Inverse eigenvalues theorem

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

Proof of the Inverse Eigenvalues Theorem

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

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

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