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Proving a transformation is linear

  1. Feb 28, 2015 #1
    1. The problem statement, all variables and given/known data
    0%5Cbegin%7Bbmatrix%7D%202x_1%20-%203x_2%5C%5C%202x_2%5C%5C%204x_1%20+%203%20%5Cend%7Bbmatrix%7D.gif
    If T is linear, show that it is linear by finding a standard matrix A for T so that:
    gif.gif
    Also show that this equation holds for the matrix you have found. If T is not linear, prove that T is not linear by showing that it does not fit the definition of a linear transformation
    2. Relevant equations
    Definition of a linear transformation:
    [itex] T(\vec{u}+\vec{v}) = T(\vec{u})+T(\vec{v}) [/itex]
    [itex] T(c\vec{u}) = cT(\vec{u}) [/itex]

    3. The attempt at a solution
    First I let $$ \vec{e}_{1} =
    \begin{bmatrix}
    1\\
    0
    \\0

    \end{bmatrix} $$
    $$ \vec{e}_{2} =
    \begin{bmatrix}
    0\\
    1
    \\0

    \end{bmatrix}$$
    $$ \vec{e}_{3} =
    \begin{bmatrix}
    0\\
    0
    \\1

    \end{bmatrix}$$
    However, when I go to separate
    $$
    \vec{b} = \begin{bmatrix}
    2x_1 - 3x_2\\
    2x_2\\
    4x_1 + 3
    \end{bmatrix}$$ I am not sure how to handle the constant, i.e, I am not sure how to rewrite as [itex] A\vec{x} [/itex]. I think once I figure that out I should be able to do the rest of problem
     
  2. jcsd
  3. Feb 28, 2015 #2

    Orodruin

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    It is not linear ... Just find a counter example to the definitions of what a linear transformation is.
     
  4. Feb 28, 2015 #3

    wabbit

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    Gold Member

    (Edit : deleted, duplicate response)
     
    Last edited: Feb 28, 2015
  5. Feb 28, 2015 #4
    Ok, I didn't realize that was all I needed to do. I have attached a picture of my work because it would be kind of long to write out in latex:
    hsPShzu.jpg
     
  6. Feb 28, 2015 #5

    Dick

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    You could simplify that a lot. If T is linear, then T(0)=0, since T(0x)=0T(x). Is it for your given T?
     
  7. Feb 28, 2015 #6
    wow I hadn't thought of that, sort of makes the problem trivial lol...anyway thanks for the help :)
     
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