Following the hint, we have A − 1 A x = A − 1 λ x . {\displaystyle A^{-1}Ax=A^{-1}\lambda x.} That is, x = λ A − 1 x {\displaystyle x=\lambda A^{-1}x} . Dividing both sides by λ {\displaystyle \lambda } , we have 1 λ x = A − 1 x {\displaystyle {\frac {1}{\lambda }}x=A^{-1}x} . Thus, 1 λ {\displaystyle {\frac {1}{\lambda }}} is an eigenvalue of A − 1 {\displaystyle A^{-1}} with eigenvector x {\displaystyle {x}} . The eigenvlalue-eigenvector pair is ( 1 λ , x ) {\displaystyle \color {blue}({\frac {1}{\lambda }},x)} .
