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For two Stokes flows in the same fluid domain with the same dynamic viscosity, symmetry of the Newtonian fluid stress tensor and incompressibility give the local reciprocal identityIn index notation, its remaining cross-gradient terms are both $2\mu e:e^{\prime }$. On the remote sphere the integrated reciprocal flux tends to zero: the velocity is $O(R^{-1})$, the stress tensor is $O(R^{-2})$ and the area is $O(R^2)$. On the body, the no-slip boundary condition makes the two velocities the constants $\mathbf{U},{\mathbf{U}}^{\prime }$. The Lorentz reciprocal theorem therefore givesBy the Linearity of Stokes flow, the sum of the two velocity-pressure fields satisfies the boundary conditions for body velocity $\mathbf{U}+{\mathbf{U}}^{\prime }$. Uniqueness of the exterior Stokes flow identifies it with the third solution. Since the stress tensor and its surface integral are linear too, $\mathbf{F}(\mathbf{U}+{\mathbf{U}}^{\prime })=\mathbf{F}(\mathbf{U})+\mathbf{F}({\mathbf{U}}^{\prime })$; scalar multiplication follows in the same way. Thus $\mathbf{F}=-A\mathbf{U}$ for a fixed-orientation hydrodynamic resistance matrix $A$. Substituting arbitrary $\mathbf{U},{\mathbf{U}}^{\prime }$ in the reciprocal identity yieldsThe minus sign is natural because viscous dissipation gives $-\mathbf{F}\cdot \mathbf{U}=2\mu \int e:edV>0$ for nonzero rigid translation in an otherwise stationary fluid; $A$ is a positive-definite matrix.

Since $\mathbf{a}$ is constant, direct differentiation gives $(\mathbf{r}\cdot \nabla )\mathbf{a}=0$, $(\mathbf{a}\cdot \nabla )\mathbf{r}=\mathbf{a}$, $\nabla \cdot \mathbf{r}=3$ and $\nabla \cdot \mathbf{a}=0$. The stated right-hand side is therefore $2\mathbf{a}$. Independently, index notation givesThus $\boxed{\nabla \times (\mathbf{a}\times \mathbf{r})=2\mathbf{a}}$. A conservative vector field has zero curl, so the field is not conservative unless $\mathbf{a}=0$. For a closed planar circuit, its circulation is $2\mathbf{a}\cdot \mathbf{A}$ by Stokes theorem; the zero-vector exception is conservative with a constant potential.

Since $M^T=I-\mathbf{a}{\mathbf{b}}^T=I-N$, nilpotence already gives $(I+N)(I-N)=I$. Explicitly, in index notation,Thus $\boxed{L{M}^T=I}$, or equivalently $M^T=L^{-1}$. Here ${\delta }_{ij}$ is the Kronecker delta.