Assuming a PDE is given by the Hesse-operator $$ -\nabla \otimes \nabla \lambda = \mathbf{F} \, , $$ then testing the left-hand-side with a symmetric tensor $\mathbf{S}$ leads to $$ -\langle \mathbf{S}, \nabla \otimes \nabla \lambda \rangle = \langle \mathrm{div} \, \mathbf{S}, \nabla \lambda \rangle \, , $$ (ignoring boundary terms) which is a non-symmetric bilinear form.
(1) Is there a way to (in general) derive a symmetric bilinear form from the Hesse-operator?
If (1) is not possible; then assuming we know the following relations for the field $$ \lambda = \mathrm{tr} \, \mathbf{T} \, , \qquad \mathbf{T}=\mathbf{T}^T \,, \qquad \mathrm{div \, div} \,\mathbf{T} = \Delta \mathrm{tr} \, \mathbf{T} \, , $$ (2) is it reasonable to assume that the test function $\mathbf{S}$ implies the same relations such that $$ \langle \mathrm{div} \, \mathbf{S}, \nabla \lambda \rangle = -\langle \mathrm{div \, div} \, \mathbf{S}, \mathrm{tr} \, \mathbf{T} \rangle = -\langle \Delta \mathrm{tr} \, \mathbf{S}, \mathrm{tr} \, \mathbf{T} \rangle = \langle \nabla \mathrm{tr} \, \mathbf{S}, \nabla \mathrm{tr} \, \mathbf{T} \rangle \, , $$ (again, neglecting boundary terms) which is a symmetric bilinear form with respect to $\{ \mathbf{S}, \mathbf{T} \}$ ?