Generalized Eigenvectors

Not every square matrix is diagonalizable. Yet every operator on a finite-dimensional complex vector space is still built entirely from a handful of scalars: its eigenvalues. The gap between these two facts is filled by generalized eigenvectors. Consider the matrix with a single eigenvalue λ\lambda but a Jordan block structure. It has "too few" ordinary eigenvectors to span the space. So we ask a weaker question. Instead of vectors killed outright by T−λIT - \lambda I, what about vectors killed by some power (T−λI)k(T-\lambda I)^k? This small relaxation turns out to be exactly enough. The idea matters because it rescues the entire diagonalization program from operators that resist it. Where an eigenvector satisfies Tv=λvTv = \lambda v, a generalized eigenvector satisfies this only "eventually." This happens after repeated application of T−λIT - \lambda I. Tracking how these vectors form an ascending chain of null spaces reveals a stabilization phenomenon. This phenomenon is tied intimately to the Minimal Polynomial and to the algebraic multiplicity of λ\lambda from the Characteristic Polynomial. The big picture is that generalized eigenvectors let us decompose any finite-dimensional complex vector space. It becomes a direct sum of subspaces, one per eigenvalue. Each is invariant under TT and each has dimension equal to the algebraic multiplicity. On each such subspace, TT restricted looks like λI\lambda I plus a nilpotent piece. This is a dramatic simplification even when true diagonalization is impossible. This is the bridge connecting Geometric vs Algebraic Multiplicity to a genuine structural decomposition. This article develops the ascending chain of null spaces ker⁡(T−λI)⊆ker⁡(T−λI)2⊆⋯\ker(T-\lambda I) \subseteq \ker(T - \lambda I)^2 \subseteq \cdots. It proves that this chain must stabilize. It identifies the stable value with the generalized eigenspace. It also proves the key dimension theorem that makes the whole theory work. It sets up, but does not complete, the full Generalized Eigenspace Decomposition and the Jordan Canonical Form.

Key takeaways
  • A generalized eigenvector for eigenvalue λ\lambda is a nonzero vv with (T−λI)kv=0(T-\lambda I)^k v = 0 for some positive integer kk.
  • The null spaces ker⁡(T−λI)j\ker(T-\lambda I)^j form an ascending chain that strictly increases until it stabilizes, then stays fixed forever.
  • Stabilization happens by step j=n=dim⁡Vj = n = \dim V at the latest; the stable subspace is the generalized eigenspace G(λ,T)G(\lambda, T).
  • dim⁡G(λ,T)\dim G(\lambda,T) equals the algebraic multiplicity of λ\lambda in the characteristic polynomial — this is the central dimension theorem.
  • Ordinary eigenspaces can be too small to span VV; generalized eigenspaces are never too small — together they span VV (over C\mathbb{C}).
  • Key idea: replacing "killed by T−λIT-\lambda I" with "eventually killed by some power of T−λIT - \lambda I" recovers enough vectors to reconstruct the whole space.

Where this lives today

Generalized eigenspaces are the working engine behind solving linear systems of ordinary differential equations x˙=Ax\dot{x} = Ax when AA is not diagonalizable: the general solution involves terms like tkeλtvt^k e^{\lambda t}v precisely when vv is a generalized eigenvector of rank k+1k+1, and these polynomial-times-exponential terms are unavoidable exactly when Jordan blocks of size greater than 11 appear. In control theory and stability analysis of dynamical systems, the size of Jordan blocks at eigenvalues on the stability boundary determines whether solutions grow polynomially, a distinction generalized eigenvectors make precise. In numerical linear algebra, the sensitivity of eigenvalue computations to perturbation is governed by whether nontrivial generalized eigenvectors exist (defective matrices are notoriously ill-conditioned for eigenvalue algorithms). In quantum mechanics and the theory of Jordan blocks for non-Hermitian (non-normal) operators, generalized eigenvectors describe resonance and decay phenomena that ordinary eigenvectors cannot capture. Finally, the entire theory is the direct staircase to the Jordan Canonical Form, which packages generalized eigenspaces and their internal nilpotent structure into the cleanest possible matrix representative of a similarity class.

:::quiz[Why does the counterexample with the backward-shift operator DD on C[x]\mathbb{C}[x] fail to produce a finite-dimensional generalized eigenspace?]

  • Because 00 is not an eigenvalue of DD
  • Because C[x]\mathbb{C}[x] is infinite-dimensional, so the pigeonhole bound on chain length has no finite nn to invoke ✓
  • Because DD has no eigenvectors at all
  • Because DD is not linear

Ask the tutor: How does the direct sum of all the generalized eigenspaces combine to give the full Jordan Canonical Form of an operator?

DefinitionGeneralized Eigenvector and Generalized Eigenspace

Let VV be a finite-dimensional vector space over a field F\mathbb{F} (take F=C\mathbb{F} = \mathbb{C} for the deepest results), and let T:V→VT: V \to V be a linear operator. Let λ∈F\lambda \in \mathbb{F} be an eigenvalue of TT. A vector v∈Vv \in V is a generalized eigenvector of TT corresponding to λ\lambda if v≠0v \neq 0 and

(T−λI)kv=0(T - \lambda I)^k v = 0

for some positive integer kk. The generalized eigenspace of TT corresponding to λ\lambda, denoted G(λ,T)G(\lambda, T), is the set of all generalized eigenvectors corresponding to λ\lambda together with the zero vector:

G(λ,T)={ v∈V:(T−λI)kv=0 for some k∈Z+ }=⋃k=1∞ker⁡(T−λI)k.G(\lambda, T) = \{\, v \in V : (T-\lambda I)^k v = 0 \text{ for some } k \in \mathbb{Z}^{+} \,\} = \bigcup_{k=1}^{\infty} \ker(T-\lambda I)^k.

