Consider a positive-definite matrix $A \in \mathbb{R}^{n \times n}$ and a positive semi-definite matrix $B \in \mathbb{R}^{n \times n}$, and consider the $i$-th largest eigenvalue of $(A + B)^{-1} B$. I know all eigenvalues of this matrix are less than 1, and is it possible to obtain a non-trivial upper bound on $\lambda_i(A + B)^{-1} B)$ that is less than 1?
I've tried Weyl's inequality and a conclusion from Bathias' "Matrix Analysis" that says, for any two operators $A, B$ on Hilbert space $H$ with dimension $n$, for all $i, j$ such that $i+j \leq n+1, \lambda_{i+j-1}(A B) \leq \lambda_i(A) \lambda_j(B)$, where $\lambda_i(A)$ is the $i$-th largest eigenvalue of $A$. But the bound I obtain in terms of the eigenvalues of $A$, $B$ is trivial - the bound I obtain may be larger than 1.
Any help is greatly appreiciated.
Let $B=\pmatrix{B_1&0\\ 0&0}$ and $A=\pmatrix{X&Y^T\\ Y&Z}$, where $B_1$ is an $r\times r$ positive definite matrix. Then $$ (A+B)^{-1}B =\pmatrix{(B_1+S)^{-1}&\ast\\ \ast&\ast}\pmatrix{B_1&0\\ 0&0} =\pmatrix{(I+SB_1^{-1})^{-1}&0\\ \ast&0}, $$ where $S=X-Y^TZ^{-1}Y$ the Schur complement of $Z$ in $A$. Therefore, when $i,j,k\in\{1,2,\ldots,r\}$ and $i+j=k+1$, \begin{align} \lambda^{\downarrow}_k\left((A+B)^{-1}B\right) &=\frac{1}{1+\lambda^{\uparrow}_k(SB_1^{-1})} =\dfrac{1}{1+\dfrac{1}{\lambda^{\downarrow}_k(S^{-1}B_1)}}\\ &\le\dfrac{1}{1+\dfrac{1}{\lambda^{\downarrow}_i(S^{-1})\lambda^{\downarrow}_j(B_1)}}\\ &\le\dfrac{1}{1+\dfrac{1}{\lambda^{\downarrow}_i(A^{-1})\lambda^{\downarrow}_j(B_1)}} =\dfrac{1}{1+\dfrac{\lambda^{\uparrow}_i(A)}{\lambda^{\downarrow}_j(B)}} \end{align} where the first inequality is the one mentioned in your question and the second one is due to the fact that $S^{-1}$ is a principal submatrix of $A^{-1}$.