Generalized Eigenspaces and Invariant Subspaces

From an Eigenvector to a Generalized Eigenspace

For an eigenvalue \(\lambda\), an eigenvector is a nonzero vector \(\mathbf{v}\) satisfying \((\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}\). A generalized eigenvector may not satisfy this equation after one application of \(\mathbf{A}-\lambda\mathbf{I}\), but repeated applications eventually send it to zero. This difference matters: a whole collection of such vectors can form a subspace that is preserved by \(\mathbf{A}\), even when the vectors are not ordinary eigenvectors.

Let \(\mathbf{A}\in\mathbb{F}^{n\times n}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\lambda\in\mathbb{F}\). For a positive integer \(m\), the power \((\mathbf{A}-\lambda\mathbf{I})^m\) means the product of \(m\) copies of \(\mathbf{A}-\lambda\mathbf{I}\). The kernel of a matrix is the set of vectors it sends to zero.

A subspace \(W\) is invariant under \(\mathbf{A}\) if \(\mathbf{A}\mathbf{x}\in W\) for every \(\mathbf{x}\in W\). If this condition holds, the restriction \(\mathbf{A}|_W\) is the linear map from \(W\) to \(W\) defined by \(\mathbf{x}\mapsto\mathbf{A}\mathbf{x}\). An operator \(N:W\to W\) is nilpotent if \(N^k\) is the zero map for some positive integer \(k\).

The Invariance and Nilpotence Theorem

Proof. Write \(\mathbf{B}=\mathbf{A}-\lambda\mathbf{I}\). The matrices \(\mathbf{A}\) and \(\mathbf{B}\) commute, since \(\mathbf{A}\mathbf{B}=\mathbf{A}^2-\lambda\mathbf{A}=\mathbf{B}\mathbf{A}\). Therefore \(\mathbf{B}^m\mathbf{A}=\mathbf{A}\mathbf{B}^m\): this follows by repeatedly moving \(\mathbf{A}\) past one factor \(\mathbf{B}\). If \(\mathbf{x}\in\mathcal{G}_{\lambda,m}\), then \(\mathbf{B}^m\mathbf{x}=\mathbf{0}\), so \(\mathbf{B}^m(\mathbf{A}\mathbf{x})=\mathbf{A}\mathbf{B}^m\mathbf{x}=\mathbf{A}\mathbf{0}=\mathbf{0}\). Thus \(\mathbf{A}\mathbf{x}\in\mathcal{G}_{\lambda,m}\), proving invariance. The same reasoning shows \(\mathbf{B}\mathbf{x}\in\mathcal{G}_{\lambda,m}\), so the restriction \(N=\mathbf{B}|_{\mathcal{G}_{\lambda,m}}\) maps the subspace to itself. On any \(\mathbf{x}\) in that subspace, \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}+\mathbf{B}\mathbf{x}\); hence \(\mathbf{A}|_{\mathcal{G}_{\lambda,m}}=\lambda\operatorname{id}_{\mathcal{G}_{\lambda,m}}+N\). Finally, \(N^m\mathbf{x}=\mathbf{B}^m\mathbf{x}=\mathbf{0}\) for every \(\mathbf{x}\in\mathcal{G}_{\lambda,m}\), so \(N\) is nilpotent. These conclusions also hold if the subspace is just \(\{\mathbf{0}\}\). \(\square\)

The scalar part \(\lambda\operatorname{id}\) describes the eigenvalue action, while \(N\) records the additional shift within the generalized eigenspace. The theorem does not assert that \(N\) is zero: it may send a generalized eigenvector to a nonzero eigenvector. What it does assert is that repeating this shift enough times must give zero. Also, the theorem is about a restricted map on a subspace; its matrix should be written using a basis for that subspace, not assumed to be the full matrix \(\mathbf{A}\).

Proof. If \(\mathbf{v}\in E_\lambda\), then \(\mathbf{B}\mathbf{v}=\mathbf{0}\), where \(\mathbf{B}=\mathbf{A}-\lambda\mathbf{I}\). Applying \(\mathbf{B}\) another \(m-1\) times still gives \(\mathbf{B}^m\mathbf{v}=\mathbf{0}\), so \(\mathbf{v}\in\ker\mathbf{B}^m=\mathcal{G}_{\lambda,m}\). The zero vector is included as well, since every linear map sends it to zero. \(\square\)

Computing a Proper Generalized Eigenspace

When the Generalized Eigenspace Is the Whole Space

The Exponent Determines Which Kernel You Get

The theorem applies for every \(m\geq1\), but different exponents can give different subspaces. The kernel powers form an increasing sequence: \(\ker\mathbf{B}\subseteq\ker\mathbf{B}^2\subseteq\cdots\). Indeed, if \(\mathbf{B}^m\mathbf{x}=\mathbf{0}\), then \(\mathbf{B}^{m+1}\mathbf{x}=\mathbf{B}\mathbf{0}=\mathbf{0}\). Thus choosing an exponent too small may capture only the ordinary eigenvectors, not all the generalized eigenvectors for that eigenvalue.

Common Mistake: Using Only the First Power

A frequent error is to decide that a vector is not in the generalized eigenspace because \((\mathbf{A}-\lambda\mathbf{I})\mathbf{x}\ne\mathbf{0}\). That test decides membership in the eigenspace, not necessarily membership in a kernel of a higher power. In the last example, the incorrect calculation is “\(\mathbf{B}\mathbf{e}_2=\mathbf{e}_1\ne\mathbf{0}\), so \(\mathbf{e}_2\notin\mathcal{G}_{4,2}\).” The conclusion does not follow: the definition requires applying \(\mathbf{B}\) twice. In fact, \(\mathbf{B}^2\mathbf{e}_2=\mathbf{B}\mathbf{e}_1=\mathbf{0}\), so \(\mathbf{e}_2\in\mathcal{G}_{4,2}\). The correct rule is to check the specific power in the definition.

Another point to track is the space on which an operator acts. The matrix \(\mathbf{A}\) acts on \(\mathbb{F}^n\), but \(\mathbf{A}|_{\mathcal{G}_{\lambda,m}}\) has domain and codomain \(\mathcal{G}_{\lambda,m}\). To display its matrix, choose a basis of that subspace and compute the image of each basis vector. The first worked example shows why this matters: the restricted matrix is \(2\times2\), although the original matrix is \(3\times3\).

The decomposition of the restricted map is useful because it separates two kinds of action. Multiplication by \(\lambda\) is the scalar behavior shared across the generalized eigenspace; the nilpotent part describes how generalized eigenvectors link to one another. The theorem justifies studying that smaller invariant subspace on its own. It does not require the subspace to be the whole ambient space, and it does not say that every vector in it is an eigenvector.