For arbitrary $A\in\mathbb{C}^{3\times 3}$, is it always possible to find a unitary matrix $Q$ such that $QA$ has zero diagonals? Namely, to make $QA$ in the following form: $$ QA=\begin{bmatrix} 0 & x_{12} & x_{13}\\ x_{21} & 0 & x_{23}\\ x_{31} & x_{32} & 0\\ \end{bmatrix} $$
Here is my thought: In QR factorization, we apply a sequence of unitary matrices to $A$ to produce more and more zero elements. For example, using Givens rotation, we introduce 1 zero each time, while using Householder we introduce a column of zeros each time. For the current problem, we can always apply Givens rotation to make the 1st and 2nd diagonals zero, i.e., $\exists Q' s.t.$ $$ Q'A=\begin{bmatrix} 0 & x_{12} & x_{13}\\ x_{21} & 0 & x_{23}\\ x_{31} & x_{32} & x_{33}\\ \end{bmatrix} $$
However, I could not prove, neither disprove, that we can also make the 3rd diagonal zero.
Thanks for any comments.
Consider $A = \pmatrix{1 & 1 & 1\cr 1 & 1 & 1\cr 1 & 1 & 1\cr}$. Then all columns of $QA$ are equal. In order for the diagonal elements to be $0$, we'd need $QA = 0$. But obviously this can't happen for a unitary matrix $Q$.