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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: $$t^* := \inf_{0 \le t < \infty} \{t :\text{ $A-tB$ has $0$ as an eigenvalue}\} \implies A-t^* B \text{ is positive semi-definite}$$

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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.

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$A-tB=B^{1/2}( B^{-1/2}AB^{-1/2} - tI )B^{1/2}$, 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 $\rho(AB^{-1})$.

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