This is a linear subspace of VV, since each ker⁡(T−λI)k\ker(T - \lambda I)^k is a subspace and they are nested increasingly (proved below), so their union is a subspace.

Intuition

Picture the chain of subspaces ker⁡(T−λI)⊆ker⁡(T−λI)2⊆ker⁡(T−λI)3⊆⋯\ker(T-\lambda I) \subseteq \ker(T-\lambda I)^2 \subseteq \ker(T-\lambda I)^3 \subseteq \cdots. The first term is the ordinary eigenspace: vectors that TT sends exactly to λ\lambda times themselves. The second term contains vectors that are not eigenvectors, but which T−λIT - \lambda I maps into the eigenspace. Applying T−λIT - \lambda I repeatedly to such a vector eventually annihilates it, just more slowly. Think of T−λIT - \lambda I as a "clock" that ticks a vector toward zero; ordinary eigenvectors die in one tick, generalized eigenvectors die in finitely many ticks. A single 2×22\times 2 Jordan block (λ10λ)\begin{pmatrix}\lambda & 1\\ 0 & \lambda\end{pmatrix} makes this concrete. The vector e1e_1 is an honest eigenvector: (T−λI)e1=0(T-\lambda I)e_1 = 0. The vector e2e_2 is not an eigenvector — (T−λI)e2=e1≠0(T - \lambda I)e_2 = e_1 \ne 0 — but (T−λI)2e2=(T−λI)e1=0(T-\lambda I)^2 e_2 = (T-\lambda I)e_1 = 0. So e2e_2 needs two ticks to die; it is a generalized eigenvector of "rank 2." The eigenspace ker⁡(T−λI)\ker(T-\lambda I) is only one-dimensional (spanned by e1e_1), too small to span C2\mathbb{C}^2, but the generalized eigenspace ker⁡(T−λI)2\ker(T-\lambda I)^2 is all of C2\mathbb{C}^2.

Figure
The ascending chain of null spaces for a 2×2 Jordan block stabilizes after one strict jump
Chain: ker(T-λI)^j inside V = ℂ², eigenvalue λ V = ℂ² (dim 2) span(e₁) = ker(T−λI)^1, dim 1 ker(T−λI)^2 = ℂ², dim 2 (stabilized) strict increase dim ker(T−λI)^1 = 1 dim ker(T−λI)^2 = 2 dim ker(T−λI)^3 = 2 (equal, stable) Stabilizes at j = 2 = dim V

worked-examples-preface

Before the theorem, note the two facts the picture suggests: the chain of null spaces cannot increase forever inside a finite-dimensional space, and once two consecutive terms of the chain are equal, every later term equals them too. Both facts are proved rigorously below as part of establishing that G(λ,T)G(\lambda,T) is achieved at a finite, bounded stage.

LemmaStabilization of the null-space chain

Let VV be finite-dimensional with dim⁡V=n\dim V = n, let S:V→VS: V \to V be linear, and consider the chain

{0}=ker⁡S0⊆ker⁡S1⊆ker⁡S2⊆⋯ .\{0\} = \ker S^0 \subseteq \ker S^1 \subseteq \ker S^2 \subseteq \cdots.

Then:

(a) ker⁡Sj⊆ker⁡Sj+1\ker S^j \subseteq \ker S^{j+1} for every j≥0j \ge 0.

(b) If ker⁡Sm=ker⁡Sm+1\ker S^m = \ker S^{m+1} for some m≥0m \ge 0, then ker⁡Sj=ker⁡Sm\ker S^j = \ker S^m for all j≥mj \ge m.

(c) There exists m≤nm \le n such that ker⁡Sm=ker⁡Sm+1\ker S^m = \ker S^{m+1}, and consequently ker⁡Sj=ker⁡Sm\ker S^j = \ker S^m for all j≥mj \ge m.

Proof

(a) If Sjv=0S^j v = 0 then Sj+1v=S(Sjv)=S(0)=0S^{j+1} v = S(S^j v) = S(0) = 0. So ker⁡Sj⊆ker⁡Sj+1\ker S^j \subseteq \ker S^{j+1}.

(b) We show by induction on j≥mj \ge m that ker⁡Sj=ker⁡Sj+1\ker S^j = \ker S^{j+1}; the base case j=mj = m is the hypothesis. Assume ker⁡Sj=ker⁡Sj+1\ker S^j = \ker S^{j+1} for some j≥mj \ge m; we show ker⁡Sj+1=ker⁡Sj+2\ker S^{j+1} = \ker S^{j+2}. By part (a), ker⁡Sj+1⊆ker⁡Sj+2\ker S^{j+1} \subseteq \ker S^{j+2}, so we only need the reverse inclusion. Let v∈ker⁡Sj+2v \in \ker S^{j+2}, so Sj+2v=0S^{j+2}v = 0, i.e. Sj+1(Sv)=0S^{j+1}(Sv) = 0, i.e. Sv∈ker⁡Sj+1=ker⁡SjSv \in \ker S^{j+1} = \ker S^j (inductive hypothesis). Then Sj(Sv)=0S^j(Sv) = 0, i.e. Sj+1v=0S^{j+1} v = 0, i.e. v∈ker⁡Sj+1v \in \ker S^{j+1}. This proves ker⁡Sj+2⊆ker⁡Sj+1\ker S^{j+2} \subseteq \ker S^{j+1}, hence equality, completing the induction. Thus ker⁡Sj=ker⁡Sm\ker S^j = \ker S^m for all j≥mj \ge m.

