Finding minimum $\alpha > 0$ so that $\det(A - \alpha B) = 0$ for positive definite $A,B$ Given two positive definite symmetric matrices $A,B$, I'd like to find the minimum $\alpha > 0$ such that $A - \alpha B$ is singular, i.e., the threshold where $A - \alpha B$ is no longer positive definite. An algorithmic approach is ok if it's too hard to come up with a formula.

A little different, there are a number of ways, including Cholesky decomposition, to write $$ B = C^T C,  $$ so that, with $G = C^{-1},$ we have
$$ G^T B G = I.  $$ Then solve
$$  \det \left( G^T A G - \alpha I \right) = 0.  $$ As $C$ is upper triangular, finding $G$ is not difficult. If you use some non-Cholesky method and $C$ is not triangular, then you have a little extra work finding $G.$
The point being that $G^T A G$ is still symmetric positive definite and eigenvalues are all real and just  a bit easier to deal with. 
