# Semidefinite Hermitian Matrices

If $A$ and $B$ are positive semidefinite matrices, we write $A\geq B$ in case $A-B$ is positive semidefinite. Throughout this problem, $A$ and $B$ are Hermitian matrices. Show that

• $C^{\ast}AC\geq C^{\ast}BC$ for all $C\in M_{n}(\mathbb{C)}$
• If $A\geq B$ and if $B$ is invertible, then so is $A$ and $B^{-1}\geq A^{-1}$.

I don't know how to start with either of the problems. For the first one, I have $$C^*A^*C=AC^2,$$ so $$AC^2\geq BC^2.$$ I don't know if this is helpful.

• By $C^\ast$ do you mean what is usually written $C^\dagger$, the hermitian adjoint of $C$, or $\bar C$, its complex conjugate? Note $C^\dagger = (\bar C)^T$. Regards. – Robert Lewis Dec 12 '14 at 18:35
• @RobertLewis I do mean the complex conjugate. – Dia McThrees Dec 12 '14 at 18:36

Your approach for the first assumes that $A$ and $C$ commute. So no, your approach here is not helpful in this case.
Hint: consider $C^*(A-B)C$. Why is this matrix Hermitian? Why is this matrix positive semidefinite?
Possible approach for the second problem: first, verify that $A$ must also be invertible. Then, note that $A \geq B \iff x^*Ax \geq x^*Bx$ for all $x \in \Bbb C^n$.
• Why is $C^*(A-B)C$ hermitian? – Dia McThrees Dec 12 '14 at 18:40
• @DiaMcThrees as for the question: show that $(C^*(A-B)C)^* = C^*(A-B)C$. Remember that $(PQ)^* = Q^*P^*$ – Omnomnomnom Dec 12 '14 at 18:42
• Thank you. And $C^*(A-B)C$ is positive semidefinite because $A-B$ is, right? – Dia McThrees Dec 12 '14 at 18:44
• Right. That is generally given as a theorem, but you should be comfortable with the proof of the fact that $M$ is PSD implies that $CMC^*$ is also PSD. – Omnomnomnom Dec 12 '14 at 18:46