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Prove whether or not a transformation is one to one.

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Let T: R3 -> M22 by
    T\begin{bmatrix}
    a \\
    b \\
    c
    \label{T}
    \end{bmatrix}
    =
    \begin{bmatrix}
    a-b & b-c \\
    a+b & b+c\\
    \end{bmatrix}

    Is this transformation one-to-one?

    2. Relevant equations



    3. The attempt at a solution
    I am not really certain how to go about this. A nudge in the right direction?
     
  2. jcsd
  3. Mar 6, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    A linear transformation is one-to-one if it's kernel is {0}. Try computing the kernel.
     
  4. Mar 6, 2013 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    First, use "tex" and "/tex" at the beginning and end of an entire equation, not for individual parts:
    [tex]T\begin{bmatrix}a \\ b \\ c \end{bmatrix}= \begin{bmatrix}a- b & b- c \\ a+ b & b+ c\end{bmatrix}[/tex].

    Now, can we assume that you understand what "one to one" means? T is one to one if Tu= Tv only if u= v. Taking
    [tex]u= \begin{bmatrix}a \\ b \\ c\end{bmatrix}[/tex]
    and
    [tex]v=\begin{bmatrix}x \\ y \\ z\end{bmatrix}[/tex]
    then Tu= Tv means that
    [tex]\begin{bmatrix}a- b & b- c \\ a+ b & b+ c\end{bmatrix}= \begin{bmatrix}x- y & y- z \\ x+ y & y+ z\end{bmatrix}[/tex]
    which is the same as a- b= x- y, b- c= y- z, a+ b= x+ y, and b+ c= y+ z.

    Can you prove from those equations that a= x, b= y, and c= z?
    (Dick's method is equivalent to this, taking x= y= z= 0.)
     
  5. Mar 6, 2013 #4
    So if T(u) = T(v) then 0 = T(u)-T(v) = T(u-v) so u-v is in kerT and u-v=0 so u=v?
    But only if the kerT=0.
     
    Last edited: Mar 6, 2013
  6. Mar 6, 2013 #5

    Dick

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    Science Advisor
    Homework Helper

    That's it exactly. So just find ker(T).
     
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