Yes, I think any scaling should be linear. Proof: I just drew some lines in Microsoft Paint. Say we have points a=(1,1) and b=(1,2). Then, a+b = (2,3). We want to scale y by 2 and x by 1.5. This gives us points Sa = (1.5,2) and Sb = (1.5, 4). Then, Sa + Sb = (3, 6). If we go back and scale a+b (2,3), then we get (3,6). Therefore, f(x+y) = f(x) + f(y). For cf(x) = f(cx), consider vector a+b again. Assume c = 4. 4f(a+b) = 4(3,6) = (12,24). f(4*(a+b)) = f(4(2,3)) = f((8,12)) = (12,24). So, cf(x) = f(cx).
Is non-uniform scale also a linear transform?
Yes, I think any scaling should be linear. Proof: I just drew some lines in Microsoft Paint. Say we have points a=(1,1) and b=(1,2). Then, a+b = (2,3). We want to scale y by 2 and x by 1.5. This gives us points Sa = (1.5,2) and Sb = (1.5, 4). Then, Sa + Sb = (3, 6). If we go back and scale a+b (2,3), then we get (3,6). Therefore, f(x+y) = f(x) + f(y). For cf(x) = f(cx), consider vector a+b again. Assume c = 4. 4f(a+b) = 4(3,6) = (12,24). f(4*(a+b)) = f(4(2,3)) = f((8,12)) = (12,24). So, cf(x) = f(cx).