(c) Suppose, for contradiction, that no such m≤nm \le n exists, i.e. ker⁡S0,ker⁡S1,…,ker⁡Sn\ker S^0, \ker S^1, \ldots, \ker S^n are pairwise distinct as consecutive pairs — that is, ker⁡Sj⊊ker⁡Sj+1\ker S^{j} \subsetneq \ker S^{j+1} (strict inclusion) for every j=0,1,…,n−1j = 0, 1, \ldots, n-1. A strict inclusion of subspaces of a finite-dimensional space forces a strict increase in dimension: if W⊊W′W \subsetneq W' are subspaces of a finite-dimensional space then dim⁡W<dim⁡W′\dim W < \dim W', since a basis of WW can be extended to a basis of W′W' using at least one further vector (this extension exists because W≠W′W \ne W' means some vector of W′W' is not in WW, hence not in the span of the basis of WW, so it can be adjoined to a linearly independent set). Applying this at each step,

0=dim⁡ker⁡S0<dim⁡ker⁡S1<dim⁡ker⁡S2<⋯<dim⁡ker⁡Sn.0 = \dim \ker S^0 < \dim \ker S^1 < \dim \ker S^2 < \cdots < \dim \ker S^n.

This is a strictly increasing sequence of n+1n+1 nonnegative integers, so dim⁡ker⁡Sn≥n\dim \ker S^n \ge n. But ker⁡Sn⊆V\ker S^n \subseteq V and dim⁡V=n\dim V = n, forcing dim⁡ker⁡Sn=n\dim \ker S^n = n, i.e. ker⁡Sn=V\ker S^n = V. Then also dim⁡ker⁡Sn+1≤n=dim⁡ker⁡Sn≤dim⁡ker⁡Sn+1\dim \ker S^{n+1} \le n = \dim \ker S^n \le \dim \ker S^{n+1} (the last inequality from part (a), which gives ker⁡Sn⊆ker⁡Sn+1⊆V\ker S^n \subseteq \ker S^{n+1} \subseteq V), forcing ker⁡Sn=ker⁡Sn+1\ker S^n = \ker S^{n+1}. This contradicts the assumption that all inclusions up through index n−1n-1 are strict only if we needed j=nj=n to be strict too — but we assumed strictness for j=0,…,n−1j = 0,\dots,n-1 only, so no contradiction arises there; instead we have directly exhibited m=nm = n with ker⁡Sn=ker⁡Sn+1\ker S^n = \ker S^{n+1}, which is exactly the existence claim with m≤nm \le n. Combined with part (b), ker⁡Sj=ker⁡Sn\ker S^j = \ker S^n for all j≥nj \ge n. ■\blacksquare

∎
TheoremGeneralized Eigenspace = Eventual Stable Null Space, and the Dimension Theorem

Let VV be a vector space over C\mathbb{C} with dim⁡V=n<∞\dim V = n < \infty, let T:V→VT: V \to V be linear, and let λ\lambda be an eigenvalue of TT with algebraic multiplicity aλa_\lambda (the multiplicity of λ\lambda as a root of the characteristic polynomial det⁡(xI−T)\det(xI - T)). Then:

(a) ker⁡(T−λI)n=ker⁡(T−λI)n+1=ker⁡(T−λI)n+2=⋯\ker(T-\lambda I)^n = \ker(T - \lambda I)^{n+1} = \ker(T-\lambda I)^{n+2} = \cdots, and this common subspace equals G(λ,T)G(\lambda, T); that is,

G(λ,T)=ker⁡(T−λI)n.G(\lambda,T) = \ker(T-\lambda I)^n.

(b) G(λ,T)G(\lambda, T) is invariant under TT: T(G(λ,T))⊆G(λ,T)T(G(\lambda,T)) \subseteq G(\lambda,T).

(c) dim⁡G(λ,T)=aλ\dim G(\lambda, T) = a_\lambda.

Proof

We use the technique of decomposing VV as an internal direct sum of TT-invariant complementary subspaces, together with basic facts about the characteristic polynomial under block-triangular matrix representations.

(a) Apply the Stabilization Lemma above with S=T−λIS = T - \lambda I: the chain ker⁡S0⊆ker⁡S1⊆⋯\ker S^0 \subseteq \ker S^1 \subseteq \cdots stabilizes at some m≤nm \le n, and stays fixed at ker⁡Sm\ker S^m for all j≥mj \ge m. In particular it has stabilized by index nn, so ker⁡Sn=ker⁡Sn+1=⋯\ker S^n = \ker S^{n+1} = \cdots. Since G(λ,T)=⋃k≥1ker⁡SkG(\lambda, T) = \bigcup_{k \ge 1} \ker S^k and the chain is eventually constant equal to ker⁡Sn\ker S^n for all indices ≥n\ge n (and the terms before nn are all contained in ker⁡Sn\ker S^n by the ascending property), the union equals ker⁡Sn\ker S^n exactly:

G(λ,T)=⋃k=1∞ker⁡Sk=ker⁡Sn.G(\lambda,T) = \bigcup_{k=1}^{\infty}\ker S^k = \ker S^n.

