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If the minimal polynomial splits, $V$ is the direct sum of the stabilised generalised eigenspaces

Statement

Let $T : V \to V$ be an endomorphism of a finite-dimensional vector space over an arbitrary field $F$ ; finite-dimensionality is what makes $μ_{T}$ available at all, the minimal polynomial being defined only in that case. Suppose its minimal polynomial splits over $F$ as

$μ_{T} = \prod_{i < r} (x - λ_{i})^{e_{i}},$

with distinct $λ_{i} \in F$ and $e_{i} \geq 1$ . Then

$V = ⨁_{i < r} G_{λ_{i}}^{(e_{i})} (T) .$

For every $k \geq e_{i}$ , $\ker (T - λ_{i} I)^{k} = G_{λ_{i}}^{(e_{i})} (T)$ , so these are the stabilised generalised eigenspaces. The zero space corresponds to the empty product and empty direct sum.

Facts & Assumptions

Given: The displayed split factorisation of $μ_{T}$ over $F$ .

[L1]

A nonzero polynomial splits over $F$ when it is a product of linear factors in $F [x]$ , with repetitions allowed (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

[L2]

For an endomorphism of a finite-dimensional space, the irreducible-power factors of its minimal polynomial give a direct sum of their invariant kernels, and each restriction has exactly the corresponding factor as minimal polynomial (Primary decomposition: the irreducible-power factors of $μ_{T}$ split $V$ into their invariant kernels).

[L3]

Coprime polynomials satisfy a Bézout identity (Bézout identity and the Euclidean algorithm for polynomials over a field).

[L4]

The generalised eigenspace of exponent $e$ is $G_{λ}^{(e)} (T) = \ker (T - λ I)^{e}$ (Primary components $\ker q (T)^{e}$ and generalised eigenspaces $G_{λ}^{(e)} (T) = \ker (T - λ I)^{e}$ ).

Proof

technique · direct

1.1L1L2L4

By [L1], the irreducible factors are the distinct linear polynomials $x - λ_{i}$ . Applying [L2] gives the displayed direct sum, and [L4] identifies its $i$ -th summand with $G_{λ_{i}}^{(e_{i})} (T)$ .

2.1step 1.1L2L3

Fix $i$ and $k \geq e_{i}$ . The kernel of $(T - λ_{i} I)^{k}$ contains the $i$ -th summand. On every other primary summand, $(x - λ_{i})^{k}$ is coprime to $(x - λ_{j})^{e_{j}}$ ; evaluating a Bézout identity from [L3] shows $(T - λ_{i} I)^{k}$ is invertible there. Hence its kernel contains no vector from the other summands.

3.1step 2.1L2∎

Thus the kernel at every $k \geq e_{i}$ is exactly $G_{λ_{i}}^{(e_{i})} (T)$ . If the factorisation is empty, [L2] gives $V = 0$ .

Depends on

Used by

Split characteristic polynomials decompose into generalised eigenspaces of the algebraic multiplicities Proposition

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Anthony W. Knapp, Basic Algebra, 2nd ed., Ch. V, §5, Theorem 5.19 (standard reference, not scraped)
Sheldon Axler, Linear Algebra Done Right, 4th ed., §8B, Theorem 8.22 (complex-field special case) (standard reference, not scraped)

If the minimal polynomial splits, VV is the direct sum of the stabilised generalised eigenspaces