Why is Engel's theorem equivalent to every element being represented by strictly upper triangular matrices? I've come across 2 definitions of Engels Theorem, I was wondering if somebody could help me understand why they are equivalent.
Engels Theorem (i): A Lie Algebra $L$ is nilpotent if and only if for all $x \in L$ the linear map $ad(x): L \rightarrow L$ is nilpotent.
Engels Theorem (ii): Suppose that $L$ is a lie subalgebra of $gl(V)$ such that every element of $L$ is a nilpotent linear transformation of $V$. There there is a basis of $V$ in which every element of $L$ is represented by a strictly upper triangular matrix.
I just get confused on what is the differece between a Lie Algebra being nilpotent and all its elements being nilpotent? Version $(i)$ has to do with the Lie algebra being nilpotent where as version $2$ has to do with every element being nilpotent.
I'd appreciate some clarification and general insight!! Thanks!
 A: It may help to think about nilpotency in terms of the lower central series
$$L=L^{(1)}\supset L^{(2)}\supset L^{(3)}\supset\dots,\quad L^{(i+1)}=[L,L^{(i)}].$$
Being nilpotent exactly means that there is some $s\in\mathbb{N}$ such that $L^{(s)}=0$.
For the Lie algebra of all strictly upper triangular matrices, the lower central series is very concrete, with each successive term $L^{(i)}$ having more and more zeros above the diagonal. For instance, in the Lie algebra of all $4\times 4$ strictly upper triangular matrices, the lower central series consists of matrices of the forms
$$
\begin{bmatrix}0&\star&\star&\star\\0&0&\star&\star\\0&0&0&\star\\0&0&0&0\end{bmatrix}
\supset
\begin{bmatrix}0&0&\star&\star\\0&0&0&\star\\0&0&0&0\\0&0&0&0\end{bmatrix}
\supset
\begin{bmatrix}0&0&0&\star\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}
\supset
\begin{bmatrix}0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}
$$
Variant (ii) of Engel's theorem can then be seen as a statement about the lower central series. It guarantees that the lower central series must eventually reach zero, since the lower central series for the (possibly larger) Lie algebra of all upper triangular matrices reaches zero. Hence such a Lie algebra is a nilpotent.
A: The second version of Engel's theorem implies the first one by an induction. The induction step uses that, if every $x$ is a nilpotent endomorphism, there exists a nonzero $v\in V$ such that $\mathfrak{g}.v=0$. This property is saying that there is a flag of subspaces
$$
0=V_0\subset V_1\subset \cdots \subset V_n=V
$$
with $\dim (V_i)=i$ and $\mathfrak{g}.v_i\subset V_{i-1}$ for $i=1,\ldots ,n$. So there is a basis such that the matrices of elements in $\mathfrak{g}$ are all strictly upper-triangular.
If we consider the adjoint representation of an abstract Lie algebra $L$, then $ad(L)$ is a subalgebra of $\mathfrak{gl}(L)$. Then $L$ nilpotent means that all matrices $ad(x)$ are strictly upper-triangular with respect to this flag basis.
A: Why these are different: you can have a nilpotent Lie algebra without all of its elements being nilpotent. A stupid way to do this would be to take a subalgebra $L$ of $\mathfrak{gl}(\mathbb R^n)$ generated by all powers of a matrix $A$. These all commute, so $\operatorname{ad}(L) = 0$. But its elements aren't nilpotent.
Why version (ii) implies version (i): If $\operatorname{ad}(x)$ is nilpotent for all $x \in L$, then we can apply the second form of the theorem to $\operatorname{ad}(L)$, rather than to $L$. Once we have a basis for $\operatorname{ad}(L)$ in which everything is strictly upper triangular, then $\operatorname{ad}(L)$ is doomed to be nilpotent, and then so is $L$.
