Strain tensor of rigid body If we have a particle that satisfies rigid body motion then its velocity $u$ is of the form $$u=a\times x+b$$ Why is its strain tensor zero? The strain tensor is given by $$D(u)=\frac{1}{2} (\nabla u +\nabla u^t)$$ _See the answers for my attempt._

$u$ can be written in the form $$u=Ax+b$$ where $A$ is a skew symmetric matrix. Then $\
abla u=A$ and consequently $$D(u)=\frac{1}{2} (A+A^t)=0$$