1. ## Positive semidefinite matrices

Let P and Q be Hermitian positive definite matrices. We prove that for all $x \in \textbf{C}^{n}$ ,
$x^{*}Px \leq x^{*}Qx$ if and only if $x^{*}Q^{-1}x \leq x^{*}P^{-1}x$.

I suppose we need to show that $(Q-P)$ is positive semidefinite iff $(P^{-1} - Q^{-1})$ is also positive semidefinite.
But I am not sure how to proceed.

2. Thanks, I read the post and it makes sense but I still have a few questions:

when you say A has a positive square root $A^{1/2}$, do you mean $A^{1/2}$ is positive definite?
And why do we need it to be positive?

3. Also, how do you prove that if $T^{*}T \leq I$ then $TT^{*} \leq I$?

4. Originally Posted by math8
Thanks, I read the post and it makes sense but I still have a few questions:

when you say A has a positive square root $A^{1/2}$, do you mean $A^{1/2}$ is positive definite?
And why do we need it to be positive?
Yes, that is what it means; and no, for this proof it's not important to take the positive square root of A. In general, a positive definite matrix has many square roots (but only one of them is positive definite), and for this proof any square root would do.

Originally Posted by math8
Also, how do you prove that if $T^{*}T \leq I$ then $TT^{*} \leq I$?
The only way I know to prove that is to use the norm of an operator. Define $\|T\| = \sup\{\|Tx\|:\|x\|\leqslant1\}$, where $\|x\| = (x^*x)^{1/2}$. Then $\|T^*T\|=\|T\|^2$ and $\|T^*\|=\|T\|$. It follows that

$T^{*}T \leqslant I\ \Leftrightarrow\ \|T^*T\| \leqslant1\ \Leftrightarrow\ \|T\|\leqslant1\ \Leftrightarrow\ \|T^*\|\leqslant1\ \Leftrightarrow\ \|TT^*\|\leqslant1\ \Leftrightarrow\ TT^{*} \leqslant I$.