lambda)$ of size $3$ next to a Jordan chain $\boldsymbol{v}_3\to\boldsymbol{v}_2\to\boldsymbol{v}_1$ under $\boldsymbol{A}-\lambda\boldsymbol{I}$$\lambda \mathbf{I}$, which then sends the eigenvector $\boldsymbol{v}_1$ to $\boldsymbol{0}$. Only $\boldsymbol{v}_1$ is an eigenvector; the others satisfy $\boldsymbol{A}\boldsymbol{v}_2=\lambda\boldsymbol{v}_2+\boldsymbol{v}_1$ and $\boldsymbol{A}\boldsymbol{v}_3=\lambda\boldsymbol{v}_3+\boldsymbol{v}_2$${\mathbf{v}}_2$. In the basis $(\boldsymbol{v}_1,\boldsymbol{v}_2,\boldsymbol{v}_3)$ the columns of the matrix of $\boldsymbol{A}$ are the coordinates of these images, $(\lambda,0,0)$, $(1,\lambda,0)$ and $(0,1,\lambda)$, which is exactly $\boldsymbol{J}(\lambda)$. Each $1$ above the diagonal records one step along the chain.

Subtracting $\lambda\boldsymbol{I}$ leaves only the $1$s above the diagonal, and $\boldsymbol{J}(\lambda)-\lambda\boldsymbol{I}$ shifts: it sends the third basis vector to the second, the second to the first and the first to $\boldsymbol{0}$. Hence $(\boldsymbol{J}(\lambda)-\lambda\boldsymbol{I})^3=\boldsymbol{0}$ for a block of size $3$. Its kernel is one-dimensional, so a block of size $k$ has algebraic multiplicity $k$ but geometric multiplicity $1$. A block of size $1$ is just $(\lambda)$, with an ordinary eigenvector.

The order of the chain matters. Listing the basis as $(\boldsymbol{v}_3,\boldsymbol{v}_2,\boldsymbol{v}_1)$ instead puts the $1$s below the diagonal; the convention here starts the basis with the eigenvector.

Key Results¶

• A Jordan block is upper triangular and has one eigenvalue on its diagonal.
• The length of a Jordan chain corresponds to the size of the associated Jordan block.

Examples¶

• $\begin{pmatrix}4&1\\0&4\end{pmatrix}$ is a $2\times2$ Jordan block.
• If $\boldsymbol{v}_1\neq\boldsymbol{0}$$0$, $(\boldsymbol{A}-\lambda \boldsymbol{I})\boldsymbol{v}_2=\boldsymbol{v}_1$ and $(\boldsymbol{A}-\lambda \boldsymbol{I})\boldsymbol{v}_1=\boldsymbol{0}$$0$, then $(\boldsymbol{v}_1,\boldsymbol{v}_2)$ is a Jordan chain of length $2$.

Prerequisites¶

• eigenvalue
• generalized eigenvectors

Related Concepts¶

• Algebraic and Geometric Multiplicity
• Diagonalizable Matrix
• Kernel

Sources¶

• Math II: Linear Algebra · Week 6

Aliases: Jordan blocks, Jordan chains Keywords: Jordan block, Jordan cell, Jordan chain Symbols: lambda

Last updated: 27 October 2026 11:52