(b) Let v∈G(λ,T)=ker⁡Snv \in G(\lambda,T) = \ker S^n, so Snv=0S^n v = 0 where S=T−λIS = T - \lambda I. Since TT and SS commute (as S=T−λIS = T - \lambda I is a polynomial in TT), we have

Sn(Tv)=T(Snv)=T(0)=0,S^n(Tv) = T(S^n v) = T(0) = 0,

so Tv∈ker⁡Sn=G(λ,T)Tv \in \ker S^n = G(\lambda,T). Hence G(λ,T)G(\lambda,T) is TT-invariant.

(c) This is the substantive claim. Let g=dim⁡G(λ,T)g = \dim G(\lambda,T). We show g=aλg = a_\lambda by exhibiting a basis of VV adapted to G(λ,T)G(\lambda,T) and analyzing the resulting block form of the matrix of TT. Step 1: Construct a TT-invariant complement. We claim there is a subspace W⊆VW \subseteq V with T(W)⊆WT(W) \subseteq W and V=G(λ,T)⊕WV = G(\lambda,T) \oplus W. To see this, first note G(λ,T)∩ker⁡TG(\lambda,T) \cap \ker T-type intersection arguments are unnecessary; instead argue directly via the operator Sn=(T−λI)nS^n = (T-\lambda I)^n. Consider U=ran⁡(Sn)=(T−λI)n(V)U = \operatorname{ran}(S^n) = (T-\lambda I)^n(V), the image of SnS^n. By the rank-nullity theorem applied to the linear map Sn:V→VS^n: V \to V,

dim⁡ker⁡Sn+dim⁡ran⁡Sn=n,i.e.g+dim⁡U=n.\dim \ker S^n + \dim \operatorname{ran} S^n = n, \quad\text{i.e.}\quad g + \dim U = n.

We claim V=G(λ,T)⊕UV = G(\lambda,T) \oplus U, i.e. ker⁡Sn⊕ran⁡Sn=V\ker S^n \oplus \operatorname{ran} S^n = V. Since the dimensions already sum to n=dim⁡Vn = \dim V, it suffices to show ker⁡Sn∩ran⁡Sn={0}\ker S^n \cap \operatorname{ran} S^n = \{0\}. Let v∈ker⁡Sn∩ran⁡Snv \in \ker S^n \cap \operatorname{ran} S^n. Write v=Snuv = S^n u for some u∈Vu \in V. Since Snv=0S^n v = 0, we get S2nu=0S^{2n} u = 0, so u∈ker⁡S2nu \in \ker S^{2n}. But by part (a) (stabilization at index nn), ker⁡S2n=ker⁡Sn\ker S^{2n} = \ker S^n, so u∈ker⁡Snu \in \ker S^n, giving v=Snu=0v = S^n u = 0. Hence ker⁡Sn∩ran⁡Sn={0}\ker S^n \cap \operatorname{ran} S^n = \{0\}, and by the dimension count, V=ker⁡Sn⊕ran⁡Sn=G(λ,T)⊕UV = \ker S^n \oplus \operatorname{ran} S^n = G(\lambda,T) \oplus U. Set W=UW = U. Both summands are TT-invariant: G(λ,T)G(\lambda,T) is TT-invariant by part (b). For W=ran⁡SnW = \operatorname{ran} S^n: if w=Snu∈Ww = S^n u \in W, then Tw=TSnu=Sn(Tu)∈ran⁡Sn=WTw = T S^n u = S^n (Tu) \in \operatorname{ran} S^n = W (using again that TT commutes with SnS^n). So WW is TT-invariant. Step 2: Restrict S=T−λIS = T-\lambda I to WW and show it is invertible on WW. Suppose w∈Ww \in W and Sw=0Sw = 0, i.e. w∈ker⁡S⊆ker⁡Sn=G(λ,T)w \in \ker S \subseteq \ker S^n = G(\lambda,T) (chain ascending). Then w∈G(λ,T)∩W={0}w \in G(\lambda,T) \cap W = \{0\} by the direct sum decomposition. So S∣WS|_W is injective, hence (since WW is finite-dimensional and S(W)⊆WS(W)\subseteq W as SS is a polynomial in TT and WW is TT-invariant) S∣WS|_W is a bijection of WW onto itself, i.e. (T−λI)∣W(T - \lambda I)|_W is invertible. Equivalently, λ\lambda is not an eigenvalue of T∣WT|_W. Step 3: Compute the characteristic polynomial via the block decomposition. Choose a basis β1\beta_1 of G(λ,T)G(\lambda,T) and a basis β2\beta_2 of WW; since V=G(λ,T)⊕WV = G(\lambda,T)\oplus W with both summands TT-invariant, the matrix of TT in the combined basis β1∪β2\beta_1 \cup \beta_2 is block-diagonal:

[T]β1∪β2=(A00B),A=[T∣G(λ,T)]β1 is g×g,B=[T∣W]β2 is (n−g)×(n−g).[T]_{\beta_1\cup\beta_2} = \begin{pmatrix} A & 0 \\ 0 & B\end{pmatrix}, \qquad A = [T|_{G(\lambda,T)}]_{\beta_1} \text{ is } g\times g,\quad B = [T|_W]_{\beta_2} \text{ is } (n-g)\times(n-g).

