Linear transformations with functions

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The discussion centers on the linear transformation T: R^2→R^2 defined by T(x,y) = (x^2,y), which is identified as not being a linear transformation due to its failure to satisfy linearity conditions. The user seeks the preimage of the function f(x) = 2x + 1, interpreted as pairs (x, 2x + 1). The solution involves finding pairs (x, y) such that T(x, y) = (x^2, y) equals (a, 2a + 1) for some a. The preimage is determined to be the set of points in R^2 where y equals either 2√x + 1 or -2√x + 1. This highlights the complexity of working with transformations that involve non-linear components.
Xyius
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For the linear transformation,

T: R^2\rightarrow R^2, T(x,y) = (x^2,y)

find the preimage of..
f(x)= 2x+1

I have no trouble with these types of problems when it comes to vectors that aren't functions. Any help would be appreciated! Thanks!

~Matt
 
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This is no linear transformation, since T(α v) = T(α(x, y)) = T(αx, αy) = (α^2x^2, αy), which does not equal αT(v), for some v from R^2 and some α from R.
 
Xyius said:
For the linear transformation,

T: R^2\rightarrow R^2, T(x,y) = (x^2,y)

find the preimage of..
f(x)= 2x+1

I have no trouble with these types of problems when it comes to vectors that aren't functions. Any help would be appreciated! Thanks!

~Matt
Since T is mapping pairs of numbers into pairs of numbers, I assume that by "f(x)= 2x+1" you mean the set of pairs (x, 2x+1)[/math]

So you are looking for (x, y) such that T(x, y)= (x^2, y)= (a, 2a+1) for some number a. Okay, since x^2= a, x can be either \sqrt{a} or -\sqrt{a}. And, of course, y= 2a+ 1. That is, the preimage is the set in R^2 \{(x,y)| y= 2\sqrt{x}+ 1 or y= -2\sqrt{x}+ 1\}.
 
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