What is the the derivative of inverse skew-symetric matrix $\frac{\partial (W^{-1})}{\partial \omega_i}$ where $W=[\omega]_\times$, $\omega\in\mathbb{R}^3$ and $\omega_i$ is the $i$-th element of $\omega$; $[\omega]_\times=\begin{bmatrix}0&-\omega_3&\omega_2\\ \omega_3 & 0 & -\omega_1\\ -\omega_2 &\omega_1 & 0\end{bmatrix}$.

The wiki page on matrix calculus about Identites in differential form says: $$ d(X^{-1})=-X^{-1}dXX^{-1} $$

Is it correct to say that: $$ \frac{\partial (W^{-1})}{\partial \omega_i}=-W^{-1}\frac{\partial (W)}{\partial \omega_i}W^{-1}=-W^{-1}[e_i]_\times W^{-1}, $$ where $e_i$ is the $i$-th column of an identity matrix of $3\times3$.

  • 1
    $\begingroup$ Yes, that looks correct. For anyone not familiar with the $[\omega]_\times$ notation, you should state that $$W=\sum_{i=1}^3 \omega_i\,[e_i]_\times$$ $\endgroup$ – greg Jan 8 '17 at 16:52
  • $\begingroup$ Hi, I just realise the odd skew-symmetric matrix cannot be invertible. :( $\endgroup$ – Shuda Li Jan 8 '17 at 17:09

Your Answer

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

Browse other questions tagged or ask your own question.