Identity augmentation in quantum controllability ($\mathfrak{g}+\mathbb{R}iI=\mathfrak{u}(N),\mathrm{Tr}\mathfrak{g}\ne 0\Rightarrow \mathfrak{g}=\mathfrak{u}(N)$)

For $N\ge 2$, suppose a skew-Hermitian Lie algebra becomes all of $\mathfrak{u}(N)$ after adding the scalar identity direction. Scalar matrices commute, so $[\mathfrak{g},\mathfrak{g}]=[\mathfrak{u}(N),\mathfrak{u}(N)]=\mathfrak{su}(N)$ is contained in $\mathfrak{g}$. If $\mathfrak{g}$ contains any nonzero-trace generator, subtracting its traceless part supplies $iI$, and $\mathfrak{g}$ was already all of $\mathfrak{u}(N)$. This permits using identity-augmented commutator calculations without assuming an additional physical identity control.