= Solution
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 identity
$$
\nabla\cdot(\boldsymbol\sigma\mathbf u'-\boldsymbol\sigma'\mathbf u)=0.
$$
In <index notation>, its remaining cross-gradient terms are both $2\mu e:e'$. 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 $\mathbf U,\mathbf U'$. The <Lorentz reciprocal theorem> therefore gives
$$
\boxed{\mathbf F\cdot\mathbf U'=\mathbf F'\cdot\mathbf U.}
$$
By the <Linearity of Stokes flow>, the sum of the two velocity-pressure fields satisfies the boundary conditions for body <velocity> $\mathbf U+\mathbf U'$. Uniqueness of the exterior <Stokes flow> identifies it with the third solution. Since the <stress tensor> and its surface <integral> are linear too, $\mathbf F(\mathbf U+\mathbf U')=\mathbf F(\mathbf U)+\mathbf F(\mathbf U')$; scalar multiplication follows in the same way. Thus $\mathbf F=-A\mathbf U$ for a fixed-orientation <hydrodynamic resistance matrix> $A$. Substituting arbitrary $\mathbf U,\mathbf U'$ in the reciprocal identity yields
$$
\boxed{A=A^T,\qquad\mathbf F=-A\mathbf U.}
$$
The minus sign is natural because <viscous dissipation> gives $-\mathbf F\cdot\mathbf U=2\mu\int e:e\,dV>0$ for nonzero rigid translation in an otherwise stationary fluid; $A$ is a <positive-definite matrix>.
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