How to prove the eigenvalues of $(A+B)^{-1}A$ are within $[0, 1)$, when $A$ is positive semidefinite and $B$ is positive definite?

Question

The eigenvalue $\lambda_i$ of $(A + B)^{-1}A$ is within $[0,1)$, where $A$ is positive semidefinite and $B$ is positive definite. How to prove this?

Thank you in advance.

Answer 1

Let $(A+B)^{-1}A x=\lambda x$ with $x \neq 0$. Then $Ax=\lambda (Ax+Bx)$ which gives $(1-\lambda) \langle Ax, x \rangle =\lambda \langle Bx, x \rangle$. Can you finish?

Answer 2

I figured it out.

$(A+B)^{-1}Ax = \lambda x$ and multiply $x^T(A+B)$ on the left to get:

$x^TAx = \lambda x^T(A + B)x$ and now it's easy to see the range of $\lambda$.