positive dimension, and so the quotient is a nonzero abelian Lie algebra, which certainly contains an ideal of codimension one and by the ideal correspondence, it corresponds to an ideal of codimension one in . Step 3: There exists some linear functional in such that is nonzero. This follows from the inductive hypothesis (it is easy to check that the eigenvalues determine a linear functional). Step 4: is a -invariant subspace. (Note this step proves a general fact and does not involve solvability.) Let , , then we need to prove . If then it's obvious, so assume and set recursively . Let and be the largest such that are linearly independent. Then we'll prove that they generate U and thus is a basis of U. Indeed, assume by contradiction that it's not the case and let be the smallest such that , then obviously . Since are linearly dependent, is a linear combination of . Applying the map it follows that is a linear combination of . Since by the minimality of m each of these vectors is a linear combination of , so is , and we get the desired contradiction. We'll prove by induction that for every and there exist elements of the base field such that and The case is straightforward since . Now assume that we have proved the claim for some and all elements of and let . Since is an ideal, it's , and thus and the induction step follows. This implies that for every the subspace U is an invariant subspace of X and the matrix of the restricted map in the basis is upper triangular with diagonal elements equal to , hence . Applying this with instead of X gives . On the other hand, U is also obviously an invariant subspace of Y, and so since commutators have zero trace, and thus . Since is invertible (because of the assumption on the characteristic of the base field), and and so . Step 5: Finish up the proof by finding a common eigenvector. Write where L is a one-dimensional vector subspace. Since the base field is algebraically closed, there exists an eigenvector in for some (thus every) nonzero element of L. Since that vector is also eigenvector for each element of , the proof is complete. Consequences The theorem applies in particular to the adjoint representation of a (finite-dimensional) solvable Lie