text: Q:
  
  How can I prove $[r][\omega]^{2}r = -[\omega][r]^2\omega$?
  
  A derivation I am reading from a book requires me to prove
  $[r][\omega]^{2}r = -[\omega][r]^2\omega$ .
  Now this was part of a larger derivation and hence the book skipped a few intermediary steps and I am not able to reach the RHS from LHS or vice versa.
  Here, r and $\omega \in \Re^3$ i.e. they are 3 dimensional vectors. $[r]$ represents the skew symmetric matrix representation of the vector $r$ : $\begin{align}
  [r] = 
  \begin{bmatrix}
   0 & -r_3 & r_2 \\ 
  r_3 & 0 &-r_1 \\
   -r_2 & r_1 & 0 
  \end{bmatrix}
  \end{align}   $
  This representation has the benefit that the cross product $\vec r \times \vec \omega$ is the same as the matrix and vector multiplication $[r]\cdot\omega$
  A few results using this notation: for $a,b \in \Re^3$
  
  $[a]b = -[b]a$
  $[a] = -[a]^T$
  $[a][b] = ([b][a])^T = [a]^T[b]^T $
  
  Steps the textbook performed to derive the result:
  $[r][\omega]^2r = -[r]^T[\omega]^T[r]\omega = -[\omega][r]^2\omega$
  My attempt:
  $[r][\omega]^2r = [r][\omega][\omega]r = -[r][\omega][r]\omega = -[r]^T[\omega]^T[r]\omega$ = ?(How to proceed)
  Is there anything I am missing? An insight in the right direction will be very helpful.
  Thanks in advance!
  
  A:
  
  It's easy to prove the identity using the vector triple product formula $a\times(b\times c)=(a\cdot c)b-(a\cdot b)c$:
  \begin{aligned}
  &r\times\left[\omega\times(\omega\times r)\right]+\omega\times\left[r\times(r\times\omega)\right]\\
  &=r\times\left[(\omega\cdot r)\omega-(\omega\cdot\omega)r\right]+\omega\times\left[(r\cdot\omega)r-(r\cdot r)\omega\right]\\
  &=(\omega\cdot r)r\times\omega-(\omega\cdot\omega)r\times r+(r\cdot\omega)\omega\times r-(r\cdot r)\omega\times\omega\\
  &=0.
  \end{aligned}
  The identity in question can also be viewed as a special case of Hall-Witt identity, which someone happened to have asked about yesterday.
  
  