Consider the Hermitian matrices:
$$A=\begin{bmatrix}
0 & -0.7011-0.0912i & 0 \\
-0.7011+0.0912i & 0 & 0.6702-0.2255i\\
0 & 0.6702+0.2255i & 0
\end{bmatrix}\\
B=\begin{bmatrix}
0 & -0.0912+0.7011i & 0 \\
-0.0912-0.7011i & 0 & -0.2255-0.6702i\\
0 & -0.2255+0.6702i & 0
\end{bmatrix}\\ $$
I am interested in knowing if it is possible to find a diagonal unitary matrix, $U$, that transforms $A$ and $B$ (individually) so that one matrix is purely real and the other purely imaginary? To be clear, I need $U$ to transform:
$$A'=UAU^{\dagger}\\B'=UBU^{\dagger}$$
so that one of $A'$ and $B'$ is purely real and the other purely imaginary (this transformation may not be possible, but that is what I would like to find out).
Best Answer
I find that a helpful way to think of these diagonal similarities is to note that $$ \operatorname{diag}(u) A \operatorname{diag}(u)^\dagger = A \circ (uu^\dagger), \tag{1} $$ where $\circ$ denotes the Hadamard (entrywise) product.
With (1), we can see that because for all entries of $B$ we have either $B_{jk} \in \{\pm i A_{jk}\}$, it follows that $[UBU^\dagger]_{ij} \in \{\pm i UAU^\dagger\}_{ij}$. So, it suffices to find $U = \operatorname{diag}(u)$ such that $UAU^\dagger$ has real entries.
Using (1), we have can see that we want $u$ such that $$ a_{12} u_1 u_2^* \in \Bbb R, \quad a_{23} u_2 u_{3}^* \in \Bbb R, $$ where $z^*$ denotes the complex conjugate of $z$. One way to ensure this is to make it so that $$ u_1^*u_2 = \frac{a_{12}}{|a_{12}|}, \quad u_2^* u_3 = \frac{a_{23}}{|a_{23}|}. $$ Setting $u_1 = 1$ and applying these equations yields $$ u_2 = \frac{a_{12}}{|a_{12}|}, \qquad u_3 = \frac{a_{12}a_{23}}{|a_{12}a_{23}|}. $$