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For a real operator, nonreal generalised eigenspaces of the complexification occur in conjugate pairs

Statement

Let T:VV be an endomorphism of a real vector space, let σ=σcan be the canonical conjugation of VC, and let λC and e1. Then

σ(Gλ(e)(TC))=Gλ(e)(TC).

Thus for nonreal λ the generalised eigenspaces of TC for λ and for λ are interchanged by the real-linear involution σ, and they have the same real dimension.

Facts & Assumptions

Given: A real vector space V, an endomorphism T:VV, a complex scalar λ, and an exponent e1.

[L1]

A conjugation is conjugate-linear and an involution (Conjugations and real structures on a complex vector space).

[L2]

The complexification of a real operator commutes with the canonical conjugation, because it comes from a real operator (A complex-linear operator comes from a real operator exactly when it commutes with the chosen conjugation).

[L3]

The generalised eigenspace of exponent e is Gλ(e)(TC)=ker(TCλI)e (Primary components kerq(T)e and generalised eigenspaces Gλ(e)(T)=ker(TλI)e).

Proof

technique · direct
1.1

The operator TC commutes with σ by [L2], and σ is an R-linear involution, hence a bijection, with σ(zw)=zσw by [L1].

L1L2
2.1

The powers commute with σ up to conjugation of the scalar: for wVC, σ(TCwλw)=TCσwλσw by step 1.1 and [L1]; iterating this e times gives σ(TCλI)e=(TCλI)eσ.

step 1.1L1algebra
3.1

If wker(TCλI)e, then (TCλI)eσw=σ(TCλI)ew=σ(0)=0, so σwGλ(e)(TC) by [L3].

step 2.1L3
3.2

Conversely, if wGλ(e)(TC), then w:=σw satisfies (TCλI)ew=σ(TCλI)ew=0, so wGλ(e)(TC); since σ is an involution, the two inclusions combine to the equality σ(Gλ(e)(TC))=Gλ(e)(TC).

step 2.1L1L3
4.1

Because σ is real-linear and bijective by step 1.1, the two generalised eigenspaces have the same real dimension; for nonreal λ they form the conjugate pair interchanged by σ.

step 1.1step 3.2

Depends on

Used by

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