Prove that any eigenvectors of $A^{-1}$ must be in some subspace of $adj(A)$
Since $A^{-1}$=$\frac{1}{detA}ajdA$, so if $\lambda$ is eigenvalue of $A^{-1}$ so it exist $x\not=0$ such that $A^{-1}x=\lambda x$ then $\frac{1}{detA}ajdAx=\lambda x$ since $\frac{1}{det}$ is some scalar then $adjAx=detA\lambda x$ and if I put $\lambda´=detA\lambda$ then $adjAx=\lambda^´ x$, so they have the same eigenvectors, I think that is easy way to show but what you think?