Given three positive semi-definite matrices $A, B, C$. Show $\operatorname{Tr}(AB) \leq \operatorname{Tr}(AC)$ where $B \preceq C$?
This inequality is the matrix form of multiplying a positive number to both sides of an equality.
My attempt:
Since $A$ is P.S.D $A=A^{1/2}A^{1/2}$ so $\text{Tr}(AB)=\text{Tr}(A^{1/2}BA^{1/2})$, I need to show $\text{Tr}(A^{1/2}BA^{1/2}) \leq \text{Tr}(A^{1/2}CA^{1/2})$ using $B \preceq C$ which I stuck.
Also, if you can show it differently please add that method as well but please first complete my answer.
Best Answer
Your first step is good. Since $$ x^H A^{1/2}BA^{1/2} x=(A^{1/2} x)^HB(A^{1/2} x)\leq (A^{1/2}x)^H C(A^{1/2} x)=x^{H}A^{1/2}C A^{1/2} x, $$ we have $A^{1/2}BA^{1/2}\preceq A^{1/2}CA^{1/2}$. Thus $$ \mathrm{Tr}(A^{1/2}BA^{1/2})=\sum_{k=1}^n e_k^H A^{1/2}BA^{1/2} e_k\leq \sum_{k=1}^n e_k^H A^{1/2}CA^{1/2} e_k=\mathrm{Tr}(A^{1/2}CA^{1/2}), $$ where $(e_1,\dots,e_n)$ is an orthonormal basis of $\mathbb{C}^n$.