Abstract Linear Algebra 37

TL;DR

The fitting index D is the smallest J such that the generalized eigenspace chain stabilizes: ker((A−ΛI)^D) = ker((A−ΛI)^{D+1}).

Briefing Cornell Notes

Briefing

Fitting index is introduced as the first point in the Jordan-chain ladder where the dimensions stop growing—an invariant that pinpoints when generalized eigenspaces become “stable.” Starting with a complex square matrix A and a fixed eigenvalue Λ, the construction uses N = A − ΛI. For each natural number J, the generalized eigenspace E^J is defined as ker(N^J), and the corresponding range subspace R^J is defined as ran(N^J). These subspaces form nested chains: E^0 ⊆ E^1 ⊆ E^2 ⊆ … and R^0 ⊇ R^1 ⊇ R^2 ⊇ … . Because dimensions are bounded by the ambient space dimension n (where A is n×n), the chain cannot keep strictly increasing forever.

The fitting index D is defined using the E-chain: it is the smallest index where no further dimension increase occurs, meaning E^D = E^{D+1}. Since each strict inclusion forces at least a one-dimensional jump and the total dimension cannot exceed n, D is guaranteed to exist and is finite. The key payoff comes from showing that the same D also marks stability in the range chain. Using the rank-nullity theorem on N^J for arbitrary J, the dimensions of ker(N^J) and ran(N^J) must add up to n. Therefore, whenever dim(E^J) jumps, dim(R^J) must jump by the same amount; and whenever dim(E^J) stays constant, dim(R^J) must also stay constant. This forces the first equality point to coincide: E^D = E^{D+1} if and only if R^D = R^{D+1}. In other words, the fitting index can be read off from either the kernel chain or the range chain.

Once that alignment is established, the ranges become the computationally convenient route. The transcript notes that applying N to ran(N^D) produces ran(N^{D+1}), and because the chain is stable at D, ran(N^{D+1}) = ran(N^D). Concretely, the map induced by N from R^D to R^{D+1} becomes a linear map from R^D to itself with equal domain and codomain dimensions. That dimension match upgrades the map from surjective to bijective, yielding an isomorphism between R^D and R^{D+1}. The argument then iterates: since N maps R^{D+1} onto R^{D+2} and R^{D+1} already equals R^D, the same equality persists. Inductively, R^{D+k} = R^D for every k ≥ 0, and the same stability transfers back to the kernel chain.

The result is a clean structural picture: the fitting index D is exactly where the generalized eigenspace chain stops changing, and from that point onward the dimensions—and thus the subspaces—remain fixed. Because D cannot exceed n, the chain reaches stability after finitely many steps. Until D, the generalized eigenspaces E^J strictly grow and therefore contain the generalized eigenvectors that will be organized in the next installment of the series.

Cornell Notes

Fix an eigenvalue Λ of a complex n×n matrix A and set N = A − ΛI. Define generalized eigenspaces E^J = ker(N^J) and range subspaces R^J = ran(N^J), which form nested chains. The fitting index D is the smallest J such that E^D = E^{D+1}; it exists because dimensions are bounded by n and each strict inclusion increases dimension. Rank-nullity forces the same stabilization point in the range chain: E^D = E^{D+1} exactly when R^D = R^{D+1}. At that index, N induces an isomorphism between R^D and R^{D+1}, and the equality continues for all larger powers, so the chain stays stable from D onward.

How is the fitting index D defined using generalized eigenspaces E^J = ker((A−ΛI)^J)?

With N = A − ΛI and E^J = ker(N^J), the fitting index D is the smallest natural number such that the chain stops growing: E^D = E^{D+1}. Equivalently, D is the first step where the dimension of ker(N^J) no longer increases. Since dim(E^J) is bounded above by n, such a D must occur after finitely many steps.

Why does stabilization in the kernel chain automatically imply stabilization in the range chain?

For each J, rank-nullity applied to N^J gives dim(ker(N^J)) + dim(ran(N^J)) = n. If dim(E^J) = dim(ker(N^J)) increases from J to J+1, then dim(ran(N^J)) must increase by the same amount to keep the sum equal to n. Likewise, if dim(E^J) stays the same, dim(ran(N^J)) cannot change. This forces the first equality point to match: E^D = E^{D+1} if and only if R^D = R^{D+1}.

What makes the range chain R^J = ran(N^J) especially useful once D is reached?

At the fitting index, the range chain is stable: R^D = R^{D+1}. Because N maps R^D onto R^{D+1}, this becomes a linear map from R^D to itself. With equal dimensions on both sides, surjectivity implies bijectivity, so N acts as an isomorphism between R^D and R^{D+1}.

How does the isomorphism at index D lead to stability for all larger powers?

Since R^{D+1} = R^D, applying N again gives N(R^{D+1}) = R^{D+2}. But N(R^{D+1}) is exactly R^{D+2}, and because the previous step already identifies R^{D+1} with R^D, the same stabilization logic repeats. Inductively, R^{D+k} = R^D for every k ≥ 0, and the kernel chain stabilizes at the same D as well.

Why must the fitting index D be finite and bounded by n?

Each strict inclusion E^J ⊂ E^{J+1} forces a dimension increase of at least one. But dim(E^J) can never exceed the ambient dimension n. Therefore, after at most n steps of strict growth, the chain must reach an index where no further increase occurs—so D exists and satisfies D ≤ n.

Review Questions

Given N = A − ΛI, what does the equality E^D = E^{D+1} tell you about ker(N^J) as J increases past D?
Explain how rank-nullity applied to N^J forces the same fitting index for the kernel chain and the range chain.
Why does N induce an isomorphism on R^D once R^D = R^{D+1} holds?

Key Points

1
The fitting index D is the smallest J such that the generalized eigenspace chain stabilizes: ker((A−ΛI)^D) = ker((A−ΛI)^{D+1}).
2
Generalized eigenspaces E^J = ker(N^J) and ranges R^J = ran(N^J) form nested chains with dimensions bounded by n.
3
Rank-nullity applied to N^J links the kernel and range dimensions, forcing the stabilization point to coincide for both chains.
4
At the fitting index, N maps R^D onto R^{D+1} with equal dimensions, so the induced map is an isomorphism.
5
Stability at D propagates forward: R^{D+k} = R^D and E^{D+k} = E^D for all k ≥ 0.
6
The fitting index must be finite because each strict step increases dimension by at least one, and the maximum dimension is n.

Highlights

The fitting index D is defined as the first place where the generalized eigenspace dimensions stop increasing: E^D = E^{D+1}.

Rank-nullity forces the same D to appear in the range chain: R^D = R^{D+1} exactly when E^D = E^{D+1}.

Once stability hits, N becomes an isomorphism on the stabilized range R^D, and that stability persists for all higher powers.

Because each strict inclusion increases dimension and dim(E^J) ≤ n, D cannot exceed n.

Topics

Fitting Index
Generalized Eigenspaces
Jordan Chains
Rank-Nullity
Range Stabilization

Abstract Linear Algebra 37 | Fitting Index