one simple property of ANY linear transformation T is that: T(0) = 0 (where these are 0-vectors, not 0-scalars).
here is the proof:
T(0) = T(0+0) (since 0 = 0+0)
= T(0) + T(0) (since T is linear).
subtracting T(0) from both sides, we get:
0 = T(0) - T(0) = [T(0) + T(0)] - T(0) = T(0) + [T(0) - T(0)] = T(0) + 0 = T(0).
now let's look at YOUR function:
T(x,y) = (2x-3y,x+4,5y).
the 0-vector of R^{2} is (0,0), and T(0,0) = (2*0-3*0,0+4,5*0) = (0,4,0) ≠ (0,0,0), so....
I tried it with arbitrary values actually. I let u=(x1,x2) and v=(y1,y2). So T(u+v)=T(x1+y1,x2+y2). So it then becomes T=(2x1+2y1-3x2-3y2,x1+y1+4,5x2+5y2 which is the same as T(u)+T(v)=(2x1-3x2,x1,5x2)+(2y1-3y2,y1+4,5y2) but i dont know where to take it from here? As you can see T(u) /= T(v),
My bad I think I figured it out. Since T(u+v)=T(u)+T(v), im supposed to evaluate each vector separately on the RHS of the equation. So vector u=[x1,x2] and v=[y1,y2]. T(u)=[2x1-3x2,x1+4,5x2] and T(v)=[2y1-3y2,y1+4,5y2] so T(u)+T(v)=[2x1-3x2+2y1-3y2,x1+y1+8.,5x2+5y2]. And that does not equal the LHS of the equation making this not a linear transformation. I know when writing vectors im not supposed to use comma's to separate the entries but I
don't know latex yet.