In physics, when considering the motion of a system with $N$ degrees of freedom described by vector $x$, the linearized equations of motion take the form $$M \ddot{x} = - K x.$$ Here, $M$ is a symmetric (in most cases diagonal), positive definite matrix, and $K$ is a symmetric, (in general) indefinite matrix. Using the standard ansatz $x(t) \propto e^{i\omega t}$, we have $\ddot{x} = -\omega^2 x$, which, in turn, leads to the eigenvalue equation $$\omega^2 M x = K x.$$
Usually, this problem is solved by simultaneously diagonalizing both $M$ and $K$. However, given that $M$ is positive definite, wouldn't it make more sense to write the eigenvalue problem as $$M^{-1} K x = \omega^2 x$$ and solve it the usual way? Wouldn't the solutions (eigenvalues and eigenvectors) necessarily be the same?