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Linear Algebra (Linear Transformation)

  1. Jul 11, 2009 #1
    1. The problem statement, all variables and given/known data
    True or False:
    [tex]
    T(x,y)=(2x+5y,-x+2)\text{ is a linear transformation from }\mathbb{R}^2\text{ to }\mathbb{R}^2.
    [/tex]


    2. Relevant equations
    None


    3. The attempt at a solution
    I thought the answer was true, but the correct answer is false. Here is my reasoning for true:

    T depends only on x and y and the transformation depends only on x and y, so it must be in the same space.
     
  2. jcsd
  3. Jul 11, 2009 #2
    Your reasoning is not detailed enough. Do you know how to check if a transformation is linear?
     
  4. Jul 11, 2009 #3
    I think I do. If a transformation is linear, then:

    [tex]T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})[/tex]
    and
    [tex]T(a\mathbf{u})=aT(\mathbf{u})[/tex]
     
  5. Jul 11, 2009 #4
    You need to show your answer though. A function [itex]T:V\to W[/itex] is a linear transformation if for all [itex]\vec u, \vec v\in V[/itex] and for all scalars [itex]r\in\mathds{R}[/itex],
    [tex]T(\vec u+\vec v) = T(\vec u)+T(\vec v) [/tex]
    [tex]T(r\vec u) = rT(\vec u)[/tex]
    (You can actually combine these two requirements into one, but I think it is usually best to leave them separate early on.)

    In your case, let [itex]\vec u=(x,y)[/itex] and [itex]\vec v = (u,v)[/itex]. (Don't get confused between my re-use of u and v.) You need to try and show that
    [tex]T(\vec u+\vec v) = T(x+u,y+v) = T(x,y)+T(u,v) = T(\vec u)+T(\vec v)[/tex]
    [tex] T(r\vec u) = T(rx,ry) = rT(x,y) = rT(\vec u)[/itex]

    If you can show this, then T is a linear transformation. If either of these two statements are not true, then T is not a linear transformation.
     
  6. Jul 11, 2009 #5
    Do I pick arbitrary values for u and v and then test them, or is there a more systematic way to go about it?
     
  7. Jul 11, 2009 #6
    You pick arbitrary values just as I mentioned at the end of my last post.
     
  8. Jul 11, 2009 #7
    Okay, so this is obviously not a linear transformation because the first component involves both an x and a y, correct?.
     
  9. Jul 11, 2009 #8
    I'm not sure what you mean by an x and a y. It's not a linear transformation because T(u+v) is not equal to T(u)+T(v). You probably need to illustrate that by picking two arbitrary vectors, and performing the transformations. You will then see that they are not equal.
     
  10. Jul 11, 2009 #9
    I forgot, you need to show a counterexample.
     
  11. Jul 11, 2009 #10
    I get that they are equal. What am I doing wrong?

    [tex]
    \text{Let }\overset{\rightharpoonup }{u}=\left(
    \begin{array}{c}
    1 \\
    0
    \end{array}
    \right)\text{ and }\overset{\rightharpoonup }{u}=\left(
    \begin{array}{c}
    0 \\
    1
    \end{array}
    \right)
    [/tex]

    [tex]
    T(u+v)=\left(
    \begin{array}{c}
    2+5 \\
    2-1
    \end{array}
    \right)=\left(
    \begin{array}{c}
    7 \\
    1
    \end{array}
    \right)
    [/tex]

    [tex]
    T(u)+T(v)=\left(
    \begin{array}{c}
    2 \\
    2-1
    \end{array}
    \right)+\left(
    \begin{array}{c}
    5 \\
    0
    \end{array}
    \right)=\left(
    \begin{array}{c}
    7 \\
    1
    \end{array}
    \right)
    [/tex]
     
  12. Jul 11, 2009 #11
    Okay, I just happened to pick a case that works. Here is one that doesn't:

    [tex]
    \text{Let }\overset{\rightharpoonup }{u}=\left(
    \begin{array}{c}
    2 \\
    0
    \end{array}
    \right)\text{and }\overset{\rightharpoonup }{u}=\left(
    \begin{array}{c}
    0 \\
    1
    \end{array}
    \right)
    [/tex]

    [tex]
    T(u+v)=\left(
    \begin{array}{c}
    4+5 \\
    -2+2
    \end{array}
    \right)=\left(
    \begin{array}{c}
    9 \\
    0
    \end{array}
    \right)
    [/tex]

    [tex]
    T(u)+T(v)=\left(
    \begin{array}{c}
    4 \\
    -2+2
    \end{array}
    \right)+\left(
    \begin{array}{c}
    5 \\
    2
    \end{array}
    \right)=\left(
    \begin{array}{c}
    9 \\
    2
    \end{array}
    \right)
    [/tex]

    Thanks for your help everyone!
     
  13. Jul 11, 2009 #12

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You might want to check the highlighted value ...
     
  14. Jul 11, 2009 #13
    Oops, you are right. Is this correct?

    [tex]
    \text{Let }\overset{\rightharpoonup }{u}=\left(
    \begin{array}{c}
    1 \\
    0
    \end{array}
    \right)\text{ and }\overset{\rightharpoonup }{u}=\left(
    \begin{array}{c}
    0 \\
    1
    \end{array}
    \right)
    [/tex]

    [tex]
    T(u+v)=\left(
    \begin{array}{c}
    2+5 \\
    2-1
    \end{array}
    \right)=\left(
    \begin{array}{c}
    7 \\
    1
    \end{array}
    \right)
    [/tex]

    [tex]
    T(u)+T(v)=\left(
    \begin{array}{c}
    2 \\
    2-1
    \end{array}
    \right)+\left(
    \begin{array}{c}
    5 \\
    2
    \end{array}
    \right)=\left(
    \begin{array}{c}
    7 \\
    3
    \end{array}
    \right)
    [/tex]
     
  15. Jul 11, 2009 #14

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Much better :approve:

    P.S. Nice use of latex
     
  16. Jul 11, 2009 #15
    Thanks!
     
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