Connection one form transformation law

The coefficient column of $e_j^{\mathrm{\prime }}$ in frame $e$ is the $j$th column of $A$. Applying the local formula to every column gives $\mathrm{\nabla }{e}^{\mathrm{\prime }}=e(dA+\omega A)$. By definition in frame $e^{\mathrm{\prime }}$, it is also $e^{\mathrm{\prime }}{\omega }^{\mathrm{\prime }}=eA{\omega }^{\mathrm{\prime }}$. Equality of the coefficients in a basis gives $A{\omega }^{\mathrm{\prime }}=dA+\omega A$.

Left multiplication by $A^{-1}$ gives the asserted formula. For $A=I$ it returns ${\omega }^{\mathrm{\prime }}=\omega$; constant $A$ gives conjugation; in rank one it gives ${\omega }^{\mathrm{\prime }}=\omega +A^{-1}dA$, valid for either sign of the nonzero scalar $A$. In rank zero all matrices are empty and the identity is unique. Empty overlaps impose no condition. Smooth invertibility is needed at every point, so singular frame changes are not admitted.