 Solution

An isotropic tensor has unchanged components under every rotation of the Cartesian axes. Under proper rotations in three dimensions, rank one has no nonzero isotropic tensor, rank two has $\boxed{T_{ij}=A{\delta }_{ij}}$, and rank three has $\boxed{T_{ijk}=B{ϵ}_{ijk}}$. Here $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. If invariance under reflections is also required for a true tensor, the rank-three case vanishes; $ϵ$ is then a pseudotensor.

The velocity gradient tensor for the given field isIts symmetric part, the strain-rate tensor if $\mathbf{u}$ is interpreted as a velocity, isThe antisymmetric part has the specified ordering $D_{ik}-e_{ik}={ϵ}_{ijk}{\omega }_j$, withThe index ordering matters for this sign; it agrees with a local rotation velocity $\mathbf{\omega }\times \mathbf{x}$. This is the axial part of the scalar, symmetric-traceless and axial tensor decomposition, retaining the trace inside the symmetric part here.

At the specified point, $e=\left(\begin{array}{ccc}9& 6& 0\\ 6& 0& 0\\ 0& 0& 4\end{array}\right)$. The upper block has characteristic polynomial ${\lambda }^2-9\lambda -36=(\lambda -12)(\lambda +3)$. Thus the principal values and unit eigenvectors areThese mutually orthogonal directions are the principal strain rates' axes for this local field; reversing an eigenvector does not change its axis.