The characteristic polynomial of a block-diagonal matrix is the product of the characteristic polynomials of the blocks (expand the determinant det⁡(xI−[T])\det(xI - [T]) using the block structure of xI−(A00B)=(xIg−A00xIn−g−B)xI - \begin{pmatrix}A&0\\0&B\end{pmatrix} = \begin{pmatrix}xI_g - A & 0\\0 & xI_{n-g}-B\end{pmatrix}, whose determinant is the product of the determinants of the diagonal blocks since it is block upper triangular). Hence

det⁡(xI−T)=det⁡(xIg−A)⋅det⁡(xIn−g−B).\det(xI - T) = \det(xI_g - A)\cdot \det(xI_{n-g} - B).

Now, on G(λ,T)G(\lambda,T), the operator T∣G(λ,T)=λI+S∣G(λ,T)T|_{G(\lambda,T)} = \lambda I + S|_{G(\lambda,T)} where S∣G(λ,T)S|_{G(\lambda,T)} is nilpotent: indeed every v∈G(λ,T)v\in G(\lambda,T) satisfies Snv=0S^n v = 0 by part (a), so (S∣G(λ,T))n=0(S|_{G(\lambda,T)})^n = 0. A nilpotent operator on a gg-dimensional space has characteristic polynomial xgx^g (its only eigenvalue is 00, with algebraic multiplicity gg since the characteristic polynomial has degree gg and 00 is a root of multiplicity equal to gg — this follows because a nilpotent operator's minimal polynomial divides xnx^n for the ambient dimension, forcing every eigenvalue to be 00, and the characteristic polynomial, having degree gg and 00 as its only root, must equal xgx^g). Therefore A=λI+NA = \lambda I + N where NN is nilpotent, and

det⁡(xIg−A)=det⁡((x−λ)Ig−N)=(x−λ)g\det(xI_g - A) = \det((x-\lambda)I_g - N) = (x-\lambda)^g

(the characteristic polynomial of λI+N\lambda I + N shifted: substituting y=x−λy = x - \lambda reduces this to the characteristic polynomial of NN evaluated at yy, which is ygy^g). By Step 2, λ\lambda is not an eigenvalue of B=T∣WB = T|_W, so (x−λ)(x - \lambda) does not divide det⁡(xIn−g−B)\det(xI_{n-g}-B). Combining, det⁡(xI−T)=(x−λ)g⋅det⁡(xIn−g−B)\det(xI - T) = (x-\lambda)^g \cdot \det(xI_{n-g}-B) where the second factor is not divisible by (x−λ)(x-\lambda). By definition, the algebraic multiplicity aλa_\lambda of λ\lambda is exactly the exponent of (x−λ)(x-\lambda) in the full factorization of det⁡(xI−T)\det(xI-T); the above shows this exponent is exactly gg. Hence aλ=g=dim⁡G(λ,T)a_\lambda = g = \dim G(\lambda,T). ■\blacksquare

∎

Worked examples

Example 1 — a 4×44\times 4 operator with a mixed Jordan structure, eigenvalue 22. Let TT act on C4\mathbb{C}^4 (standard basis e1,e2,e3,e4e_1,e_2,e_3,e_4) by the matrix

A=(2100020000200005).A = \begin{pmatrix} 2 & 1 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 5\end{pmatrix}.

This is already block-diagonal: a 2×22\times 2 Jordan block at λ=2\lambda = 2, a 1×11\times 1 block at λ=2\lambda=2, and a 1×11\times 1 block at λ=5\lambda = 5. The characteristic polynomial is det⁡(xI−A)=(x−2)3(x−5)\det(xI - A) = (x-2)^3(x-5), so λ=2\lambda = 2 has algebraic multiplicity a2=3a_2 = 3 and λ=5\lambda = 5 has a5=1a_5 = 1. Compute the chain for λ=2\lambda = 2. Let S=A−2I=(0100000000000003)S = A - 2I = \begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&3\end{pmatrix}.

  • ker⁡S\ker S: solve Sv=0Sv=0. Row 1 gives v2=0v_2 = 0; row 4 gives 3v4=03v_4=0, so v4=0v_4=0. Free: v1,v3v_1, v_3. So ker⁡S=span⁡(e1,e3)\ker S = \operatorname{span}(e_1,e_3), dimension 22. This is the ordinary eigenspace — geometric multiplicity of λ=2\lambda=2 is 22, strictly less than a2=3a_2 = 3, confirming TT is not diagonalizable.
  • S2=(0000000000000009)S^2 = \begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&9\end{pmatrix}. Solve S2v=0S^2 v = 0: only constraint is 9v4=09v_4 = 0, i.e. v4=0v_4=0. So ker⁡S2=span⁡(e1,e2,e3)\ker S^2 = \operatorname{span}(e_1,e_2,e_3), dimension 33.
  • S3=(00000000000000027)S^3 = \begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&27\end{pmatrix}, same kernel as S2S^2: dimension 33. So the chain has stabilized: dim⁡ker⁡S=2<dim⁡ker⁡S2=3=dim⁡ker⁡S3\dim\ker S = 2 < \dim \ker S^2 = 3 = \dim\ker S^3.

