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One more linear transformation

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data

    M22 ---> R is a linear transformation.

    given:

    T[ 1 0 ] = 1
    ,,[ 0 0 ]

    T[ 1 1 ] = 2
    ,,[ 0 0 ]

    T[ 1 1 ] = 3
    ,,[ 1 0 ]

    T[ 1 1 ] = 4
    ,,[ 1 1 ]

    find
    T[ 1 3 ]
    ,,[ 4 2 ]

    and
    T[ a b ]
    ,,[ c d ]

    2. Relevant equations

    none.

    3. The attempt at a solution

    I don't know if it is valid to "add" the transformations. I'm tempted to add the given linear transformations like 2T + 2T - 1T - 2T which gives
    T[ 1 3 ]
    ,,[ 4 2 ]
    which I believe it is 10. Is that valid?

    T[ a b ]
    ,,[ c d ]
    Now this one is really making me stuck. It looks like something recursive but I can't wrap my head around it. Is there something special to this?
     
  2. jcsd
  3. Mar 5, 2009 #2
    If you subtract the third equation from the fourth, you get
    [tex]T\[
    \left(
    \begin{array}{cc}
    0 & 0 \\
    0 & 1 \end{array}
    \right)=1\] [/tex]

    Similarly you get

    [tex]T\[
    \left(
    \begin{array}{cc}
    0 & 0 \\
    1 & 0 \end{array}
    \right) = T
    \left(
    \begin{array}{cc}
    0 & 1 \\
    0 & 0 \end{array}
    \right) = 1\][/tex]

    Correct. For linear transformations, T(aA+bB)=aT(A)+bT(B).

    Should be obvious by now.
     
  4. Mar 5, 2009 #3
    Wow, that's amazing that you saw through the pattern in < 15 minutes. It's overly dead obvious now. Thank you so very much! :smile:
     
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