I'm confused about Engel's Theorem (or one specific version of it):
If ${\displaystyle {\mathfrak {g}}\subset {\mathfrak {gl}}(V)}$ is a Lie subalgebra such that every ${\displaystyle X\in {\mathfrak {g}}}$ is a nilpotent endomorphism and if $V$ has positive dimension, then there exists a nonzero vector $v$ in $V$ such that ${\displaystyle X(v)=0}$ for each $X$ in ${\displaystyle {\mathfrak {g}}}$.
Since the vector space in question is finite dimensional we can choose a basis of $V$ and the elements of $\mathfrak{gl}(V)$ then become $n\times n$ matrices (assuming $V\cong \mathbb{K}^n$ for some field $\mathbb{K}$), denote this space as $\mathfrak{gl}_n(\mathbb{K})$.
Let now $\mathfrak{g}\subset \mathfrak{gl}_n(\mathbb{K})$. In Humphreys book, page 12 (directly after the title "3.3 Proof of Engel's Theorem"), he states:
Recall that single nilpotent linear transformation alsways has at least one eigenvector, corresponding to its unique eigenvalue 0. This is just the case $\mathrm{dim}(\mathfrak{g})=1$ of the following theorem [above stated thorem].
I don't understand how the first sentence describes the same as the second, or to be more precise: Why does setting $\mathrm{dim}(\mathfrak{g})=1$ for $A\in\mathfrak{g}\subset \mathfrak{gl}_n(\mathbb{K})$ make $A$ a nilpotent matrix. With the restriction of $\mathrm{dim}(\mathfrak{g})=1$ I would have expected $A$ to be an element of $\mathbb{K}$, but according to these lecture notes (see proof of Thm. 3.4) if $\mathrm{dim}(\mathfrak{g}) = 1$, then $\mathfrak{g} = \mathbb{K}A$ for $A\in gl_V$, where $gl_V$ is the space of linear operators on $V$ aka $A$ is a $n\times n$ invertible matrix.
It seems that I'm misunderstanding something very fundamental here...
After talking to a TA, I think I've got it.. The Lie algebra $\mathfrak{gl}_n(\mathbb{K})$ is generated by a set $\{b_k\}_{k=1}^{n^2}$. We can choose the $b_k:=e_{ij}$ for $1\le i,j\le n$, where $e_{ij}$ has a $1$ at position $(i,j)$ and zeros else.
If we now consider $\mathfrak{g}\subset \mathfrak{gl}_n(\mathbb{K})$ and demand $\mathrm{dim}(\mathfrak{g})=1$, we will have only one of these $e_{ij}$ as basis (assuming $\mathfrak{g}$ is chosen in such a way that it actually can be spanned by only one of these basis vectors and not a linear combination), aka $\mathfrak{g}=\mathbb{K}e_{ij}\cong \mathbb{K}A$ for $A\in gl_V$, as claimed by the lecture notes.