= Solution
In <spontaneous symmetry breaking>, the <action> respects a <symmetry> while a selected <vacuum state> is invariant only under a proper <subgroup>. A nonzero <vacuum expectation value> can identify that choice. For a local <gauge theory>, gauge transformations are redundancies rather than transformations between physically different states; a gauge-fixed scalar expectation value describes the <Higgs mechanism> and its massive excitations. It should not be interpreted as breaking an observable local redundancy.
Write the <adjoint scalar field> as $\Phi=\Phi_0+\varphi$, with Hermitian <gauge generators> normalized by $\operatorname{tr}(t^at^b)=\delta^{ab}/2$. Since $\Phi_0$ is constant, the part of the <covariant derivative> linear in the <gauge field> is $D_\mu\Phi_0=igA_\mu^a[t^a,\Phi_0]$. Both $t^a$ and $\Phi_0$ are Hermitian, so their <commutator> is anti-Hermitian. The scalar kinetic term therefore contributes
$$
\mathcal L_{\rm mass}
=-g^2A_\mu^aA^{b\mu}\operatorname{tr}([t^a,\Phi_0][t^b,\Phi_0])
=\frac12A_\mu^a(m^2)^{ab}A^{b\mu}.
$$
Matching coefficients gives the <adjoint scalar gauge-boson mass matrix>
$$
\boxed{(m^2)^{ab}=-2g^2\operatorname{tr}([t^a,\Phi_0][t^b,\Phi_0]).}
$$
It is positive semidefinite: for real coefficients $x_a$, its <quadratic form> is $2g^2\operatorname{tr}(C^\dagger C)\ge0$, where $C=[x_at^a,\Phi_0]$. Its null space consists of generators commuting with the vacuum, the <Lie algebra> of its <centralizer>.
For a diagonal vacuum $\Phi_0=\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)$, the <matrix unit> $E_{ij}$ obeys $[E_{ij},\Phi_0]=(\lambda_j-\lambda_i)E_{ij}$. The two normalized Hermitian generators built from $E_{ij},E_{ji}$ each acquire
$$
m_{ij}^2=g^2(\lambda_i-\lambda_j)^2.
$$
The diagonal generators $t^3,t^8$ commute with every such vacuum and remain massless. Orthogonality of these root generators also shows that the <mass> matrix is diagonal in the supplied <Gell-Mann matrices> basis. Take $g>0$ and $v\ne0$ below; $v=0$ restores the full <SU(3)> group.
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