I have a trouble to prove the following
let $I_n$ be the $n\times n$ identity matrix and $A$ is $n\times n$ positive semi-definite matrix and $b$ is a positive scalar.
Now, can we say $I_n-(I_n+bA)^{-1}$ is also positive definite?
We may assume A is non singular. then how can I prove this?
On
At the risk of setting too much abstraction against the OP: The answer remains Yes in the infinite-dimensional situation, i.e., $\,A\,$ is a bounded linear operator acting on a Hilbert space, and moreover if no eigenvectors at all exist.
$C=bA\,$ is then a positive definite operator, and the further proof reads as if we were dealing with positive reals: $$ 0<C\;\implies\; I<I+C\;\implies\; (I+C)^{-1}<I\;\implies\; -I<-(I+C)^{-1} \\[1.5ex] \implies\;0<I-(I+C)^{-1} $$
Ok, This seems to mean it is true. Let $(\lambda, x)$ be an eigenpair of the matrix $A$, with $\lambda > 0$. Then we know that $$ (I_n + b A)^{-1}x = \frac{1}{1 + b \lambda} x. $$ Also, we then see $$ [I_n - (I_n + b A)^{-1}]x = \left(1 - \frac{1}{1 + b \lambda} \right) x. $$ Since $b, \lambda > 0$ we see that the eigenvalue is also positive. Since eigenpair was arbitrary, this proves the claim.