Solving $\lambda U^{\dagger}V -\bar{\lambda} V^{\dagger}U = A$

Question

Given an arbitrary $A \in \mathfrak{su}(n)$, is it always possible to solve for $U,V \in SU(n)$ and $\lambda \in \mathbb{C}$ such that $\lambda U^{\dagger}V -\bar{\lambda} V^{\dagger}U = A$?

Update: There was a slip in the question.

Answer 1

The answer is affirmative. By a change of basis, we may assume that $\frac1{\|A\|}A=\operatorname{diag}(iy_1,\, iy_2,\, \ldots,\, iy_n)$, where each $y_j$ is a real number between $-1$ and $1$. Therefore each $y_j=2i\sin\theta_j$ for some real $\theta_j$. Let $W=\operatorname{diag}\left(e^{i\theta_1},\ldots,e^{i\theta_n}\right)$. Then $W-W^\dagger=\frac1{\|A\|}A$. Let $\phi$ be the average of all $\theta_j$s. Take $\lambda=e^{i\phi}\|A\|,\ U=I$ and $V=e^{-i\phi}W$. Now we are done.

Answer 2

Since $\| U^\dagger V - V^\dagger U \| \le 2$ (where $\|.\|$ is the operator norm, say), the answer is: no, not always.