Strain tensor of rigid body

Question

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.

Answer 1

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

Answer 2

The strain tensor is zero because an inertial (Galilean) observer moving in the same frame as the body wouldn’t see the object “deform” at all.

If we declare the strain to be a quantity that is frame-independent in classical mechanics, then it has to be zero in all frames when it is zero in one frame.