= Solution
Use a mass-normalized <galactic distribution function> $F$, so that $\rho=\int F\,d^3v$ and $\rho\langle v_iv_j\rangle=\int v_iv_jF\,d^3v$. Multiplying the stationary <Collisionless Boltzmann equation> by $v_j$ and integrating over <velocity space> gives
$$
\partial_i\int v_iv_jF\,d^3v-\Phi_{,i}\int v_j\partial_{v_i}F\,d^3v=0.
$$
Assume the <galactic distribution function> decays sufficiently rapidly for the velocity boundary term to vanish. <Integration by parts> then gives $\int v_j\partial_{v_i}F\,d^3v=-\delta_{ij}\rho$, and therefore the <Jeans equations> are
$$
\boxed{\partial_i(\rho\langle v_iv_j\rangle)=-\rho\Phi_{,j}.}
$$
The second moment includes both ordered motion and <velocity dispersion>; no assumption of zero mean velocity was needed.
To obtain the <tensor virial theorem>, multiply the $j$th <Jeans equation> by $x_i$ and integrate over position. For an isolated, finite system with a vanishing spatial surface term,
$$
-\int\rho\langle v_iv_j\rangle\,d^3x=-\int\rho x_i\Phi_{,j}\,d^3x=W_{ij}.
$$
Since the left side is $-2K_{ij}$, \b[$2K_{ij}+W_{ij}=0$]. This convention puts all stellar second moments into $K_{ij}$; it does not separate ordered and random <kinetic energies>.
For a self-gravitating system, the trace $W$ is the <Newtonian gravitational potential energy>. Indeed, symmetrizing the pair integral gives
$$
W=-\frac G2\iint\frac{\rho(\mathbf x)\rho(\mathbf x')}{|\mathbf x-\mathbf x'|}\,d^3x\,d^3x'=\frac12\int\rho\Phi\,d^3x.
$$
Thus the trace of the <tensor virial theorem> is $2K+W=0$, and the total energy obeys
$$
\boxed{E=K+W=-K=W/2.}
$$
These identities require self-gravity without an additional external potential or an omitted confining boundary pressure.
Let $\langle v_I^2\rangle$ now denote a mass-weighted average over the entire initial system. Then $K_I=M_I\langle v_I^2\rangle/2$. The <gravitational radius> is defined by $R_I=-GM_I^2/W_I$, so the <virial theorem> immediately yields
$$
\boxed{E_I=-\frac12M_I\langle v_I^2\rangle=-\frac{GM_I^2}{2R_I}.}
$$
The <gravitational radius> measures total binding energy, rather than a particular geometric edge or half-mass radius.
For the accreted systems define the mass-weighted internal mean-square speed by $M_A\langle v_A^2\rangle=\sum_sM_s\langle v_s^2\rangle$. If each satellite initially satisfies the <virial theorem>, its internal energy contributes to $E_A=-M_A\langle v_A^2\rangle/2$. In <parabolic dry-merger energy accounting>, the orbital energy at large separation is zero. Assume a <dry galaxy merger>, no loss of mass or energy through escaping stars, no external work, and a final relaxed system satisfying the <virial theorem>. Then <conservation of energy> gives $E_F=E_I+E_A$ and $M_F=M_I+M_A$. Energy transferred by <Chandrasekhar dynamical friction> remains part of the total energy under these assumptions. Consequently,
$$
E_F=-\frac12M_I\langle v_I^2\rangle(1+\epsilon\eta),\qquad \eta=\frac{M_A}{M_I},\quad\epsilon=\frac{\langle v_A^2\rangle}{\langle v_I^2\rangle}.
$$
Applying the final <virial theorem> and dividing by the initial relation gives
$$
\boxed{\frac{\langle v_F^2\rangle}{\langle v_I^2\rangle}=\frac{1+\epsilon\eta}{1+\eta},\qquad\frac{R_F}{R_I}=\frac{(1+\eta)^2}{1+\epsilon\eta}.}
$$
For a mass doubling, $\eta=1$. A merger of two identical equilibrated galaxies has $\epsilon=1$, hence \b[$R_F/R_I=2$]. For accretion through many <minor galaxy mergers> whose satellites have much smaller internal mean-square speeds, $\epsilon\ll1$, hence \b[$R_F/R_I\simeq4$], tending to $4$ in the cold-satellite limit. Equivalently, nearly fixed total energy makes the <gravitational radius> scale as $M^2$ during <cold minor-merger size growth>. Small satellite mass alone does not logically imply $\epsilon\ll1$: the factor four also requires this weak-binding assumption. Likewise, equal masses require comparable internal binding to give the factor two. These are the physical limits behind the two stated merger comparisons.
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