Inverse eigenvalues theorem

In numerical analysis and linear algebra, the Inverse eigenvalues theorem states that, given a matrix A that is nonsingular, with eigenvalue ${\displaystyle \lambda >0}$, ${\displaystyle \lambda }$ is an eigenvalue of ${\displaystyle A}$ if and only if ${\displaystyle \lambda ^{-1}}$ is an eigenvalue of ${\displaystyle A^{-1}}$.

Proof of the Inverse Eigenvalues Theorem

Suppose that ${\displaystyle \lambda }$  is an eigenvalue of A. Then there exists a non-zero vector ${\displaystyle x\in R^{n}}$  such that ${\displaystyle Ax=\lambda x}$ . Therefore:

${\displaystyle x=I_{n}x=A^{-1}Ax=A^{-1}\lambda x=\lambda A^{-1}x}$ 

Since A is non-singular, null(A) = {0} and so ${\displaystyle \lambda \neq 0}$ . Therefore we may multiply both sides of the above equation by ${\displaystyle \lambda ^{-1}}$  to get that ${\displaystyle A^{-1}x=\lambda ^{-1}x}$ ; i.e., ${\displaystyle \lambda ^{-1}}$  is an eigenvalue of ${\displaystyle A^{-1}}$ . By repeating the previous argument but with A replaced by ${\displaystyle A^{-1}}$  we see that if ${\displaystyle \lambda ^{-1}}$  is an eigenvalue of ${\displaystyle A^{-1}}$  then ${\displaystyle \lambda }$  is an eigenvalue of A.