PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-16

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Split characteristic polynomials decompose into generalised eigenspaces of the algebraic multiplicities

Statement

Let $T$ be an endomorphism of a finite-dimensional $F$ -vector space whose characteristic polynomial splits over $F$ . If $e_{λ}$ is the exponent of $x - λ$ in $μ_{T}$ , then $V = ⨁_{λ} G_{λ}, G_{λ} : = \ker (T - λ I)^{e_{λ}},$ where $λ$ ranges over the eigenvalues. Each $G_{λ}$ is $T$ -invariant, $N_{λ} : = (T - λ I) ∣_{G_{λ}}$ is nilpotent of index $e_{λ}$ , and $χ_{T ∣_{G_{λ}}} = (x - λ)^{\dim G_{λ}} .$ Consequently $\dim G_{λ}$ is the algebraic multiplicity of $λ$ in $χ_{T}$ . For $V = 0$ , the sum and eigenvalue set are empty.

Facts & Assumptions

Given: A finite-dimensional endomorphism $T$ whose characteristic polynomial splits over $F$ .

[L1]

The minimal and characteristic polynomials have the same monic irreducible factors (The minimal and characteristic polynomials have exactly the same monic irreducible factors).

[L2]

If $μ_{T} = \prod_{i < r} (x - λ_{i})^{e_{i}}$ splits with distinct $λ_{i}$ and $e_{i} \geq 1$ , then $V = ⨁_{i < r} G_{λ_{i}}^{(e_{i})} (T)$ , and for every $k \geq e_{i}$ , $\ker (T - λ_{i} I)^{k} = G_{λ_{i}}^{(e_{i})} (T)$ (If the minimal polynomial splits, $V$ is the direct sum of the stabilised generalised eigenspaces).

[L3]

Writing $μ_{T} = \prod_{i < r} q_{i}^{e_{i}}$ with the $q_{i}$ distinct monic irreducibles, the subspaces $V_{i} : = \ker q_{i} (T)^{e_{i}}$ are $T$ -invariant, $V = ⨁_{i < r} V_{i}$ , and the minimal polynomial of $T ∣_{V_{i}}$ is exactly $q_{i}^{e_{i}}$ (Primary decomposition: the irreducible-power factors of $μ_{T}$ split $V$ into their invariant kernels).

[L4]

Characteristic polynomials multiply across an invariant subspace and its quotient (For invariant $W$ , $χ_{T} = χ_{T ∣_{W}} χ_{\overset{ˉ}{T}}$ ).

[L5]

For an endomorphism of a nonzero $m$ -dimensional space, nilpotency is equivalent to $μ_{N} = x^{s}$ for some $1 \leq s \leq m$ and to $χ_{N} = x^{m}$ , and in that case $s$ is the nilpotency index (Characterisations of a nilpotent endomorphism).

Proof

technique · direct

1.1L1L2L3

Fact [L1] makes $μ_{T}$ split over $F$ , so its distinct monic irreducible factors are the $x - λ$ with $λ$ an eigenvalue and $G_{λ} = \ker (T - λ I)^{e_{λ}}$ . Fact [L2] gives the displayed invariant direct sum, and [L3] applied to the same factorisation gives that each $T ∣_{G_{λ}}$ has minimal polynomial exactly $(x - λ)^{e_{λ}}$ .

2.1step 1.1L5algebra

Fix an eigenvalue $λ$ . Since $λ$ is an eigenvalue, $\ker (T - λ I) \neq 0$ , and $\ker (T - λ I) \subseteq G_{λ}$ because $e_{λ} \geq 1$ ; hence $G_{λ}$ is nonzero and [L5] applies to it. On $G_{λ}$ , $N_{λ} = T ∣_{G_{λ}} - λ I$ has minimal polynomial $x^{e_{λ}}$ by step 1.1, so [L5] makes $N_{λ}$ nilpotent of index $e_{λ}$ and gives $χ_{N_{λ}} = x^{\dim G_{λ}}$ . Translating the scalar variable yields $χ_{T ∣_{G_{λ}}} = (x - λ)^{\dim G_{λ}}$ .

3.1step 1.1step 2.1L4∎

Repeated use of [L4] on the invariant direct sum multiplies these restricted characteristic polynomials to obtain $χ_{T} = \prod_{λ} (x - λ)^{\dim G_{λ}}$ , so the exponent is exactly the algebraic multiplicity. The empty product gives the zero-space case.

Depends on

Dependency tree · 2 levels

Sources

S. Axler, Linear Algebra Done Right, 4th ed., Section 8A