 Solution

Let $Q$ be the orthogonal matrix taking old components to new ones, so $x_i^{\prime }=Q_{ip}{x}_p$. With repeated indices summed, the tensor component transformation law isAn isotropic tensor has the same components after every proper rotation. Since $Q_{ip}{Q}_{jp}={\delta }_{ij}$, the Kronecker delta is invariant. Applying this identity to each factor shows that $a{\delta }_{ij}$ and every linear combination $b{\delta }_{ij}{\delta }_{kl}+c{\delta }_{ik}{\delta }_{jl}+d{\delta }_{il}{\delta }_{jk}$ are invariant as well.

To show that an isotropic second-rank tensor must be scalar, start with any matrix $t$, without assuming symmetry. Invariance under quarter-turns about axes 1 and 2 implies invariance under their squares, the half-turnsThe first forces $t_{12},t_{21},t_{13},t_{31}$ to vanish. The second also forces $t_{23},t_{32}$ to vanish. The matrix is therefore diagonal. A quarter-turn about axis 1 interchanges its 22 and 33 entries; one about axis 2 interchanges its 11 and 33 entries. All three must agree, giving $t=aI$.

A rotation preserves the ball $V_R$, $|x{|}^2$ and the volume element. Changing integration variables by that rotation shows that $A$ and $B$ are isotropic tensors. Write $A_{ij}=a{\delta }_{ij}$. Taking its tensor contraction givesso $a=4\pi R^7/21$. Every permutation of the indices of $B$ leaves its integrand unchanged. Its complete symmetry therefore makes all three isotropic coefficients equal, say $d$:Double contraction gives $15d=B_{iikk}={\int }_{V_R}|x{|}^{4}dV=4\pi R^7/7$. Thus the isotropic fourth moment of a ball and its radial contraction are

Put $d=4\pi R^7/105$, so that $a=5d$. Contract $B$ with the prescribed unit vector twice:This has the spectrum of a scalar matrix plus a rank-one projector. In the direction $n$, the outer product $n{n}^T$ acts as the identity; on the plane perpendicular to $n$, it acts as zero. ThereforeAny nonzero multiple of $n$ is a parallel eigenvector, and any nonzero vector in the perpendicular plane is a transverse eigenvector. The transverse eigenvalue has multiplicity two; an orthonormal basis of that plane completes a set of principal axes.