Simultaneous diagonalization of two symmetric matrices vs. diagonalization of one nonsymmetric matrix 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?
 A: First, although it may appear as though $M^{-1} K$ is not necessarily diagonalizable because it isn't symmetric, it is diagonalizable because it is similar to a symmetric matrix. To see this, note that because $M$ is positive definite, it has a positive definite square root $M^{½}$ and
$$
M^{-1} K = M^{-½} (M^{-½} K M^{-½}) M^{½} = M^{-½} \tilde{K} M^{½},
$$
i.e., $M^{-1} K$ is similar to the symmetric matrix $\tilde{K}$.
So yes, the seemingly-nonsymmetric eigenvalue problem $M^{-1} K x = \omega^2 x$ would give the same solutions as the generalized eigenvalue problem $Kx = \omega^2 M x$.
One way to keep everything symmetric is to multiply $Kx = \omega^2 M x$ through with $M^{-½}$:
$$
M^{-½} K x = \omega^2 M^{½} x,
$$
and perform the change of variable $\tilde{x} = M^{½} x$:
$$
\tilde{K} \tilde{x} = \omega^2 \tilde{x}.
$$
(Note that you could replace $M^{½}$ everywhere with the Cholesky factor of $M$.)
There exists a large variety of generalized eigenvalue solvers. Some proceed iteratively and do not need to compute $M^{-1}$, $M^{½}$ or $M^{-½}$ to solve the problem—they only require products with $M$ and $K$. Others perform factorizations of certain matrices. Given the nature of your particular problem, one may be a better fit than others, but it is difficult to tell without more information. It's not because $M$ is positive definite that transforming the problem is a good idea. The entire class of symmetric-definite generalized eigenvalue problems deals with the case where $M$ is positive definite and it contains many numerical methods designed to exploit somewhat the structure of $K$ and $M$.
