So what can we say about the relationship between eigenvalues and eigenvectors of square invertible $A$ and its inverse $A^{-1}$?
We know that $A$ is invertible iff all its eigenvalues are nonzero, thus we have $Ax=\lambda x$ iff $A^{-1}x = \frac{1}{\lambda}x$.
But is this all? Does anyone see anything more? (Guessing you do…)
Best Answer
One conclusion you can make is that all eigenvectors of $A$ are eigenvectors of $A^{-1}$ as well, and vice versa. As you noted, the corresponding eigenvalues (for the same eigenvector) are inverses of one another.