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

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SUMMARY

The discussion centers on determining the linearity of the transformation defined by the function f(x,y) = xy, which maps R² to R¹. Participants clarify that for a transformation to be linear, it must satisfy the properties T(V1 + V2) = T(V1) + T(V2) and T(cV1) = cT(V1). The conclusion reached is that f(x,y) = xy does not satisfy these properties, as demonstrated through specific examples and calculations, confirming that the transformation is not linear.

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  • #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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