Sorry for boring you my friends. I have haunted by a problem of relation between matrix product and cross product.
I would like to demonstrate the following equation: $$ (\Omega\cdot r)^T(\Omega\cdot r)=\omega^T\cdot I \cdot \omega$$ with: $$\Omega= \left[ \begin {array}{ccc} 0&-\omega_{{3}}&\omega_{{2}} \\ \omega_{{3}}&0&-\omega_{{1}}\\ -\omega_{{2}}&\omega_{{1}}&0\end {array} \right] , $$
$$I= \left[ \begin {array}{ccc} {y}^{2}+{z}^{2}&-xy&-xz \\ -xy&{x}^{2}+{z}^{2}&-yz\\ -xz&- yz&{x}^{2}+{y}^{2}\end {array} \right] , $$ $$r= \left[ \begin {array}{c} x\\ y\\ z\end {array} \right] , $$ and $$ \omega = \left[ \begin {array}{c} \omega_{{1}}\\ \omega_{{2} }\\ \omega_{{3}}\end {array} \right] $$ Actually, I could demonstrate this relation in a ugly way that I developed the left side of equation and re-assembled in the form of right side. Meanwhile, I found out there is a relation between the matrix product and the cross product as: $$\left[ \begin {array}{ccc} 0&-\omega_{{3}}&\omega_{{2}} \\ \omega_{{3}}&0&-\omega_{{1}}\\ -\omega_{{2}}&\omega_{{1}}&0\end {array} \right]\cdot\left[ \begin {array}{c} x\\ y\\ z\end {array} \right] = \left[ \begin {array}{c} \omega_{{1}}\\ \omega_{{2} }\\ \omega_{{3}}\end {array} \right] \wedge \left[ \begin {array}{c} x\\ y\\ z\end {array} \right]$$, thus the left side of the equation could take the form of: $$(\omega\wedge r)^T(\omega\wedge r)$$ It looks a little more similar to the right side. Eventually, I failed to demonstrate it in a pretty way.
Thank you in advance for taking a look. Have a nice holiday.
I don't know if this is any "prettier" than what you have derived, but you could use the the Levi-Civita tensor to expand the cross products.
The LHS becomes $$\eqalign{ (w\times r)^T(w\times r) &= (wr^T:\epsilon)\cdot(\epsilon:wr^T) \cr &= wr^T:(\epsilon\cdot\epsilon):wr^T \cr &= wr^T : (\alpha-\beta) : wr^T \cr &= wr^T : wr^T - wr^T : rw^T \cr &= w^2r^2 - (r^Tw)^2 \cr }$$ where colon denotes the Frobenius (or double-dot) product, and $(\alpha, \beta)$ are isotropic 4th order tensors with certain ("epsilon-delta") properties $$\eqalign{ \alpha : M &= M \cr \beta : M &= M^T \cr }$$
Note that the moment of inertia tensor can be written as $$\eqalign{ I &= r^2E-rr^T \cr }$$ where E is the identity tensor.
Now the RHS can be written as $$\eqalign{ I:ww^T &= (r^2E-rr^T):ww^T \cr &= w^2r^2 - (r^Tw)^2 \cr }$$ which is identical to the LHS.