= Solution
For eigenvalues $(v,v,-2v)$, the first two entries coincide. Consequently the $12$ generators $t^1,t^2$, as well as $t^3,t^8$, commute with the vacuum. The $13$ and $23$ eigenvalue differences both have magnitude $3|v|$, giving
$$
\boxed{m_{1,2,3,8}=0,\qquad m_{4,5,6,7}=3g|v|.}
$$
There are four massless and four massive <gauge bosons>. The <centralizer of a Hermitian matrix in SU(n)> preserves the degenerate two-dimensional <eigenspace>, so the unbroken group is
$$
\boxed{H=S(U(2)\times U(1))\cong\frac{SU(2)\times U(1)}{\mathbb Z_2}\cong U(2).}
$$
At the <Lie algebra> level this is $\mathfrak{su}(2)\oplus\mathfrak u(1)$, generated by $t^{1,2,3}$ and $t^8$. The finite quotient specifies the actual subgroup inside <SU(3)>; simply writing $SU(2)\times U(1)$ suppresses that global identification. The four real massive fields combine into a complex <SU(2)> doublet and its conjugate. Unbroken <SU(2)> rotations relate its two components, explaining the equal masses of the $13$ and $23$ sectors in this <SU(3) gauge theory with an adjoint Higgs field>.
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