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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

1.1

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 $G_{λ_{i}}^{(e_{i})} (T)$ .

L1L2L4

2.1

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.

step 1.1L2L3

3.1

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

step 2.1L2∎

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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