Let matrices $A, B$ be positive semidefinite. Can we prove that $A(I+BA)^{-1}$ is positive semidefinite?
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2 Answers
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By density, it is enough to prove the property when $A$ is positive definite. Then $$A(I+BA)^{-1}=A^{1/2}(I+A^{1/2}BA^{1/2})^{-1}A^{1/2}$$ is congruent to $(I+A^{1/2}BA^{1/2})^{-1}$, which itself is positive definite because $I+A^{1/2}BA^{1/2}\succeq I$.
answered Feb 26, 2018 at 12:26
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Consider the change of basis $C^tA(I+BA)^{-1}C$ for $C=I+BA$. One gets $A+ABA$, which is positive semidefinite being a some of two positive semidefinite.
The most difficult part is to show that $I+BA$ is invertible. If not, then let $v$ be in its kernel. By taking the scalar product with $Av$, one gets that $v$ must be in the kernel of $A$, which implies $v=0$.
