I am given two $n\times n$ complex matrices $A$ and $B$. $A> 0$ ($A$ is positive definite) and $B$ is Hermitian. By the properties of positive definite matrices, it follows that: $A=T^*T$ for some invertible matrix $T$. My question is: why is $(T^{-1})^{*}B(T^{-1})\geq 0 $? (i.e why is $(T^{-1})^{*}B(T^{-1})$ positive semidefinite?).
I tried the following: for any $x\in \mathbb{C}^{n}$: $x^*(T^{-1})^{*}B(T^{-1})x=\left \langle B(T^{-1})x,(T^{-1})x \right \rangle=\left \langle By,y \right \rangle$ where $y=(T^{-1})x$. I need to prove that $\left \langle By,y \right \rangle\geq 0$, but I can't see how since $B$ is Hermitian and not positive semidefinite. Any help? Thanks.