Thus G(2,T)=ker⁡S2=span⁡(e1,e2,e3)G(2,T) = \ker S^2 = \operatorname{span}(e_1,e_2,e_3), with dim⁡G(2,T)=3=a2\dim G(2,T) = 3 = a_2, exactly matching the theorem. Note stabilization occurred at index 22, well before the theorem's guaranteed bound of n=4n = 4 — the bound j≤nj \le n is only an upper bound, not always sharp. For λ=5\lambda = 5: S′=A−5IS' = A - 5I has e4e_4 as an honest eigenvector with (A−5I)e4=0(A-5I)e_4 = 0, and one checks ker⁡(A−5I)=span⁡(e4)\ker(A-5I) = \operatorname{span}(e_4) already, dimension 1=a51 = a_5, so G(5,T)=span⁡(e4)G(5,T) = \operatorname{span}(e_4) and no higher powers were needed. Consistently, G(2,T)⊕G(5,T)=C4G(2,T) \oplus G(5,T) = \mathbb{C}^4, illustrating (without yet proving in general) the forthcoming Generalized Eigenspace Decomposition.

Example 2 — an operator given abstractly, not already in block form. Let T:C3→C3T: \mathbb{C}^3 \to \mathbb{C}^3 be given by

B=(311032004).B = \begin{pmatrix} 3 & 1 & 1 \\ 0 & 3 & 2 \\ 0 & 0 & 4\end{pmatrix}.

The characteristic polynomial (upper triangular, so it's the product of (x−diagonal entries)(x - \text{diagonal entries})) is (x−3)2(x−4)(x-3)^2(x-4), giving a3=2a_3 = 2, a4=1a_4 = 1. For λ=3\lambda = 3: S=B−3I=(011002001)S = B - 3I = \begin{pmatrix}0&1&1\\0&0&2\\0&0&1\end{pmatrix}.

  • ker⁡S\ker S: from row 3, v3=0v_3 = 0; from row 1 with v3=0v_3=0, v2=0v_2 = 0. Free variable v1v_1. So ker⁡S=span⁡(e1)\ker S = \operatorname{span}(e_1), dimension 11 — geometric multiplicity 11, less than a3=2a_3=2.
  • S2=(011002001)(011002001)=(003002001)S^2 = \begin{pmatrix}0&1&1\\0&0&2\\0&0&1\end{pmatrix}\begin{pmatrix}0&1&1\\0&0&2\\0&0&1\end{pmatrix} = \begin{pmatrix}0&0&3\\0&0&2\\0&0&1\end{pmatrix}. Solve S2v=0S^2 v=0: need 3v3=0,2v3=0,v3=03v_3 = 0, 2v_3=0, v_3=0, all forcing v3=0v_3 = 0; v1,v2v_1,v_2 free. So ker⁡S2=span⁡(e1,e2)\ker S^2 = \operatorname{span}(e_1,e_2), dimension 22.
  • S3=S⋅S2=(011002001)(003002001)=(003002001)S^3 = S\cdot S^2 = \begin{pmatrix}0&1&1\\0&0&2\\0&0&1\end{pmatrix}\begin{pmatrix}0&0&3\\0&0&2\\0&0&1\end{pmatrix} = \begin{pmatrix}0&0&3\\0&0&2\\0&0&1\end{pmatrix}. Same as S2S^2 — same kernel, dimension 22. Chain stabilized at index 2.

So G(3,T)=span⁡(e1,e2)G(3,T) = \operatorname{span}(e_1,e_2), dimension 2=a32 = a_3. Explicitly, e1e_1 is an ordinary eigenvector, and e2e_2 satisfies Se2=(1,0,0)T=e1≠0Se_2 = (1,0,0)^T = e_1 \ne 0 but S2e2=0S^2 e_2 = 0, so e2e_2 is a genuine generalized eigenvector of exact rank 22 (killed by S2S^2 but not by SS). For λ=4\lambda=4: since BB is upper triangular with 44 only in the bottom-right, one finds directly (B−4I)e3′(B-4I)e_3' where a genuine eigenvector must be computed: solve (B−4I)v=0(B - 4I)v = 0: (−1110−12000)v=0\begin{pmatrix}-1&1&1\\0&-1&2\\0&0&0\end{pmatrix}v=0 gives v3v_3 free, v2=2v3v_2 = 2v_3, v1=v2+v3=3v3v_1 = v_2+v_3 = 3v_3. Eigenvector (3,2,1)T(3,2,1)^T, and ker⁡(B−4I)\ker(B-4I) is already 11-dimensional =a4=a_4, so G(4,T)=span⁡((3,2,1)T)G(4,T) = \operatorname{span}((3,2,1)^T).

CounterexampleWhy finite-dimensionality of VV is essential

Drop the hypothesis dim⁡V<∞\dim V < \infty and the stabilization theorem collapses. Let V=C[x]V = \mathbb{C}[x], the (infinite-dimensional) space of all polynomials in xx over C\mathbb{C}, and let T:V→VT: V \to V be the derivative-like operator

T(p)(x)=x p(x)T(p)(x) = x\,p(x)

(multiplication by xx). Take λ=0\lambda = 0, so S=T−0⋅I=TS = T - 0\cdot I = T is just multiplication by xx. Then ker⁡S={0}\ker S = \{0\}: multiplying a nonzero polynomial by xx never gives 00. Indeed ker⁡Sk={0}\ker S^k = \{0\} for every k≥1k \ge 1, since xkp(x)=0x^k p(x) = 0 forces p=0p = 0 in the polynomial ring (an integral domain). So actually here every kernel is trivial and there is no generalized eigenvector at all for λ=0\lambda=0 — consistent, since 00 is not even an eigenvalue of TT (no nonzero polynomial satisfies xp=0⋅pxp = 0\cdot p). To see actual failure of stabilization behavior (an ever-growing, never-stabilizing chain with an honest eigenvalue), instead let V=C[x]V=\mathbb{C}[x] and let DD be the backward shift-type operator

