# Show 1/lambda = eigenvalue of A^-1

If $A\in \mathbb{K}^{n\times n}$ is nonsingular and $\lambda \in \mathbb{K}$ is an eigenvalue of $A$ then, $\lambda\neq 0$ (why?) and there exists $0\neq x \in \mathbb{K}^n$ column vector such that $Ax=\lambda x$ . Now, multiply both sides by $A^{-1}$ .