Chapter 7 - Eigenvalues and Eigenvectors - 7.1 The Eigenvalue/Eigenvector Problem - Problems - Page 445: 41

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Since $A$ is invertible, hence $A^{-1}$ exists. We obtain $Av=\lambda v\\ \rightarrow A^{-1}(Av)=A^{-1}(\lambda v) \\ \rightarrow A^{-1}(Av)=\lambda (A^{-1} v) \\ \rightarrow v=\lambda A^{-1} v\\ \rightarrow A^{-1} v=\frac{1}{\lambda}v$ Hence $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$.