Generalized Eigenvector¶

Definition¶

A generalized eigenvector of a square matrix $\boldsymbol{A}$ for eigenvalue $\lambda$ is a nonzero vector $\boldsymbol{v}$ such that $(\boldsymbol{A}-\lambda \boldsymbol{I})^k \boldsymbol{v}=\boldsymbol{0}$ for some positive integer $k$.

Interpretation¶

The identity matrix makes $\boldsymbol{A}-\lambda\boldsymbol{I}$ subtract $\lambda\boldsymbol{v}$ from the image of $\boldsymbol{v}$. The defining equation places $\boldsymbol{v}$ in the kernel of a power of this shifted matrix, even when it is not in the kernel of the first power.

Generalized eigenvectors replace missing ordinary eigenvectors for defective matrices.

A nonzero vector $\boldsymbol{v}$ is a generalized eigenvector for $\lambda$ if $(\boldsymbol{A}-\lambda \boldsymbol{I})^k\boldsymbol{v}=\boldsymbol{0}$ for some positive integer $k$. The vector $\boldsymbol{v}_1$ is an ordinary eigenvector, while the chain $\boldsymbol{v}_3\to \boldsymbol{v}_2\to \boldsymbol{v}_1\to\boldsymbol{0}$ supplies the further vectors that a defective matrix needs, besides its eigenvectors, to form a basis of the whole space.

Key Results¶

• Every ordinary eigenvector is a generalized eigenvector with $k=1$$1$.
• Generalized eigenvectors are used to construct Jordan normal form.

Examples¶

• If $\boldsymbol{v}\neq\boldsymbol{0}$ and $(\boldsymbol{A}-\lambda \boldsymbol{I})\boldsymbol{v}=\boldsymbol{0}$$0$, then $\boldsymbol{v}$ is an eigenvector and a generalized eigenvector.
• If $(\boldsymbol{A}-\lambda \boldsymbol{I})^2\boldsymbol{v}=\boldsymbol{0}$ but $(\boldsymbol{A}-\lambda \boldsymbol{I})\boldsymbol{v}\neq\boldsymbol{0}$$0$, then $\boldsymbol{v}$ is generalized but not ordinary.

Prerequisites¶

• Eigenvalue and Eigenvector
• Matrix
• Identity Matrix
• Kernel

Related Concepts¶

• Invariant Subspace

Sources¶

• Math II: Linear Algebra · Week 6

Aliases: generalised eigenvector, root vector Symbols: lambda

Last updated: 27 October 2026 11:52