Clarification about eigenvalues

Question

Let's consider $(K-\omega^2M)u=0$ where K and M are square matrices positive definite and u is a vector.

Non trivial solution are given by the values of $\omega^2$ such that $\det (K-\omega^2M)=0$. Why do these values correspond to the eigenvalues of $M^{-1}K?$ I have only have heard about eigenvalues in the form Ax=b; where x and b are vectors...

Answer 1

If $M^{-1}$ exists then simply multiply $(K-\omega^2 M)u=0$ on the left by $M^{-1}$ to get $(M^{-1}K-\omega^2 I)u=0$.