Linear Transformation of R2 to R1: Determining Linearity of f(x,y)=xy

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The discussion centers on determining whether the transformation f(x,y) = xy is linear. Participants emphasize the need to verify the two properties of linear transformations: additivity and homogeneity. It is clarified that f(u + v) does not equal f(u) + f(v), and f(cu) does not equal cf(u), demonstrating that the transformation fails to meet the criteria for linearity. A specific example using numerical values is provided to illustrate that (u1 + v1)(u2 + v2) is not equal to u1u2 + v1v2, confirming the non-linearity of the transformation. Ultimately, the conclusion is reached that f(x,y) = xy is not a linear transformation.
  • #31
Yep! I'm going to do exactly that! Thanks a lot for your help @Mark44
 
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  • #32
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
 
  • #33
says said:
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)
says said:
cu1⋅cu2 ≠ c*u1*u2
because c2u1⋅u2 ≠ c*u1*u2 unless c = 0 or c = 1.
 

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