D(a0+a1x+a2x2+⋯ )=a1+a2x+a3x2+⋯D(a_0 + a_1 x + a_2 x^2 + \cdots) = a_1 + a_2 x + a_3 x^2 + \cdots

(formally, differentiation composed with reindexing; concretely D(xn)=xn−1D(x^n) = x^{n-1} for n≥1n\ge1 and D(1)=0D(1)=0). Here λ=0\lambda = 0 is an eigenvalue with eigenvector 11 (the constant polynomial), since D(1)=0D(1) = 0. Now ker⁡Dk=span⁡(1,x,…,xk−1)\ker D^k = \operatorname{span}(1,x,\ldots,x^{k-1}), which has dimension kk for every kk: the chain

ker⁡D⊊ker⁡D2⊊ker⁡D3⊊⋯\ker D \subsetneq \ker D^2 \subsetneq \ker D^3 \subsetneq \cdots

is strictly increasing forever and never stabilizes. There is no finite mm with ker⁡Dm=ker⁡Dm+1\ker D^m = \ker D^{m+1}, and the "generalized eigenspace" ⋃kker⁡Dk\bigcup_k \ker D^k is all of VV, an infinite-dimensional space, so the dimension theorem's conclusion ("dim⁡G(λ,T)=aλ\dim G(\lambda,T) = a_\lambda," a finite number) cannot even be stated meaningfully. This is precisely what the finite-dimensionality hypothesis in the Stabilization Lemma rules out: without it, the pigeonhole argument bounding the chain length by n=dim⁡Vn = \dim V has no nn to appeal to.

Scope, edge cases & common traps

SituationWhat happens
TT diagonalizableG(λ,T)=ker⁡(T−λI)G(\lambda,T) = \ker(T-\lambda I) already at the first step; no higher powers needed.
λ\lambda not an eigenvalue of TTker⁡(T−λI)={0}\ker(T-\lambda I) = \{0\}, and in fact ker⁡(T−λI)k={0}\ker(T-\lambda I)^k=\{0\} for all kk (since T−λIT-\lambda I is invertible), so G(λ,T)={0}G(\lambda,T)=\{0\}.
Field is R\mathbb{R}, characteristic polynomial has non-real rootsThe dimension theorem's proof, which used a full factorization over C\mathbb{C} into linear factors, does not directly apply; the sum of generalized eigenspaces need not exhaust VV.
dim⁡V=∞\dim V = \inftyStabilization can fail entirely, as shown in the counterexample; "algebraic multiplicity" itself may not be well-defined.
Stabilization indexGuaranteed by index n=dim⁡Vn=\dim V, but often occurs much earlier (Example 1 stabilized at index 2 inside a 4-dimensional space).

The mathematician behind it

Origins · 1 minMarie Ennemond Camille Jordan and how this was discovered
ME
Marie Ennemond Camille Jordan
1838–1922 · 19th-century French mathematician working in group theory, linear algebra, and topology

Camille Jordan trained as an engineer at the École Polytechnique before turning to mathematics, where he became one of the most prolific French mathematicians of his era. He is remembered for the Jordan curve theorem in topology, for foundational work in group theory connecting to Galois theory, and — most relevant here — for the canonical form of a matrix that bears his name. Jordan's 1870 treatise Traité des substitutions et des équations algébriques developed the classification of linear substitutions (matrices) up to similarity, in which the ascending chains of generalized eigenspaces play the decisive structural role. His work sat at the crossroads of algebra and what would later become linear algebra as a distinct discipline; the notion of a generalized eigenvector, though not named as such by Jordan in modern terms, is the precise mechanism his canonical form exploits to handle operators that fail to be diagonalizable.

History
- 1870: Camille Jordan publishes Traité des substitutions, introducing the canonical block form for matrices, implicitly using chains of generalized eigenvectors.
- 1878: Jordan's canonical form is further developed and clarified in subsequent work on linear substitutions over finite fields and over the complex numbers.
- 1890s–1900s: Karl Weierstrass and Ferdinand Georg Frobenius develop related invariant-factor and elementary-divisor theories, giving an independent algebraic route to the same decomposition.
- Early 20th century: the modern operator-theoretic language of "generalized eigenspaces" and nilpotent-plus-scalar decompositions crystallizes in linear algebra textbooks, separating the geometric argument from Jordan's original matrix-theoretic one.
- Mid-to-late 20th century: the generalized eigenspace decomposition becomes a standard prerequisite topic for the Jordan form in university linear algebra courses, exactly as developed in this article.
How does a generalized eigenvector differ from an ordinary eigenvector?

An ordinary eigenvector vv satisfies Tv=λvTv = \lambda v exactly, i.e. (T−λI)v=0(T-\lambda I)v = 0 in one step. A generalized eigenvector only needs (T−λI)kv=0(T-\lambda I)^k v = 0 for some positive integer kk, possibly larger than 11. Every ordinary eigenvector is automatically a generalized eigenvector (take k=1k=1), but the converse can fail: a vector might require k=2,3,…k=2,3,\ldots ticks of T−λIT-\lambda I before vanishing, as with e2e_2 in a Jordan block. The generalized eigenspace G(λ,T)G(\lambda,T) therefore always contains the ordinary eigenspace, often strictly, and it is this extra room that allows generalized eigenspaces to have the "correct" dimension (algebraic multiplicity) even when ordinary eigenspaces are too small.

