= Solution
\b[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 $\mathfrak g\ne0$ is <solvable Lie algebra>[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 $v\ne0$ 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 $v,xv,x^2v,\ldots$. If $m=\dim W$, the first $m$ of these vectors form a <basis>, and $W$ is $x$-invariant. We claim that for each $a\in\mathfrak a$,
$$
a x^jv=\lambda(a)x^jv+\operatorname{span}\{v,xv,\ldots,x^{j-1}v\}.
$$
For $j=0$ this is the definition of $v$. For the induction step, use $ax=x a+[a,x]$ 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 $[x,a]\in\mathfrak a$ gives
$$
0=\operatorname{tr}_W[x,a]=m\lambda([x,a]).
$$
Here <characteristic zero> is essential: $m\ne0$ in the field, so $\lambda([x,a])=0$. Now the common <weight space>
$$
E_\lambda=\{w\in V:aw=\lambda(a)w\text{ for all }a\in\mathfrak a\}
$$
is nonzero and $x$-invariant. Indeed, for $w\in E_\lambda$,
$$
a(xw)=x(aw)+[a,x]w=\lambda(a)xw+\lambda([a,x])w=\lambda(a)xw.
$$
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 $\mathfrak g=\mathfrak a+\mathbb Cx$. This proves the <Lie theorem>.
\b[For the printed matrices in characteristic $p$, $[x,y]=x$, and there is no common eigenvector.] Index the standard <basis> by $e_0,\ldots,e_{p-1}$. The cyclic entry in the PDF gives
$$
x e_j=e_{j-1}\quad(j\text{ modulo }p),\qquad ye_j=je_j.
$$
The $p$ diagonal <eigenvalues> of $y$ are distinct in $k$. Thus every <eigenvector> of $y$ is a scalar multiple of a single $e_j$. Since $p\ge2$, $xe_j$ is never a scalar multiple of $e_j$, proving the assertion even when $k$ is not an <algebraically closed field>.
For $1\le j\le p-1$,
$$
[x,y]e_j=j e_{j-1}-(j-1)e_{j-1}=e_{j-1}.
$$
At the cyclic boundary,
$$
[x,y]e_0=-(p-1)e_{p-1}=e_{p-1}.
$$
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 Lie algebra>[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 $k$.
\b[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 $j-i<r$, then
$$
[F_r,F_s]\subseteq F_{r+s},\qquad F_{\dim V}=0.
$$
Therefore 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> $\operatorname{ad}\mathfrak d=[\operatorname{ad}\mathfrak g,\operatorname{ad}\mathfrak g]$ is <nilpotent Lie algebra>[nilpotent]. Since $\ker(\operatorname{ad}|_{\mathfrak d})=\mathfrak d\cap Z(\mathfrak g)$ 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 Lie algebra>[nilpotent].
Conversely, every <nilpotent Lie algebra> is <solvable Lie algebra>[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 Lie algebra>[nilpotent], it is therefore <solvable Lie algebra>[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
$$
\boxed{\mathfrak g\text{ solvable}\quad\Longleftrightarrow\quad[\mathfrak g,\mathfrak g]\text{ nilpotent}.}
$$
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