Hello,

Please tell me what you think of the following;

v = [v1 , v2, v3]

Skew(v) =

$\displaystyle \left[ \begin{array}{ccc} 0 & -v3 & v2 \\ v3 & 0 & -v1 \\ -v2 & v1 & 0 \end{array} \right]$

$\displaystyle R\epsilon SO(3)$; $\displaystyle v \epsilon \Re^3$

Prove that $\displaystyle (R)Skew(v)R^T=Skew(Rv)$

$\displaystyle det[(R) Skew(v) (R^T)] = det[Skew(Rv)]$

$\displaystyle det(R) det[Skew(v)] det(R^T) = det[Skew(Rv)]$

Since by definition $\displaystyle det[SO(3)] = 1$;

Since by definition of square matrices$\displaystyle det(R^T) = det(R)$

$\displaystyle det[Skew(v)] = det [Skew(Rv)]$

Since by definition the determinant of a Skew matrix is 0 for nxn matrices where n is odd

$\displaystyle 0 = 0$

Proofs with 0=0 always make me a little nervous...