Consider two hermitian matrices $A$ and $B$. Suppose that there exists a unitary matrix $U$
such that $A+B$ is unitarily equivalent to $U A U^* +B$. Does this imply that there exists a unitary matrix $V$ such that $UAU^* + B = VAV^* + B$ and $VBV^* = B$?
More generally, I am interested in when $A+B$ is unitarily equivalent to $UAU^* + WBW^*$ for unitary matrices $U,W$.
I'd be happy about proof hints, counter-examples or simply links to useful references.
Update (after Kurt G.'s comment): Here is an example where $V$ exists, but it's not immediately obvious. Let $A=\sigma_x,B=\sigma_y,U=\sigma_z$, with the Pauli-matrices $\sigma_i$. Then $A+B=\sigma_x+\sigma_y$ is unitarily equivalent to $UAU^* + B=-\sigma_x + \sigma_y$. However, $U$ does not commute with $B$.
(In particular, it is not true that $U(A+B)U^* = A+B$.)
Nevertheless, the choice $V=\sigma_y$ works in this case.
In fact, $V$ also realizes the unitary equivalence between $A+B$ and $UAU^*+B$.
Best Answer
The claim is wrong. I.e., $A+B$ being equivalent to $UAU^* + B$ does in general not imply that there exists a unitary $V$ such that $VAV^*=UAU^*$ and $[V,B]=0$. A counter-example can be constructed as follows: Choose $A>0$ (in particular invertible), diagonal and with non-degenerate spectrum. Let $W$ be a unitary and choose $B=WAW^*$. Then: $$W(A+B)W^* = WAW^* + W W A W^* W^* = WWAW^*W^* + B .$$ Suppose now that a unitary $V$ as above exists. One can then show that $V= W^2 D$ with $D$ diagonal and unitary. As a consequence, $[V,B]=0$ translates to $WDW$ being diagonal and unitary. Thus it is sufficient to find a unitary $W$ which does not allow for a diagonal unitary matrix $D$ such that $WDW$ is diagonal. An example in 3 dimensions is as follows: $W$ is the cyclic shift on the canonical basis vectors of $\mathbb C^3$ acting as $W\vec e_{i}=\vec e_{i+1}$ (with $\vec e_4=\vec e_1$). Then $WDW \vec e_i = d_{i+1} \vec e_{i+2}$, where $d_i$ are the (diagonal) entries of $D$. Of course this example genralizes to any dimension larger than $2$.
As a side-remark, let me mention that the claim does hold true if $A$ and $B$ are projections. This follows from the general form of pairs of projections. In particular, this generalizes the Pauli-example discussed above, since Pauli-matrices are projections shifted by the identity.