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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.

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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 $\mathbb{C}^n$.

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  • $\begingroup$ Could you show it differently that $\text{Tr(X)}\leq\text{Tr(Y)}$ when $X \preceq Y$? $\endgroup$ – Saeed Jan 15 at 16:39
  • $\begingroup$ Well, if you know that a psd matrix has nonnegative eigenvalues, then you can use $0\preceq Y-X$ to conclude $\mathrm{Tr}(Y-X)\geq 0$. $\endgroup$ – MaoWao Jan 15 at 16:43
  • $\begingroup$ This is exactly my question. I know this. How can I get the last inequality in a shortest way without summation? $\endgroup$ – Saeed Jan 15 at 16:46
  • $\begingroup$ What is your definition of the trace? If it's the sum of eigenvalues, then see my last comment. If it's the sum of diagonal entries, then see my answer. In any case, there is gonna be some sum involved. $\endgroup$ – MaoWao Jan 15 at 16:48

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