Chapter 5 - Eigenvalues and Eigenvectors - 5.1 Exercises - Page 274: 25

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Suppose a nonzero vector x satisfies equation: $Ax=\lambda x$ Because A is invertible, we know its eigenvalues are not zero. left multiplying by A inverse, we get $A^{-1}Ax=A^{-1}\lambda x$ $Ix=A^{-1}\lambda x$ $x=A^{-1}\lambda x$ $\frac{1}{\lambda}x=A^{-1} x$. Thus, the eigenvalue of A inverse is the inverse of $\lambda$