Is the temperature rise caused by reflection of sunlight linear? I have saw a YouTube video showing large Fresnel lens can focus sunlight into a spot which can melt iron. 
I am very interesting of that, but large Fresnel lens are not that popular, so I was thinking if I could using a mirror array as the alternative solution.
The problem is I know the melting point of iron, but I don't know how to achieve that temperature using a mirror array.
So I ask if the rise in temperature caused by reflection of sunlight is linear? In that case, I could need a little calculating for how many mirror I need for building the array
 A: The lens works because it takes all the sunlight falling on its area, $A_1$, and focuses it onto a small spot $A_2$. The intensity in the spot is the intensity of the sunlight multiplied by $A_1/A_2$.
Exactly the same applies to a mirror. So provided your mirror has the same cross sectional area as the lens, and provided it can focus the light as effectively, it will produce the same temperature at the focus.
A: The rise in temperature would not be linear. As you are heating a piece of material, the energy loss due to radiation will increase (and so will the energy loss due to conduction and convection in air, but we are not going to discuss those here). Radiation losses can be estimated by a simple formula for black body radiators called the Stefan-Boltzmann law:
$j=\sigma T^4$
where $j$ is the radiation power emitted per surface area (i.e. it has units $[W/m^2]$), T is the absolute temperature and 
$\sigma={{2\pi^5}\over{15}}{{k^4}\over{c^2 h^3}}$. 
If we plug reasonable assumptions into this equation, e.g. $T=1808K$ for the melting point of pure iron, then we get that a one square meter sized black radiator at that temperature would lose 600kW of power trough radiation, alone.
Thankfully you are probably not trying to melt that much iron or we would need an industrial size smelter! So let's reduce the area to the surface area of a one cubic centimeter ingot, which is a little less than $10cm^2$. That's $0.001m^2$. So now you would only need $600kW/m^2*0.001m^2=600W$ of heating to overcome the radiation losses. 
In reality you will need more power than that because radiation losses are not your only concern, so your minimum heat flow should be $>1kW$ to melt a piece of iron of slightly less than $1cm^3$ of volume. Using a solar constant of roughly $1kW/m^2$ on a bright day your mirror will have to be $>1m^2$ in area to get close. 
How does one make a mirror like that on the cheap? With reflective foil and a little bit of air pressure. Google "mylar parabolic mirror" for some ideas. 
And now for the grand finale: DO NOT DO THIS AT HOME! 
This much sunlight focused on a spot will create a real hazard for everyone who looks at the focal point. The focal point is very hard to adjust on the right area and you would have to track the sun with a 3m or so focal length mirror positioned down to a few mm accuracy while staring into an extremely bright spot. 
Unless you are willing to build a radiation shield to keep everyone safe and devise methods to track safely, this is a very hazardous experiment from the second this mirror is fully pumped down to its parabolic shape and exposed to the sun. There will be a focal point somewhere, and if it hits a flammable object, there can be flames in seconds. If that flammable object is you, you will get third degree burns.
So while this is all very fun physics on paper, unless you are very good at engineering SAFE experiments, just let it go. If you want see iron melt, ask an iron foundry near you to let you see their equipment and watch how it's being used. 
