$$Tr(A|A^T U \Sigma U^T|) \leq Tr(AA^T U \Sigma U^T)$$ where $A \in \mathbb{R}^{m\times n}$ is a real rectangular matrix, $U \in \mathbb{R}^{n \times n}$ is a orthonormal matrix and $\Sigma$ is a nonnegative diagonal matrix. $|\cdot|$ represents the elementwise absolute value. Does this inequality holds? If so, is this a known inequality in some fields? How to prove this inequality in a precise way? Thank you very much!
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$\begingroup$ Two remarks : the matrix $U\Sigma U^\top$ can be any semi-definite positive symmetric matrix $S$. And if $A=I_n$, the right inequality is $\mathrm{Tr}(|S|) \ge \mathrm{Tr}(S)$. $\endgroup$– Christophe LeuridanCommented Sep 12, 2023 at 18:59
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The inequality fails when we take non particular cases, for example non diagonal matrices. Say $U \Sigma U^T>0$, and $S=(U \Sigma U^T)^{-1}$. The inequality is equivalent to prove $$\text{Tr}(SB^T|B|) \leq \text{Tr}(SB^TB),$$ where $B=A^TU \Sigma U^T$. Take $S=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ and $B=\begin{pmatrix}0&0\\10&-1\end{pmatrix}$
