Eigenvalues and eigenvectors of a matrix

Question

We know that if $\lambda (\neq 0)$ is an eigenvalue of a matrix $A$ corresponding to eigenvector $X$, then $\dfrac{1}{\lambda}$ is an eigenvalue of $A^{-1}$. But whether the corresponding eigenvector is $X$ again?

Answer 1

If $Av = \lambda v$, then

$$v = Iv = A^{-1} Av = A^{-1} \lambda v = \lambda A^{-1} v$$

Answer 2

Yes, it is. $Ax = \lambda x \Rightarrow x = A^{-1}\lambda x \Rightarrow \dfrac{1}{\lambda}x = A^{-1}x$.