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Switching Matrix Bases

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Let T be the linear transformation T=[1, 3, 2, 6] (the matrix has 1 and 3 on the top row and 2 and 6 on the bottom row) relative to the standard basis

    Find the matrix relative to the basis B= {(1, 2), (-3, 1)}


    2. Relevant equations



    3. The attempt at a solution
    So what I did was apply the basis vector 1 to the matrix and then apply the basis vector 2 to the matrix again and got (7, 14) and (0, 0) but I don't know where to go from here, what do I do with these 2 vectors?
     
  2. jcsd
  3. Oct 5, 2011 #2
    I am not sure, but when saying T=[1,3; 2,6] is relative to E, it means that
    T*e1^t=(1,2)
    T*e2^t=(3,6)
    Am I correct?
    If I have remembered correctly, then we want to find T' such that
    T'*(1,2)^t=(1,2)
    T'*(-3,1)^t=(3,6)
    or in other words
    [a,b;c,d]*(1,2)^t=(1,2) => (a+2b,c+2d)=(1,2)
    [a,b;c,d]*(-3,1)^t=(3,6) => (-3a+b,-3c+d)=(3,6)
    which is actually:
    [1] a+2b=1
    [2] -3a+b=3
    [3] c+2d=2
    [4] -3c+d=6
     
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