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Over the complex numbers, every finite-dimensional representation of a solvable Lie algebra has a basis in which every representing matrix is upper triangular. The Lie theorem is often stated first as the existence of a common eigenvector in every nonzero finite-dimensional Lie algebra representation of a Solvable Lie algebra. Applying that assertion successively to quotient representations gives an invariant complete flag, and hence the upper triangular form. The same proof works over any algebraically closed field of characteristic zero.

We prove the common eigenvector assertion by induction on $\dim \mathfrak{g}$, writing the action as $av$. The zero Lie algebra is immediate. If $\mathfrak{g}\ne 0$ is solvable, its derived series of a Lie algebra shows that $[\mathfrak{g},\mathfrak{g}]\ne \mathfrak{g}$. Choose a codimension-one ideal of a Lie algebra $\mathfrak{a}$ containing $[\mathfrak{g},\mathfrak{g}]$, and choose $x\notin \mathfrak{a}$. By induction there are $v\ne 0$ and a linear functional $\lambda :\mathfrak{a}\to \mathbb{C}$ such that $av=\lambda (a)v$ for all $a\in \mathfrak{a}$.

Let $W$ be the span of $v,xv,x^{2}v,\dots$. If $m=\dim W$, the first $m$ of these vectors form a basis, and $W$ is $x$-invariant. We claim that for each $a\in \mathfrak{a}$,For $j=0$ this is the definition of $v$. For the induction step, use $ax=xa+[a,x]$ and $[a,x]\in \mathfrak{a}$. Applying the induction hypothesis to both $a$ and $[a,x]$ proves the claim. Consequently $W$ is $\mathfrak{a}$-invariant, and every $a\in \mathfrak{a}$ acts on $W$ by an upper triangular matrix with all diagonal entries $\lambda (a)$.

Since both $x$ and $a$ preserve $W$, the matrix trace of their commutator on $W$ is zero. The claim applied to $[x,a]\in \mathfrak{a}$ givesHere characteristic zero is essential: $m\ne 0$ in the field, so $\lambda ([x,a])=0$. Now the common weight spaceis nonzero and $x$-invariant. Indeed, for $w\in E_{\lambda }$,An endomorphism of a nonzero finite-dimensional complex vector space has an eigenvector, so choose an eigenvector of $x$ in $E_{\lambda }$. It is a common eigenvector for $\mathfrak{g}=\mathfrak{a}+\mathbb{C}x$. This proves the Lie theorem.

For the printed matrices in characteristic $p$, $[x,y]=x$, and there is no common eigenvector. Index the standard basis by $e_0,\dots ,e_{p-1}$. The cyclic entry in the PDF givesThe $p$ diagonal eigenvalues of $y$ are distinct in $k$. Thus every eigenvector of $y$ is a scalar multiple of a single $e_j$. Since $p\ge 2$, $x{e}_j$ is never a scalar multiple of $e_j$, proving the assertion even when $k$ is not an algebraically closed field.

For $1\le j\le p-1$,At the cyclic boundary,Hence $[x,y]=x$. The two-dimensional Lie subalgebra $kx+ky$ has derived algebra $kx$, whose own derived algebra is zero, so it is solvable. It nevertheless has no common eigenvector, including after extending $k$ to its algebraic closure. This is a failure of Lie theorem in positive characteristic. In the proof above, the obstruction is precisely that $\dim W=p$ can vanish as a scalar in $k$.

The derived algebra of a complex solvable Lie algebra is nilpotent. First suppose $\mathfrak{g}\subseteq \mathfrak{gl}(V)$. By the Lie theorem, put every element of $\mathfrak{g}$ in upper triangular form. The diagonal of a commutator of upper triangular matrices is zero, so $\mathfrak{d}=[\mathfrak{g},\mathfrak{g}]$ consists of strictly upper triangular matrices. The Lie algebra of all such matrices is a Nilpotent Lie algebra: if $F_r$ consists of matrices whose entries vanish whenever $j-i<r$, thenTherefore the Lower central series of a Lie algebra of $\mathfrak{d}$ reaches zero. Alternatively, every element of $\mathfrak{d}$ is a nilpotent linear map, and the Engel theorem states that a finite-dimensional Lie subalgebra consisting of nilpotent linear maps is a Nilpotent Lie algebra.

For an abstract complex Solvable Lie algebra $\mathfrak{g}$, apply the preceding result to its Adjoint representation. The Lie algebra $\mathrm{ad}\mathfrak{d}=[\mathrm{ad}\mathfrak{g},\mathrm{ad}\mathfrak{g}]$ is nilpotent. Since $\ker (\mathrm{ad}{|}_{\mathfrak{d}})=\mathfrak{d}\cap Z(\mathfrak{g})$ is central in $\mathfrak{d}$, some term of the Lower central series of a Lie algebra of $\mathfrak{d}$ lies in that central ideal; the next term is zero. Thus $\mathfrak{d}$ itself is nilpotent.

Conversely, every Nilpotent Lie algebra is solvable, since its derived series of a Lie algebra is contained term by term in its Lower central series of a Lie algebra. If $\mathfrak{d}$ is nilpotent, it is therefore solvable, and $\mathfrak{g}/\mathfrak{d}$ is Abelian. More directly, the derived series of a Lie algebra of $\mathfrak{g}$, after its first term, is the derived series of a Lie algebra of $\mathfrak{d}$. Consequently the derived algebra nilpotence criterion is