Let $A,B\in\mathbb{R}^{n\times n}$ be two positive definite matrices. Let $\{\lambda_i\}_{i=1}^n$, $\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_n$, denote the ordered spectrum of $A^{-1} B$ and assume $\lambda_1\ne \lambda_n$. Let us define $$ X_t := \frac{\lambda_1^t-\lambda_n^t}{\lambda_1-\lambda_n}B + \frac{\lambda_1\lambda_n^t-\lambda_n\lambda_1^t}{\lambda_1-\lambda_n}A, \quad t\ge 0. $$
My question. Is $X_t$ positive definite for all $t\ge 0$?
I've run some random numerical simulations and I've found a couple of counterexamples to my question. However, in the counterexamples I found, the corresponding $X_t$ is very close to be singular so I'm not completely sure if these are "real" counterexamples or if they are the result of numerical errors. Hence, I would like to analytically check this fact. Thanks for your help!
Recall that $\lambda_1>\lambda_n>0$.
Since $A^{-1}B$ is similar to $A^{-1/2}BA^{-1/2}$, we deduce that $A^{-1/2}BA^{-1/2}\geq \lambda_nI_n$, that is equivalent to $B-\lambda_n A\geq 0$. In the same way, $\lambda_1A-B\geq 0$.
We want to prove that, for every $t> 0$, $Y_t=({\lambda_1}^t-{\lambda_n}^t)B+(\lambda_1{\lambda_n}^t-\lambda_n{\lambda_1}^t)A> 0$.
Note that $Y_t={\lambda_1}^t(B-\lambda_nA)+{\lambda_n}^t(\lambda_1A-B)\geq 0$.
Assume that $x^TY_tx=0$; then $x^T(B-\lambda_nA)x=x^T(\lambda_1A-B)x=0$ and, therefore, $x\in\ker(B-\lambda_nA)\cap\ker(\lambda_1A-b)$. Thus $A^{-1}Bx=\lambda_nx=\lambda_1x$ and $x=0$; finally $Y_t>0$ as required.