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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.

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up vote 2 down vote accepted

It is not true. Take for example $A=I_n$, $B=-I_n$. Then you can also take $T=I$, and your assertion claims that $-I_n\geq0$, which is of course false.

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Since $T^{-1}$ is invertible, $(T^{-1})^*BT$ is positive semi-definite if and only if so is $B$. Indeed, $x^*(T^{-1})^*BTx=(T^{-1}x)^*BT^{-1}x$, and if $B$ is positive semi-definite, $x^*(T^{-1})^*BTx\geq 0$ for all $x$. Conversely, if $(T^{-1})^*BT$ is positive semi-definite then for a $x$, we can find $x'$ such that $T^{-1}x'=x$. Then $(x')^*(T^{-1})^*BTx'=x^*Bx\geq 0$.

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