Eigenvalues of a rank 2 tensor defined by an integral I've been given the question:
"Consider the tensor: 
$$ C_{ij}=\int_{V}{x_ix_j|\mathbf {x}|^2 + x_ix_j(\mathbf {x.n})^2} dV $$
where V is the volume of a sphere radius R centred on the origin. What are the eigenvectors and corresponding eigenvalues of this tensor?"
I assume that the vector n is one of them since it breaks the symmetry, however, I don't know how to show it. Any help would be greatly appreciated as I am completely stuck.
This is actually part of a longer question:

If it is at all useful, the answers I got for the previous part of the question are:
$$ A_{ij}=\frac {4\pi R^7}{21}\delta_{ij} $$
and  $$ B_{ijkl}= \frac {4\pi R^7}{189}(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}) $$
 A: The expression for $B_{ijk\ell}$ is off by a factor $\frac{9}{5}$
Using the same technique as in this post (i.e. assume it's isotropic, contract it with identity matrices, calculate the scalar factor) the correct value is
$$\eqalign{
B_{ijk\ell} &= \frac{4\pi R^7}{3\cdot 5\cdot 7}\bigg(
 \delta_{ij}\delta_{k\ell} + \delta_{ik}\delta_{j\ell} + \delta_{i\ell}\delta_{jk}
\bigg)
}$$
Contraction with $\delta_{k\ell}$ yields $A_{ij}$ -- which is the first half of $C_{ij}$
$$\eqalign{
A_{ij}
 &= B_{ijk\ell}\,\delta_{k\ell} 
 = \frac{4\pi R^7}{3\cdot 5\cdot 7}\bigg(5\,\delta_{ij}\bigg) 
 = \frac{4\pi R^7}{3\cdot 7}\,\delta_{ij} \\
}$$
The other half is the contraction with the constant dyadic tensor
$\,N_{k\ell}=n_kn_\ell$
$$\eqalign{
F_{ij}
 &= B_{ijk\ell}\,n_kn_\ell 
 = \frac{4\pi R^7}{3\cdot 5\cdot 7}\bigg(n_kn_k\,\delta_{ij}+n_in_j+n_jn_i\bigg) 
 = \frac{4\pi R^7}{105}\bigg(\delta_{ij}+2N_{ij}\bigg) \\
}$$
Putting the two halves together
$$C_{ij} = A_{ij} + F_{ij}
 = \frac{8\pi R^7}{105}\bigg(3\delta_{ij}+N_{ij}\bigg)
\\$$
NB:   Your calculation of $A_{ij}$ was correct.