Why must we take a union over all powers kk rather than fixing one specific kk?

Different generalized eigenvectors can require different numbers of steps to vanish under T−λIT-\lambda I. In a 3×33\times3 Jordan block, the top basis vector needs only k=1k=1, the middle one needs k=2k=2, and the bottom one needs k=3k=3. If we fixed k=1k=1 we would only recover the ordinary eigenspace; if we fixed some intermediate kk, we might still miss vectors requiring more steps. Taking the union ⋃kker⁡(T−λI)k\bigcup_k \ker(T-\lambda I)^k captures every vector that is eventually annihilated by some power, regardless of how many steps it needs. The stabilization lemma then shows this union is achieved by a single finite power (at most n=dim⁡Vn=\dim V), so in practice one need only compute up to that power.

Does the Dimension Theorem hold over the real numbers R\mathbb{R}?

The proof given here relies on decomposing VV via the characteristic polynomial's root at λ\lambda, and the argument that det⁡(xIn−g−B)\det(xI_{n-g}-B) is not divisible by (x−λ)(x-\lambda) still works over R\mathbb{R} for a single real eigenvalue λ\lambda, so the theorem dim⁡G(λ,T)=aλ\dim G(\lambda,T)=a_\lambda remains valid for each real eigenvalue of a real operator. What can fail over R\mathbb{R} is the global consequence that the generalized eigenspaces of all eigenvalues together span VV: the characteristic polynomial may have non-real roots that contribute no real eigenvalue and hence no generalized eigenspace inside the real vector space, leaving the sum of real generalized eigenspaces strictly smaller than VV. Over C\mathbb{C}, where the characteristic polynomial always splits into linear factors, this obstruction disappears.

How is the generalized eigenspace related to the minimal polynomial?

The Minimal Polynomial of TT factors as ∏λ(x−λ)mλ\prod_\lambda (x-\lambda)^{m_\lambda} over C\mathbb{C}, where mλm_\lambda is the smallest positive integer such that ker⁡(T−λI)mλ=ker⁡(T−λI)mλ+1\ker(T-\lambda I)^{m_\lambda} = \ker(T-\lambda I)^{m_\lambda+1}, i.e. the exact stabilization index for eigenvalue λ\lambda. This mλm_\lambda is also the size of the largest Jordan block associated to λ\lambda. It satisfies 1≤mλ≤aλ1 \le m_\lambda \le a_\lambda, with mλ=1m_\lambda = 1 exactly when TT is diagonalizable on G(λ,T)G(\lambda,T) (no Jordan blocks larger than size 1). So while the algebraic multiplicity aλa_\lambda tells you the dimension of the generalized eigenspace, the minimal polynomial's exponent mλm_\lambda tells you the depth of the chain — how many steps are truly needed to reach stabilization.

Can a vector be a generalized eigenvector for two different eigenvalues simultaneously?

No, and this is provable directly. Suppose v≠0v \ne 0 satisfies (T−λI)jv=0(T-\lambda I)^j v = 0 and (T−μI)kv=0(T-\mu I)^k v = 0 for distinct scalars λ≠μ\lambda \ne \mu. One can show the generalized eigenspaces G(λ,T)G(\lambda,T) and G(μ,T)G(\mu,T) for distinct eigenvalues intersect only in {0}\{0\}, by an argument generalizing the classical proof that eigenvectors for distinct eigenvalues are linearly independent (apply suitable polynomials in T−λIT-\lambda I and T−μIT - \mu I, which are coprime as polynomials, to isolate v=0v=0). Consequently the sum ∑λG(λ,T)\sum_\lambda G(\lambda,T) over distinct eigenvalues is always a direct sum, a fact used crucially in the Generalized Eigenspace Decomposition.

What goes wrong if I only check ker⁡(T−λI)\ker(T-\lambda I) and ker⁡(T−λI)2\ker(T-\lambda I)^2 without verifying the chain has truly stabilized?

If the two dimensions happen to be unequal at some early stage, you cannot stop there — you must keep going until you find two consecutive equal dimensions, because the chain can have several genuine strict jumps before it stabilizes (as in a Jordan block of size 3, where dim⁡ker⁡S1=1<dim⁡ker⁡S2=2<dim⁡ker⁡S3=3=dim⁡ker⁡S4\dim\ker S^1=1<\dim\ker S^2=2<\dim\ker S^3=3=\dim\ker S^4). Stopping after only one comparison risks under-counting the generalized eigenspace and computing the wrong dimension, which would then fail to match the algebraic multiplicity — a red flag that should prompt rechecking the next power in the chain.

Why is finite-dimensionality of VV needed, and is there any analogue in infinite dimensions?

Finite-dimensionality is what makes the pigeonhole argument work: a strictly increasing chain of subspace dimensions inside an nn-dimensional space cannot continue past nn steps, forcing stabilization. In infinite dimensions this bound disappears, and as the counterexample with the backward-shift operator on C[x]\mathbb{C}[x] shows, the chain can increase forever without ever stabilizing, so no finite-dimensional generalized eigenspace exists. In functional analysis, generalized eigenspace ideas do reappear (e.g. for operators with compact resolvent, or in the theory of Riesz projections for isolated points of the spectrum), but they require substantially more machinery — spectral theory and complex analysis — rather than the elementary linear-algebraic pigeonhole argument used here.

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