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Thm. Let $L$ be a solvable subalgebra of $gl(V)$, $V$ a finite dimensional nonzero vector space. Then $V$ contains a common eigenvector for all the endomorphisms in $L$.

The proof of this theorem is typically broken down into 4 parts, 1) find an ideal $K$ of codimension one, 2) show that common eigenvectors exist for $K$, 3) verify that $L$ stabilizes a space consisting of such eigenvectors, 4) find an eigenvector for a single $z\in L$ satisfying $L=K+Fz$.

For part 2, Humphreys says(p.16), there exists a linear functional $\lambda: K\rightarrow F$ satisfying $x.v=\lambda (x)v$, $x\in K$.

This may be trivial, but I do not recall this fact from basic linear algebra. I would like to understand this and/or get a reference for this statement. Thanks.

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The field is algebraically closed. Since $v$ is a common eigenvector for all elements of $K$, then we have $x(v)=\lambda(x)v$. If $x,y\in K$, then $(x+y)\cdot v=x\cdot v+y\cdot y=\lambda(x)v+\lambda(y)v=(\lambda(x)+\lambda(y))v $. By other hand, since $x+y\in K$ we have $(x+y)\cdot v=\lambda(x+y)v.$ Finally, if $c\in \mathbb{F}$ and $x\in K$, then $(cx)\cdot v=\lambda(cx)v.$ By other hand, $(cx)\cdot v=c(x\cdot v)=c(\lambda(x)v)=(c\lambda(x))v.$ This shows that $\lambda:K\rightarrow \mathbb{F}$ is a linear function.

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thanks for this answer. Btw, what linear algebra textbook do you suggest might prepare one for reading a lie algebra textbook? –  Edison Apr 29 '12 at 2:41
    
Hoffman and Kunze: Linear Algebra –  spohreis Apr 29 '12 at 20:09
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