# Positive semidefinite matrices

• May 4th 2010, 04:40 PM
math8
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.
• May 5th 2010, 12:57 AM
Opalg
• May 5th 2010, 07:20 AM
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?
• May 5th 2010, 07:59 AM
math8
Also, how do you prove that if $T^{*}T \leq I$ then $TT^{*} \leq I$?
• May 5th 2010, 01:43 PM
Opalg
Quote:

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.

Quote:

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

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