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I have a skew symmetric matrix $$C=\left( \begin{array}{ccc} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \\ \end{array} \right).$$. and we have the relation $C=UDU^{-1} $. D is a diagonal matrix

Question

  1. What is the value of U and D?

NB :: Any value of U and D solve my issue. Values of $a_i$ s cant be changed

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  • $\begingroup$ They aren't unique. As an extreme example, if $a_1 = a_2 = a_3 = 0,$ you can let $U$ be literally any invertible matrix... $\endgroup$ – user165670 Jul 27 '14 at 6:20
  • $\begingroup$ Any value of U and D solve my issue. Values of $a_i$ s cant be changed $\endgroup$ – Nirvana Jul 27 '14 at 6:22
  • $\begingroup$ Are $U$ and $D$ allowed to have complex entries? The eigenvalues of $C$ are $0$ and $\pm i \sqrt{a_1^2+a_2^2+a_3^2}$. $\endgroup$ – JimmyK4542 Jul 27 '14 at 6:27
  • $\begingroup$ Yes provided $ U(e^{D})U^{-1} $ should be a real matrix . $\endgroup$ – Nirvana Jul 27 '14 at 6:30
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Assuming that the $a_i$ are real then $E=iC$ is Hermitean so $E=VF{V ^{-1}}$ with $V$ unitary and $F$ diagonal with real entries (the eigenvalues of $E$). Generically $V$ and $F$ are unique (up to a phase factor in $V$). Now $C=V(-iF)V^{-1}$ so we can take $U=V$ and $D=-iF$. But then $U{e ^D}U ^{-1}$ is not a real matrix. Why do you think it is? And what do you mean with "Any values of $U$ and $D$ solve my issue "?

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