One cannot collimate light from an LED accurately without loosing a great deal of light and / or being happy with a very wide collimated beam, because the source is often quite a wide extended source (sometimes up to 1mm across). This may or may not be a helpful answer depending on exactly what you mean by collimated, i.e. how accurately you need to collimate, or, otherwise put, what the acceptable angular spread of your "collimated" output is and how much light you're willing to lose. You will need to do some calculations to find out.
Put simply: the main factors here are that if you design your collimation optics to collimate output from the centre of the chip (i.e. a light emitted from a point on the chip centre will become an on-axis plane wave at the collimator output), light from point sources at the sides of the chip will be mapped to slanted plane waves. The angular spread is then of the order of $w / f$, where $w$ is the LED chip's width and $f$ the focal length of the system. You can make this spread smaller by increasing $f$, but then the collimated beam becomes very wide and low intensity and the optics begin to get very big if you don't want to lose light. This may or may not be a problem.
The above is an imaging optics argument, but the idea is very general as we're stepping into the optical version of the second law of thermodynamics. In optical terms this is that the source's optical grasp (sometimes called optical extent or étendue) cannot be lowered by passive optical processing. Optical grasp is roughly the angular spread of a beam multiplied by the beam's width (why the simple word "optical spread" never took hold is beyond me and testament to how badly we English-speaking scientists treat our mother tongue). So you can see my imaging optics argument working here again: you can lower the angular spread in a beam at the expense of widening it. Of course you can simply throw away most of the beam if you want the beam narrow, but the system becomes highly inefficient. So you need to do some calculations to see what works for your application.
You can see the second law form more clearly with the following argument. If your LED chip were a blackbody radiator, and if you could make an arbitrarily narrow collimated beam from it, you could focus all the blackbody radiation in the collimated beam down onto a much smaller spot than the chip. The smaller spot's temperature would rise until steady state were reached, i.e. the power into the chip were equal to that radiated back. By the Stefan-Boltzmann law, if the image were smaller than the LED chip, the image would need to be at a higher temperature than the source to balance power flows, and this violates the Carnot/Clausius statement of second law of thermodynamics that heat cannot continuously, spontaneously be shifted from a lower temperature body to a higher temperature one.
Now the above doesn't rule out some fancy future active technology that can truly collimate an LED chip's output, needing work input of $k_B\,T\,\log 2$ joules for each bit of light state forgotten in accordance with the Landauer Principle form of the second law of thermodynamics. I say more about this in my answer here.
I've just recalled a question very like yours Is there any optical component that uniformizes the incoming light? and my answer to it is here.
There are several ways to approach this problem. If we can estimate the power density achieved in $W/m^2$, then the temperature that can be reached follows from the Stefan-Boltzmann law.
First method:
1) Take the total power collected, and see the size it got focused down to. You state the area of the mirror array is 0.6 m$^2$ (roughly), and with power incident on surface of earth about 1 kW, you get 600 W (consistent with the claimed 580 W from your question). If this is incident on an area of 1.14 cm$^2$, the power density is $\frac{580}{0.000114}\approx 5 MW/m^2$. A perfect black body of 1.14 cm$^2$, with this power incident on one side, and insulated perfectly from the other side, would be able to reach a temperature $T$ such that
$$\Phi = \sigma T^4$$
so:
$$T = \sqrt[4]{\frac{5 \cdot 10^6}{5.67\cdot 10^{-8}}} = 3000 K$$
(round numbers...)
However, if you tried to heat a disk (twice the area - no insulation on the back) the temperature reached would drop by $\sqrt[4]{2}$, to T = 2600 K (~2330°C). Note that the temperature doesn't drop all the way to 1500 K (~1230°C) - this is that 4th power in the Stefan-Boltzmann law shows its - ahem - power...
2) A second method would look at "how big The Sun looks" from the vantage point of the focus. When you are at the focal point of the mirrors, you "see" that it is as large as the satellite dish. That means that the heat flux increases, compared to the flux from The Sun, by the ratio of apparent areas. Which is actually the same thing as saying "you appear to be a lot closer to The Sun and can use the inverse square law to determine how much more power per unit area you experience".
