The structure constants of a Lie algebra are determined by $[T^a, T^b] = f^{ab}\,_c T^c$. As a $(2,1)$ tensor, they can however be considered as a linear map from vectors to antisymmetric tensors.
Is this map injective, for the Lie algebra $\mathfrak{su}(n)$? That is, given $f^{ab}\,_c v^c = A^{ab}$, where $A$ can be obtained from some such $v$, can we solve for $v$? And is there an explicit expression for $v$?
Edit: To be even more explicit, my question is whether or not the map $F:v \mapsto A$, where $A^{ab} = \sum_c f^{ab}\,_c v^c$, is injective. Here $v \in \mathbb{R}^{n^2-1}$ is simply some real vector, and $A$ is an antisymmetric tensor acting on the same space. ($a,b,c$ run over $1,\ldots,n^2-1$.)
Example: For $\mathfrak{su}(2)$, $f^{ab}\,_c = \epsilon^{ab}\,_c$ is the standard 3D bijection between vectors and antisymmetric tensors, which is invertible. The inverse is given simply by $\epsilon_{ab}\,^c$ up to constants.
(In higher than 3 dimensions, of course, there are more antisymmetric tensors than vectors, which suggests in a hand-waving sense that the map is 'likely' to be injective.)