In mathematics, specifically the theory of Lie algebras, Lie's theorem states that, over an algebraically closed field of characteristic zero, if is a finite-dimensional representation of a solvable Lie algebra, then there's a flag of invariant subspaces of with , meaning that for each and i. Put in another way, the theorem says there is a basis for V such that all linear transformations in are represented by upper triangular matrices. This is a generalization of the result of Frobenius that commuting matrices are simultaneously upper triangularizable, as commuting matrices generate an abelian Lie algebra, which is a fortiori solvable. A consequence of Lie's theorem is that any finite dimensional solvable Lie algebra over a field of characteristic 0 has a nilpotent derived algebra (see #Consequences). Also, to each flag in a finite-dimensional vector space V, there correspond a Borel subalgebra (that consist of linear transformations stabilizing the flag); thus, the theorem says that is contained in some Borel subalgebra of . Counter-example For algebraically closed fields of characteristic p>0 Lie's theorem holds provided the dimension of the representation is less than p (see the proof below), but can fail for representations of dimension p. An example is given by the 3-dimensional nilpotent Lie algebra spanned by 1, x, and d/dx acting on the p-dimensional vector space k[x]/(xp), which has no eigenvectors. Taking the semidirect product of this 3-dimensional Lie algebra by the p-dimensional representation (considered as an abelian Lie algebra) gives a solvable Lie algebra whose derived algebra is not nilpotent. Proof The proof is by induction on the dimension of and consists of several steps. (Note: the structure of the proof is very similar to that for Engel's theorem.) The basic case is trivial and we assume the dimension of is positive. We also assume V is not zero. For simplicity, we write . Step 1: Observe that the theorem is equivalent to the statement:<ref>{{harvnb|Serre|loc=Theorem 3{{}}}}</ref> There exists a vector in V that is an eigenvector for each linear transformation in . Indeed, the theorem says in particular that a nonzero vector spanning is a common eigenvector for all the linear transformations in . Conversely, if v is a common eigenvector, take to its span and then admits a common eigenvector in the quotient ; repeat the argument. Step 2: Find an ideal of codimension one in . Let be the derived algebra. Since is solvable and has