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Find out if it's linear transformation

  1. Oct 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Does a linear transformation ##g : \mathbb{R}^2 \rightarrow \mathbb{R}^2## so that ##g((2, -3)) = (5, -4)## and ##g((-\frac{1}{2}, \frac{3}{4})) = (0, 2)## exist?

    2. Relevant equations


    3. The attempt at a solution
    For a linear transformation to exist we need to know if those two satisfy the following conditions:
    ##g(\vec v_1 + \vec v_2) = g(\vec v_1) + g(\vec v_2)## for any ##\vec v_1, \vec v_2## in ##\mathbb{R}^2##
    ##g(\alpha \vec v) = \alpha g(\vec v)## for any scalar ##\alpha##
    So I thought I should do like this:
    I sum the two vectors ##(2, -3)## and ##(-\frac{1}{2}, \frac{3}{4})## like this ##g((2, -3) + (-\frac{1}{2}, \frac{3}{4}))## and put it equal to the sum of ##(5, -4)## and ##(0, 2)##. I will then have something like this:
    ##g((2, -3) + (-\frac{1}{2}, \frac{3}{4})) = (5, -4) + (0, 2)##
    ##g((\frac{3}{2}, -\frac{9}{4})) = (5, -2)##
    But at this point I actually don't know if it really exist or not. Am I missing something?
     
  2. jcsd
  3. Oct 16, 2016 #2

    PeroK

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    Do you notice anything about the two vectors being acted on by ##g##?
     
  4. Oct 16, 2016 #3
    Oh! If I multiply by ##-4## the second vector acted on by ##g##, I'll get the first vector! But I have to multiply the vector that is is equal to by ##-4## too. And since ##(0, -8)## is not equal to ##(5, -4)## means that the linear transformation doesn't exist, right?
     
  5. Oct 16, 2016 #4

    PeroK

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    Exactly.
     
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