How to show $\text{Tr}(AB) \leq \text{Tr}(AC)$ where $B \preceq C$?

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.

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 $$\mathbb{C}^n$$.
• Could you show it differently that $\text{Tr(X)}\leq\text{Tr(Y)}$ when $X \preceq Y$? – Saeed Jan 15 at 16:39
• Well, if you know that a psd matrix has nonnegative eigenvalues, then you can use $0\preceq Y-X$ to conclude $\mathrm{Tr}(Y-X)\geq 0$. – MaoWao Jan 15 at 16:43