The symmetry of the isotropic cartesian tensor of Newtonian fluid

Question

I'm trying to prove the symmetry of the isotropic tensor in the linear relation between the shear stress and strain rate for Newtonian fluid $$T_{ij}=\beta_{ijlm}\frac{\partial u_{l}}{\partial x_{m}}.$$ My attempt is based on the assumption that the shear stress vanishes for pure rotation, that is $$\beta_{ijlm}\frac{1}{2}\left(\frac{\partial u_{l}}{\partial x_{m}}+\frac{\partial u_{m}}{\partial x_{l}}\right)+\beta_{ijlm}\frac{1}{2}\left(\frac{\partial u_{l}}{\partial x_{m}}-\frac{\partial u_{m}}{\partial x_{l}}\right)=0,$$ since there is no elongation, we get $$\beta_{ijlm}\left(\frac{\partial u_{l}}{\partial x_{m}}-\frac{\partial u_{m}}{\partial x_{l}}\right)=0.$$ Does this imply the symmetry of $\beta_{ijlm}$?