The answer is simply that black holes grow when you throw mass at their horizons. The small region you last saw the mass (dimly and red-shifted) just outside t

The answer is simply that black holes grow when you throw mass at their horizons. The small region you last saw the mass (dimly and red-shifted) just outside the black hole ultimately ends up being on the inside, even from your perspective. Below I'm going to ignore angular momentum; black hole spin changes the details but not the central thrust of my comment. You could think about it this way: Schwarzschild is an eternal vacuum solution (to the Einstein Field Equations of General Relativity (EFEs)). Like other solutions to the EFEs, it tends to be investigated by tracking the behaviour of "test particles". Test particles don't change the EFEs themselves -- they don't have mass, they don't have any non-gravitational interactions at all, they're not physical, they're just a tool used to trace out the geometry of the spacetime. Throwing in a test particle doesn't change the eternal nature of the Schwarzschild black hole -- it always has the same mass, and the Schwarzschild radius is totally determined by that mass (so the horizon is always of constant size). That is, the test particle is not a perturbation of the Schwarzschild black hole. A significant mass would perturb the Schwarzschild metric though. Unlike test particles, that mass enters into the EFEs. The relevance is that the Schwarzschild radius is totally determined by the central mass. When the significant additional mass (the perturbing infaller) is far away from the central mass, the geometry (described by the EFE's metric tensor) still looks a lot like Schwarzchild. However, as the perturbing mass approaches and falls in, things depart from Schwarzschild for a bit, then returns to Schwarzschild but with an increased central mass. The infaller ultimately ends up at the singularity, leaving M_before < M_after, so the horizon must grow in proportion to the mass that fell in because the Schwarzschild r_s = 2GM/c^2, where M is the central mass. Distant viewers can measure the central mass in several ways; it's observable. Since throwing actual mass (rather than test particles) into Schwarzschild changes the size (and, briefly, shape) of the horizon, you could consider it as if the object you see dimming and moving verrrry slowly doesn't just come to a halt at a constant horizon: instead, the horizon "reaches out" and snatches the infaller just outside the M_before horizon. The worry that nothing actually falls in arises if one insists on keeping M constant (M_before == M_after) even as one has mass outside the centre of the black hole. Constant M is easier to work with mathematically, which leads to things like test particles or Hawking negative energy quanta, and so on. Extrapolating from those convenience uses tends to lead to confusions like "nothing initially outside can actually end up inside", which is just wrong. Finally, using perturbation methods, a compact infaller like a neutron star would raise a significant bump on the horizon, distorting it slightly from ~spherical. That distortion vanishes in a short time (even for an outside observer, who can also detect gravitational waves), and the post-infall result is a bigger spherical horizon. We have observed several neutron star-black hole mergers. The result, generically, is a more massive black hole (and a lot of gravitational radiation).