Wrong:
"since infrared waves have a shorter wavelength"
Infrared has longer wavelength than visible and visible longer wavelength than ultraviolet .
White is a term for visible light mixed wavelengths. In the plot you can see that almost half of the sun's radiated energy arrives as visible light. The white buildings reflect this visible light which otherwise,impinging on the surfaces would be absorbed and turned into infrared by the interactions, adding to the arriving infrared.
What is absorbed and what is reflected depends on the chemical bonds of the surfaces, whether the incoming radiation can excite molecular states of the materials. Infrared is in frequencies/wavelengths of the black body radiation of bodies in the temperature ranges comfortable for the human body, so they easily raise the vibrational and rotational levels of solids and liquids and the kinetic energy of gasses.
37C curve seen here practically all in infrared, and lower temperatures more so.
Thus white paint will not reflect infrared as efficiently as visible, a large part of infrared will be absorbed as also some part of visible will scatter at the surface and degrade to infrared. Infrared can be reflected by metal mirrors, from the collective fields in metals . If you put aluminum foil in front of a heater you are sheltered from most of the heat which is reflected, but some of it is absorbed as can be seen by touching the foil.
If I expose an object to EM radiation only from the infrared spectrum, will it only reflect back infrared?
Yes, but most of it will be absorbed ( except by mirror metal surfaces) because the materials have the receptors for these wavelengths. This is due to the fact that larger wavelengths have photons with less energy which cannot excite higher energy levels.The energy of the photons goes as h*c/lamda where h is plancks constant, lamda the wavelength and c the velocity of light.
Is this true for other types of EM radiation?
No.Visible and ultraviolet by scatterings degrade their energy down to infrared frequencies, depending on the material.
Is it possible to make an object that looks white and absorbs a lot of infrared radiation?
Usually most of the infrared will be absorbed except by mirror metal surfaces.
If an object reflects most of the EM radiation that it receives of a particular wavelength λ, will it also reflect most of the radiation it receives of wavelengths less than λ (and absorb most of the radiation of wavelengths larger than λ)? Is this why objects that reflect most visible light (and hence look white) also reflect most infrared radiation (since infrared waves have a shorter wavelength)?
There is no such rule. It depends on the material and its chemical bonds.
Incandescent bulb are black body emitters. Basically, something is heated enough so that it radiates most of the power put into it. The black body radiation spectrum is well known, with it being a function of temperature. The lower the temperature, the more the bulk of the radiation shifts to lower wavelengths.
Normal incadescent bulbs used for lighting are actually horribly inefficient. We don't have a material that we can heat up so hot to maximize the amount of visible light in its black body radiation spectrum. Actually, we can heat material that high, but it won't last long. We don't have a material that won't evaporate or sublime slowly enough to make a useable lightbulb with its black body radiation optimized for visible light.
Real LEBs (Light Emitting Bulbs) are a tradeoff between efficiency and lifetime. The hotter you run the fillament, the more efficient but also the shorter it lasts. This effect is quite non-linear. A little hotter makes a big difference.
There is therefore another reason to have separate heat lamps from ordinary LEBs, which is to make a more economical IR emitting bulb. The fillament operating temperature is adjusted for good IR production and less light. This does increase efficiency a little, but a ordinary LEB still emits most of its radiation as IR already. However, running the fillament cooler greatly increases bulb life and forces less tradeoffs in the design otherwise. Not only does the fillament evaporate slower, but the wire can be thicker, for example.
So dedicated heat lamps are a little more efficient at their intended purpose, but also last a lot longer and are more rugged and durable than incandescent bulbs shifted more towards producing visible light.
Best Answer
In a solid, "heat" consists of random vibrations of the atoms in that solid around their equilibrium positions. If the radiation striking that solid has a wavelength component that is close to one of those possible vibration modes, then the radiation will couple strongly with that vibratory mode and the solid will accept energy from the incident radiation and its temperature will rise.
If the incident radiation has too high a frequency (X-ray or gamma) the coupling is poor and the radiation just goes right through without interacting much. If the frequency is too low (radio frequencies lower than radar) the radiation bounces off and also doesn't interact much. This leaves certain specific frequency bands (like infrared and visible light wavelengths) where the interaction is strong.
Note that this picture is somewhat simplified in that there are frequency bands in the gigahertz range where the RF energy bounces off electrically conductive materials like metal (this gives us radar) but interacts strongly with dielectrics and materials containing water molecules (this gives us microwave ovens).
Note also as pointed out below by Frederic, molecules possess resonant modes that their constituent atoms do not and these can be excited by RF energy as well. Many of these molecular modes lie within the infrared range, giving rise to the field of IR spectroscopy.