The group $G:=\text{GL}_n(\mathbb C)$ is an open subset of $M_n(\mathbb C)$. A vector field on $G$ is a map $ X:G\to M_n(\mathbb C). $ We denote by $\ell(x)X$ the left action of $x\in G$ on such a vector field $X$, that is, we define the vector field $\ell(x)X$ by $ (\ell(x)X)(y):=xX(x^{-1}y). $ Thus, $X$ is invariant under this action if and only if $X(x)=xX(1)$ for all $x$ in $G$. Such a vector field is said to be left invariant. In particular, for each $a\in M_n(\mathbb C)$ there is a unique left invariant vector field $\widetilde a$ such that $\widetilde a(1)=a$. We have $\widetilde a(x)=xa$ for all $x$ in $G$, and $\widetilde a$ is $C^\infty$. We must check $ \left[\ \widetilde a\ ,\ \widetilde b\ \right]=\widetilde{[a,b]} $ for all $a,b\in M_n(\mathbb C)$.
Let $e_{ij}\in M_n(\mathbb C)$ be the matrix with a one in the $(i,j)$ position and zeros elsewhere. It suffices to verify that the above display holds for $a=e_{ij}$ and $b=e_{pq}$.
We have $ X_{ij}(x):=\widetilde{e_{ij}}(x)=x\ e_{ij}=\sum_{p,q}\ x_{pq}\ e_{pq}\ e_{ij}=\sum_p\ x_{pi}\ e_{pj}. $ If we write $ X_{ij}=\sum_k\ x_{ki}\ \ \frac{\partial}{\partial x_{kj}}\quad,\quad X_{pq}=\sum_r\ x_{rp}\ \ \frac{\partial}{\partial x_{rq}}\quad, $ then the Lie bracket $[X_{ij},X_{pq}]$ is just the commutator of the differential operators $X_{ij}$ and $X_{pq}$, and we get $ [\widetilde e_{ij},\widetilde e_{pq}]=[X_{ij},X_{pq}]=\delta_{jp}\ X_{iq}-\delta_{qi}\ X_{pj}=\widetilde{[e_{ij},e_{pq}]}, $ as was to be shown.