Tricky transformations

  • #1
We know that a transformation from V to W is linear if the following hold:
1.) For every x, y in V, T(x+y) = T(x) + T(y)
2.) For every x in V and for every a in R (real numbers), T(ax) = aT(x)

I need two nonlinear transformations from R to R. One must satisfy #1 above and violate #2. The other must violate #1 and satisfy #2.
 

Answers and Replies

  • #2
Hurkyl
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Sounds like homework, so I'm moving it there. What have you done so far on this problem?
 
  • #3
dextercioby
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The second one is really simple.Just take a nonlinear operator

[tex] T(x)=x^{2} [/tex]

I'll let a mathematician deal with the difficult issue.

Daniel.


EDIT:The above is wrong.I'll let a mathematician deal with the whole problem.
 
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  • #4
Hurkyl
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So are you claiming that T(ax) = aT(x) for your function?
 
  • #5
dextercioby
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Ooops,sorry,Hurkyl,didn't see it. :redface: That nasty quadratic breaks both if them...

Daniel.
 
  • #6
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I think T(u) = u + k works for #1 but not #2.

I don't know about the other one.
 
  • #7
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No, it doesn't. With that,

[tex]T(u+v) = u+v+k \neq u+v+2k = T(u) + T(v).[/tex]
 
  • #8
Hurkyl
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Here's a hint...

Suppose T(x) satisfies #1, and that you know T(x). Then, you also know T(2x) and T(3x), right? What about T(x/2)? T(47x/163)?
 
  • #9
Hurkyl
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PS, this is a standard method of attack, and it's a good way to learn what things "really" mean.

The whole point is to learn precisely what property #1 tells you, so you can find out what you can "break" so that property #2 fails. (and vice versa)
 
  • #10
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Data said:
No, it doesn't. With that,

[tex]T(u+v) = u+v+k \neq u+v+2k = T(u) + T(v).[/tex]
Data,

Touche! What was I thinking?
 
  • #11
Hurkyl
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It's a mistake we all make at least once -- the trick is to catch it before you tell anybody. :smile:
 

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