Unitary equivalence of sums of unitary equivalent hermitian matrices

Question

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 $UAU^*+B$.

Answer 1

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.

Answer 2

The Pauli matrices $$ \sigma_x=\left(\begin{matrix}0&1\\1&0\end{matrix}\right)\,,\quad\sigma_y=\left(\begin{matrix}0&-i\\i&0\end{matrix}\right)\,,\quad\sigma_z=\left(\begin{matrix}1&0\\0&-1\end{matrix}\right) $$ are unitary, Hermitian and satisfy \begin{align}\tag{1} \sigma_x^2=\sigma_y^2=\sigma_z^2=\boldsymbol{1}\,,\quad\quad\sigma_x\,\sigma_y=i\,\sigma_z\,,\quad(\text{anti symmetric in }x,y,z)\,. \end{align} The fact that the matrices square to one can also be written as $$\tag{2} \sigma_x^*=\sigma_x\,,\quad\sigma_y^*=\sigma_y\,,\quad\sigma_z^*=\sigma_z\,. $$ You are looking at \begin{align} \sigma_x+\sigma_y=\left(\begin{matrix}0&1-i\\1+i&0\end{matrix}\right)\text{ and } -\sigma_x+\sigma_y=\left(\begin{matrix}0&-1-i\\-1+i&0\end{matrix}\right)\,. \end{align} A unitary matrix that makes these two equivalent is $$ \left(\begin{matrix}1&0\\0&i\end{matrix}\right)\,. $$ Another such matrix is $$ \left(\begin{matrix}0&-i\\i&0\end{matrix}\right)\,. $$ which happens to equal $\sigma_y\,.$ If I am not mistaken, the general form of a unitary matrix that makes $\sigma_x+\sigma_y$ and $-\sigma_x+\sigma_y$ equivalent is $$ \left(\begin{matrix}a&b\\-b&ia\end{matrix}\right)\,. $$ where the complex numbers $a=a_1+ia_2,b=b_1+ib_2$ must satisfy $$ \frac{a_1}{a_2}=\frac{b_1-b_2}{b_1+b_2} $$ plus scaling such that $a_1^2+a_2^2+b_1^2+b_2^2=1\,.$ So far so good.

A bit more general:

The anti symmetry in (1) implies for each $\mu\not=\nu$ that $\sigma_\mu\,\sigma_\nu\,\sigma_\mu=-\sigma_\nu$ holds. Because of (2) it is easy that for each $\mu\not=\nu$ and each $\rho\not=\mu\,,$ $$ \sigma_\mu+\sigma_\nu\quad\text{ and }-\sigma_\mu+\sigma_\nu=\sigma_\rho\,\sigma_\mu\,\sigma_\rho+\sigma_\nu =\sigma_\rho\,\sigma_\mu\,\sigma_\rho^*+\sigma_\nu $$ are unitary equivalent, and the nicest unitary matrix that does the job is $\sigma_\nu\,.$

This holds because the Pauli matrices have so many nice properties.

Conclusion. I suspect that in the world of the general case the grass will not be very green.