Ideal sources are linear. On way to think about this is solving simple circuits in the VI plane. Since I=V/R, a resistor which is a linear element is a line through the origin with slope 1/R. If we connect this to a voltage source, we solve it graphically. The graph of the voltage source is a vertical line at the value of the voltage. The intersection of resistor line and the voltage line will give the current. Now to be linear means that add to or multiply the input, the voltage source in this case, the output will be will be added to or multiplied by that same amount. So now consider if you took 2*V. The vertical line will be twice as far from the origin and so will the current. This is true because all the components including the source are linear. If you put a diode equation on this graph doubling the voltage would not double the current. The diode is a nonlinear element. You can see that the ideal current source is also linear because its graph is a horizontal line at the given current.
Another example of this is if we put a voltage divider across an ideal voltage source. Let's say you set it up to divide by 5. If I put 5 volts in, I get 1 volts out. Now if you multiply this voltage by 3 you get 15 volts in and 3 Volts out. The output is three times as much because all the elements including the source are linear.