LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Kernel and rank sequences of powers stabilise once equality occurs

Statement

Let $T$ be an endomorphism of a finite-dimensional vector space $V$ . For every $k \geq 0$ , $\ker T^{k} \subseteq \ker T^{k + 1}, im T^{k + 1} \subseteq im T^{k},$ so the nullities weakly increase and the ranks weakly decrease. If $\ker T^{m} = \ker T^{m + 1}$ for some $m \geq 0$ —equivalently, if $rank T^{m} = rank T^{m + 1}$ —then for every $r \geq 0$ , $\ker T^{m + r} = \ker T^{m}, im T^{m + r} = im T^{m} .$

Facts & Assumptions

Given: An endomorphism $T$ of a finite-dimensional vector space $V$ .

[L1]

The kernel and image of a linear map are $\ker S = {v : S (v) = 0}$ and $im S = {S (v) : v \in V}$ (Kernel and image of a linear map).

[L2]

Rank and nullity are the dimensions of the image and kernel (Rank and nullity of a linear map with finite-dimensional domain).

[L3]

For an endomorphism $S$ of $V$ , $\dim V = nullity S + rank S$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

Proof

technique · induction

1.1

If $T^{k} v = 0$ , then $T^{k + 1} v = 0$ , and every $T^{k + 1} v$ equals $T^{k} (T v)$ ; hence the displayed kernel and image inclusions hold, and [L2] turns them into the asserted dimension inequalities.

L1L2algebra

1.2

Assume $\ker T^{m} = \ker T^{m + 1}$ . The equality $\ker T^{m + 0} = \ker T^{m}$ is the base case.

base

2.1

By [L3], equality of the two consecutive kernel dimensions is equivalent to equality of the two consecutive ranks; together with the inclusions in step 1.1, either dimension equality is equivalent to equality of the corresponding subspaces.

step 1.1L2L3

2.2

If $\ker T^{m + r} = \ker T^{m}$ and $v \in \ker T^{m + r + 1}$ , then $T^{r} v \in \ker T^{m + 1} = \ker T^{m}$ , so $v \in \ker T^{m + r}$ ; the reverse inclusion is in step 1.1, completing the induction on $r$ .

step 1.1ih

3.1

Applying [L3] to every $T^{m + r}$ shows that the later images all have the same dimension as $im T^{m}$ ; the nested image inclusions from step 1.1 therefore make them equal, completing the claim.

step 1.1step 1.2step 2.2L3discharge-induction∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 49 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

S. Axler, Linear Algebra Done Right, 4th ed., Results 8.1-8.3 (standard reference, not scraped)