Now The Sun looks like a disk that is 0.5° diameter; and with the dimensions given, your dish is equivalent to a disk with a diameter of 86 cm ($\sqrt{102\cdot73}=86.3$) and the focal distance is 138 cm (which I derived from the size of the focal spot, which is really an "image" of The Sun).
At a distance of 138 cm, a disk of 86 cm diameter "looks" 69x larger than The Sun - so it has an apparent area that is 4800x larger than The Sun - it therefore "feels the heat of 4800 suns". That is remarkably similar to the answer we got before, despite a different approach (but not really, if you look closely). So we will again get the same estimate for the temperature you can reach.
This second approach helps us understand that to get to higher temperatures, we need "The Sun to look even bigger" - that is, we need a larger dish or a smaller focal length. Making the focal length smaller will only work if the individual mirrors in the dish are small compared to the size of the focal spot - otherwise they will result in significant blurring of the focus and thus lower the power density. Simply increasing the size of the mirror array doesn't increase the power density - only the quality of the focus does. In principle the best you can do is create a giant 3D array of mirrors that make it look like The Sun is "everywhere" - a full $4\pi$ array would in principle give you the light of 50,000 suns (10x more than this mirror). Such a device would illuminate an object from all sides, and the temperature of that object would be (by the same equation as above) $\sqrt[4][10]$ x greater, or 5500 K (~5230°C). This is very close to the temperature of the surface of The Sun - and that is no surprise. If I had not used rounded values along the way (since there is a lot of estimating going on) I might have expected the answer to be 5776 K (~5500°C) - the surface temperature of The Sun, and the theoretical limit of such a device. So 5500 K (~5230°C) is "close enough for estimating".
Best Answer
The first thing you should be aware of, is the difference between temperature and heat. Heat is thermal energy (energy related to the mean speed of the molecules/atoms in your material) that a body has, and is useful because energy is conserved. So if you have heat transfer from a body to another, if you know that the first body gives the second 10 Joules, the second receives the same amount. Temperature is related to the amount of heat a body has, but it is not conserved, because different materials have different heat capacities. This means that different materials will have different temperatures when they have the same thermal energy.
In your solar heater, the receiver gets heat from the sun ($Q_\mathrm{sun}$), and loses heat to the atmosphere ($Q_\mathrm{atm}$). The steady state will be achieved when the energy flux to the body is equal to the energy flux from the body. That is, when $Q_\mathrm{sun} = Q_\mathrm{atm}$. Note that the units of $Q$ are Energy/time (e.g., Watts)
From the parameters you give, the power the receiver gets is $Q_\mathrm{sun}=P_\mathrm{sun}\,x\,r\,a$, that is, the power density of solar radiation at sea level (about 1.0 kW/m${}^2$) times the area of your dish times its reflectivity times the fraction of that that your receiver absorbs. Notice that this is (essentially) independent of the temperature of the receiver. I believe $a$ might depend a little on that, but not very much.
On the other hand, the heat your receiver loses is not so easy to calculate. It will lose heat by three different processes: radiation, conduction and convection. I think (though I'm not very familiar with this) that convection will be the most important process, but this requires that your receiver is thermally isolated form the rest of the structure of your solar condenser.
So lets say that $Q_\mathrm{atm} = Q_\mathrm{convection}$. If we use Newton's law of convection (not always valid, but a reasonable approximation here, I think) we have $Q_\mathrm{convection}= h\,S(T-y)$, where $h$ is the heat transfer coefficient and $S$ is the surface of the receiver. Notice that this does depend on the temperature of the receiver($T$). I'm not sure what your parameter $i$ is, but I think is related to the conduction of heat.
From the above equations, you can solve for $T$: $$P_\mathrm{sun}\,x\,r\,a = h\,S(T-y)$$ $$T = y + \frac{P_\mathrm{sun}\,x\,r\,a}{h\,S}$$
Regarding the most important factors in the performance, it's clear that the reflectivity is very important, not only in the visible, but in the infrared too, sine a lot of the energy of the sun's power is infrared. A metallic mirror is good for this. The absorption efficiency will be equally important, too. I think tis will be related to the reflectivity of your receiver: you want it as pitch black as possible (again, in the visible and infrared). You might want to paint it with soot, for example. I'm sure there is a lot of literature on the subject, but again, I'm not an expert on this.