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It is trivial to show that $0 \leq A \leq B \Rightarrow Tr(A^2) \leq Tr(B^2)$, but does this generally hold for all $p >$ 2 as well?

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  • $\begingroup$ They lie in $\mathbb{R}^n$ $\endgroup$ – wemblem Mar 11 '13 at 11:59
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Yes. Let $\lambda_k(X)$ denotes the $k$-th smallest eigenvalue of a Hermitian matrix $X$. Then $0\le A\le B$ implies that $\lambda_k(A)\le\lambda_k(B)$ for all $k$, because of the Courant-Fischer minimax principle. Now $\operatorname{tr}(A^p)\le\operatorname{tr}(B^p)$ is trivial.

Note, however, that it is not necessarily true that $A^p\le B^p$.

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