For an $n$-dimensional Lie algebra, is there always a matrix representation $\{M\}$ and a single vector $v$ such that $\{Mv\}$ is $n$-dimensional? For an $n$-dimensional Lie algebra $L$, is there always a matrix representation $\rho:L\to\mathfrak{gl}(V)$ and a single vector $v\in V$ such that $\{\rho(x)v\mid x\in L\}$ is an $n$-dimensional subspace of $V$?
This would necessarily be a faithful representation.
I'm focusing on Lie algebras over $\mathbb R$, but more general answers are welcome.
This might have something to do with weights or Whitehead's lemma, but I don't know enough about representation theory to be sure.
 A: Yes, as corollary of Ado's theorem.
First observe that for given $v$, the map $L_{\rho,v}:x\mapsto\rho(x)v$ is linear. The goal is to show that there exists $(\rho,v)$ for which this map is injective. Indeed choose $(\rho,v)$ for which this map has kernel $K(\rho,v)$ of minimal dimension. 
Assume by contradiction that there exists a nonzero $x$ in $K(\rho,v)$. Let $\rho'$ be a faithful finite-dimensional representation (as ensured by Ado's theorem). Then there exists $v'$ in the space of $\rho'$ such that $\rho'(x)v'\neq 0$. Hence $(\rho\oplus\rho',v\oplus v')$ contradicts the minimality.
The argument, suitably reformulated, shows that $n$ being the dimension of $\mathfrak{g}$, then $\rho^{\oplus n}$ possesses a vector with the required property for any faithful representation $\rho$. Also note that the proof is completely self-contained (i.e., doesn't use Ado's theorem) if we assume beforehand that $\mathfrak{g}$ has a faithful linear representation, which for instance is a trivial fact when it has a trivial center.
