Let $A$ and $B$ be two $n\times n$ hermitian matrices. Does $U^{*}AU+B \prec_{w} A+B$ for any unitary matrix $U$? Here the notation $``\prec_{w}"$ stands for the weak majorization, that is, $x\prec_{w} y$ if and only if $\sum\limits_{j=1}^{k}\lambda_{j}^{\downarrow}(x)\leq \sum\limits_{j=1}^{k}\lambda_{j}^{\downarrow}(y)$, for each $1\leq k\leq n$, and $\{\lambda_{j}^{\downarrow}(x)\}_{j=1}^{n}$ and $\{\lambda_{j}^{\downarrow}(y)\}_{j=1}^{n}$ are eigenvalues of $x$ and $y$ ordering in the non-increasing order respectively.
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Consider $A = \left[\begin{matrix} 1&0 \\ 0&0\end{matrix}\right]$, $B = \left[\begin{matrix} 0&0 \\ 0&1\end{matrix}\right]$ and $U = \left[\begin{matrix} 0&1 \\ 1&0\end{matrix}\right]$
Then $A+B = \left[\begin{matrix} 1&0 \\ 0&1\end{matrix}\right]$ and $U^*AU + B = \left[\begin{matrix} 0&0 \\ 0&2\end{matrix}\right]$.
Therefore, $U^*AU + B \succ A+B$ and $U^*AU + B \nprec_w A+B$.
I guess the easy answer to your question is that when $U=I$ everything is fine. In general, most unitaries will not satisfy this weak majorization order.
answered Nov 30, 2021 at 16:34
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$\begingroup$ Many thanks for your wonderful example! And I have a further question: Dose $\sum_{j=1}^{n}|\lambda_{j}(UAU^{*}+B)|=\sum_{j=1}^{n}|\lambda_{j}(A+B)|$ hold true? The example you cnstructed above do satisfy this relation. Here $(\lambda_{j}(C))_{j=1}^{n}$ are eigenvalues of a $n\times n$ hermitian matrix $C$. $\endgroup$ Commented Dec 1, 2021 at 9:12
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$\begingroup$ Ha, I think I have found out a counterexample. Let $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$, $B=\begin{pmatrix}2&0\\0&0\end{pmatrix}$ and $U=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. It is easy to verify that $\sum_{j=1}^{n}|\lambda_{j}(A+B)|\not=\sum_{j=1}^{n}|\lambda_{j}(U^{*}AU+B)|$. Thank you for your enlightening example! @Chris Ramsey $\endgroup$ Commented Dec 1, 2021 at 10:56