Given an $A \in \mathfrak{su}(n)$, is it always possible to solve for $U,V \in SU(n)$ and $\lambda \in \mathbb{C}$ such that $\lambda U^{\dagger}V -\bar{\lambda} V^{\dagger}U = A$?
Cross posted from MSE as no answer appeared.
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$\begingroup$ See answers for the crossposted question at MSE. $\endgroup$– Dietrich BurdeCommented Jul 17, 2015 at 12:26
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$\begingroup$ Please wait a bit longer before cross-posting -- say, at least a week or so. $\endgroup$– Stefan Kohl ♦Commented Jul 17, 2015 at 15:00
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1 Answer
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Since $A$ is skew hermitian, it is normal and we may assume that $A=diag(i\theta_j)$ where the $(\theta_j)_j$ are known real numbers. We choose $U=I$ and $V=diag(e^{i\alpha_j})$ where the $(\alpha_j)_j$ are unknown real numbers. The required equality can be written: for every $j$, $\lambda e^{i\alpha_j}-\bar{\lambda}e^{-i\alpha_j}=i\theta_j$. We choose $\lambda=\sup_j\{|\theta_j/2|\}+1$; then the conditions over the $(\alpha_j)_j$ become: for every $j$, $\sin(\alpha_j)=\dfrac{\theta_j}{2\lambda}$ and we are done.
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$\begingroup$ That's great thanks! Will this still be possible is $U$ is also given? I.e. given $A$ and $U$ can we always solve for $V$? $\endgroup$– BenjaminCommented Jul 16, 2015 at 23:54
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$\begingroup$ and $\lambda$ I should have said. $\endgroup$– BenjaminCommented Jul 17, 2015 at 0:06
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$\begingroup$ Actually $V \mapsto UV$ does the job if $U$ is given and $V$ is your $V$ from your answer. $\endgroup$– BenjaminCommented Jul 17, 2015 at 0:09