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Unformatted text preview: If L is one dimensional then we are
dealing with one endomorphism x of V . Since C is algebraically closed, the characteristic
polynomial of x has a root λ ∈ C and a corresponding eigenvector v . (x(v ) = λv . So,
λx = λ.) So, suppose that dim L ≥ 2.
The next step is to ﬁnd a codimension one ideal J in L. Since [LL] L, this is easy.
Take any codimension one vector subspace of L/[LL] and let J be the inverse image of
this in L.
By induction there is a nonzero vector v ∈ V and a linear map λ : J → C so that
x(v ) = λ(x)v for all x ∈ J . Let W be the set of all v ∈ V with this property (for the
same linear function λ). Let y ∈ L, y ∈ J .
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Claim: y (W ) ⊆ W . Suppose for a moment that this is true. Then, we can ﬁnd an eigenvalue λ(y ) and
eigenvector w ∈ W so that y (w) = λ(y )w. By the remark, this extended linear map
λ satisﬁes the desired equation. The Claim is proved more generally in the following
lemma.
Lemma 3.2.3. Suppose that J is an ideal in L, V is a representation of L and λ : J → F
is a linear map. Assume char F = 0. Let
W = {v ∈ V  x(v ) = λ(x)v ∀x ∈ J } Then y (W ) ⊆ W for all y ∈ L. Proof. Deﬁne Wi recursively as follows. W0 = 0 and
Wk+1 = {v ∈ V  x(v ) − λ(x)(v ) ∈ Wk ∀x ∈ J } Then W = W1 ⊆ W2 ⊆ · · · since V is ﬁnite dimensional w...
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This document was uploaded on 03/09/2014 for the course MATH 223a at Brandeis.
 Fall '11
 K.Igusa
 Algebra

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