Is f(x,y)=xy a linear transformation from R2 to R1?

[says]

Yep! I'm going to do exactly that! Thanks a lot for your help @Mark44

 

[says]

f(u + v) = f(<u1+v1,u2+v2>)=
f(u) = u1*u2
f(v) = v1*v2
f(cu) = f(<cu1,cu2>)=cu1⋅cu2(u1+v1)⋅(u2+v2) = u1u2+u1v2+v1u2+v1v2 ≠ u1*u2 + v1*v2

cu1⋅cu2 ≠ c*u1*u2

 

[Mark44]

f(u + v) = f(<u1+v1,u2+v2>)=
f(u) = u1*u2
f(v) = v1*v2
f(cu) = f(<cu1,cu2>)=cu1⋅cu2(u1+v1)⋅(u2+v2) = u1u2+u1v2+v1u2+v1v2 ≠ u1*u2 + v1*v2

This looks good. Even better would be:
f(u + v) = f(<u1+v1,u2+v2>) = (u1+v1)⋅(u2+v2) = u1u2+u1v2+v1u2+v1v2 ≠ u1*u2 + v1*v2 = f(u) + f(v)

cu1⋅cu2 ≠ c*u1*u2

because c²u1⋅u2 ≠ c*u1*u2 unless c = 0 or c = 1.