3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $-1$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.