1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear Transformation Isomorphism

  1. Mar 1, 2010 #1
    I think I've solved this problem, but the examples in my textbook are not giving me any indication as to whether my reasoning is sound.

    1. The problem statement, all variables and given/known data
    Is the transformation

    [tex]T(M) = M\left[ \begin{array}{cccc} 1 & 2 \\ 3 & 6\end{array} \right][/tex]

    from [tex]\mathbb{R}[/tex]2x2 to [tex]\mathbb{R}[/tex]2x2 linear? If it is, determine whether it is an isomorphism.

    2. Relevant equations
    T(f + g) = T(f) + T(g)
    T (kf) = k T(f)
    T-1(T(M)) = M

    3. The attempt at a solution
    T(M1+M2) = T(M1) + T(M2)

    T(kM) = k T(M

    Therefore, T(M) is a linear transformation.

    [tex]\left[ \begin{array}{cccc} 1 & 2 \\ 3 & 6\end{array} \right][/tex] is not invertible, so T is an isomorphism if Ker(T) = 0.

    [tex]\left[ \begin{array}{cccc} m_1 & m_2 \\ m_3 & m_4\end{array} \right]\left[ \begin{array}{cccc} 1 & 2 \\ 3 & 6\end{array} \right]=\left[ \begin{array}{cccc} m_1+3m_2 & 2m_1+6m_2 \\ m_3+3m_4 & 2m_3+6m_4\end{array} \right]=\left[ \begin{array}{cccc} 0 & 0 \\ 0 & 0\end{array} \right][/tex]

    Then [tex]m_1 = -3m_2[/tex], [tex]m_3 = -3m_4[/tex] and

    [tex]Ker(T)=\left[ \begin{array}{cccc} -3 & 1 \\ -3 & 1\end{array} \right]\neq \left[ \begin{array}{cccc} 0 & 0 \\ 0 & 0\end{array} \right][/tex]

    Therefore, the transformation T(M) is linear, but is not an isomorphism.

    So I guess my question is, have I done this correctly? Thanks for any help.
     
  2. jcsd
  3. Mar 1, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, I believe you have done that correctly. T(0)=0 and T([-3,1],[-3,1])=0. Definitely not 1-1.
     
  4. Mar 1, 2010 #3
    Awesome. Thanks!
     
  5. Mar 2, 2010 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Because T is not invertible, it is not an isomorphism- every isomorphism is, by definition, invertible.

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook