Let $R$ be a real orthogonal matrix, $$RR^T = I$$ and let $\Omega$ be a real skew-symmetric matrix, $$\Omega^T = -\Omega$$

Please show (or disprove, although I'm pretty sure it's true) that, $$ R \Omega = \Omega R$$

I.e. prove whether or not orthogonal matrices and skew-symmetric matrices always commute in multiplication.

Is it possible to show using only the defining properties I listed? Or perhaps it might be necessary to also use the fact that skew-symmetric matrices commute with their transposes.

  • $\begingroup$ Quick lemma for skew-symmetric matrices commute with their transposes: just left and right multiply the skew-symmetric definition by $\Omega$. $$\Omega^T := -\Omega$$ $$\Omega \Omega^T = -\Omega \Omega$$ $$\Omega^T \Omega = -\Omega \Omega$$ $$\therefore\ \ \Omega \Omega^T = \Omega^T \Omega$$ $\endgroup$ – jnez71 Feb 26 '17 at 9:48

The claim is false. Consider $$ \Omega=\left[ \begin{array}{ccc} 0&1&0\\ -1&0&0\\ 0&0&0 \end{array} \right] $$ and $$ R=\left[ \begin{array}{ccc} 1&0&0\\ 0&0&1\\ 0&1&0 \end{array} \right].$$

What you may have tried are the two by two matrices, which the commutativity holds except possibly when the orthogonal matrix has determinant $-1$.

  • $\begingroup$ That was essentially the case... Thanks! $\endgroup$ – jnez71 Feb 26 '17 at 10:38

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