Getting a positive semi-definite matrix from two positive definite matrices

Question

Suppose I have two positive-definite Hermitian matrices $A$ and $B$. Their eigenvalues are strictly positive reals.

Consider the matrices $A-tB$ for $0 \le t < \infty$. My goal is to conclude that there is some smallest $t$ such that $A-tB$ has zero as an eigenvalue, and all other eigenvalues non-negative. How do I show this (and is this even true)? I know some results about continuity (the eigenvalues of a convergent sequence of matrices also converge), but am not sure if such a $t$ exists: $$t^* := \inf_{0 \le t < \infty} \{t :\text{ $A-tB$ has $0$ as an eigenvalue}\} \implies A-t^* B \text{ is positive semi-definite}$$

Answer 1

I am not exactly sure what you are asking, but the set of positive semi-definite matrices is a convex cone in the space of hermitian matrices (viewed as a vector space), and so if you connect a matrix inside the cone to the matrix outside the cone, that line segment will intersect the boundary of the cone in exactly one point, which is the matrix you want.

Answer 2

$A-tB=B^{1/2}( B^{-1/2}AB^{-1/2} - tI )B^{1/2}$, which is congruent to $B^{-1/2}AB^{-1/2} - tI$. Therefore, by Sylvester's law of interia, $A-tB$

• is positive definite whenever $0\le t<\rho(B^{-1/2}AB^{-1/2})=\rho(AB^{-1})$,
• is positive semidefinite but singular when $t=\rho(AB^{-1})$, and
• has a negative eigenvalue when $t>\rho(AB^{-1})$.

Therefore, the only $t$ (and hence the minimum $t$) such that $A-tB\succeq0$ and $A-tB$ is singular is $\rho(AB^{-1})$.