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.



Weak majorizations for sum of two hermitian matrices
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.


