= Solution
Take real field coordinates $\phi^I$; for complex <scalar fields>, split them into real and imaginary parts. Assume a smooth <scalar potential> and positive, canonically normalized <kinetic terms> in a relativistic theory. At a <classical vacuum>, $V_I(\phi_0)=0$. The <Taylor expansion> is
$$
V(\phi_0+\xi)=V(\phi_0)+\frac12\xi^I(M^2)_{IJ}\xi^J+O(\|\xi\|^3),\qquad (M^2)_{IJ}=\left.\frac{\partial^2 V}{\partial\phi^I\partial\phi^J}\right|_{\phi_0}.
$$
Thus the <scalar mass matrix> is the <Hessian matrix>, and its <eigenvalues> give squared masses of the linearized scalar excitations.
Put $X_a(\phi)=it^a\phi$. Infinitesimal invariance of the <scalar potential> gives the identity $V_I X_a^I=0$ at every field value. Differentiate with respect to $\phi^J$ and evaluate at the <classical vacuum>:
$$
(M^2)_{JI}X_a^I(\phi_0)+V_I(\phi_0)\partial_JX_a^I(\phi_0)=0,
\qquad\boxed{M^2(it^a\phi_0)=0.}
$$
Every nonzero infinitesimal symmetry direction therefore lies in the <kernel> of the <scalar mass matrix>. This is the classical <Goldstone theorem> expressed through <Goldstone directions in the scalar mass matrix>.
Let $H_0$ be the full <stabilizer subgroup> of $\phi_0$. The linear map from the <Lie algebra> of $G$ to field space, $T\mapsto iT\phi_0$, has kernel equal to the <Lie algebra> of $H_0$. The <rank-nullity theorem> gives
$$
\boxed{\dim\operatorname{span}\{it^a\phi_0\}=\dim G-\dim H_0.}
$$
There are consequently \b[at least $\dim G-\dim H_0$ massless scalar directions], namely the tangent directions to the symmetry orbit through the <classical vacuum>. If the intended $H$ is $H_0$, these are the $\dim G-\dim H$ symmetry-required <Goldstone bosons>. Exactly that many massless modes occur if the <scalar mass matrix> is positive definite on a complement of those tangent directions. A nonsingular positive field-space kinetic metric changes normalization, but not the number of zero masses.
Two qualifications are needed for the literal assumptions. A subgroup fixing the <classical vacuum> need not be the full <stabilizer subgroup>. For example, take $G=SO(3)$ acting on a real triplet and $V=\lambda(|\phi|^2-v^2)^2/8$ with $\lambda,v>0$. At $\phi_0=ve_3$, the <scalar mass matrix> is $\lambda v^2\operatorname{diag}(0,0,1)$: there are two massless modes. Choosing $H=\{1\}$ satisfies the printed invariance condition but would incorrectly predict three. The full <stabilizer subgroup> is $SO(2)$.
Even with the full <stabilizer subgroup>, symmetry does not exclude an <accidental massless scalar>. Take $G=SO(2)$ rotating $(x,y)$, with an invariant singlet $z$, and
$$
V(x,y,z)=\frac\lambda4(x^2+y^2-v^2)^2+\kappa z^4,\qquad\lambda,\kappa>0.
$$
At $(v,0,0)$ the full continuous stabilizer is trivial, but the <scalar mass matrix> is $\operatorname{diag}(2\lambda v^2,0,0)$. One zero direction is the <Goldstone boson>; the other is an <accidental massless scalar> at quadratic order. \b[The proof establishes the symmetry-required count, not unconditional equality with the total number of massless fields.] The statement concerns global internal symmetry; gauging it changes the physical interpretation through the <Higgs mechanism>.
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