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Linear Transformation

  1. Sep 23, 2015 #1
    1. The problem statement, all variables and given/known data

    Being T: ℝ2 → ℝ2 the linear operator which matrix in relation to basis B = {(-1, 1), (0, 1)} IS:
    [T]b =
    \begin{bmatrix}
    1 & 0\\
    -3 & 1
    \end{bmatrix}

    True or False: T(x,y) = (x, 3x+y) for all x,y∈ℝ?
    2. Relevant equations


    3. The attempt at a solution3

    So first I convert (x,y) from canonical to basis B and found (-1, 2).
    Next, I calculate [T]b. (-1,2). Found that the result is (-x, 3x +2y).
    Now, from my perspective i have to convert it to canonical basis again.
    But how am I suposed to do it?
     
    Last edited: Sep 23, 2015
  2. jcsd
  3. Sep 23, 2015 #2

    RUber

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    Homework Helper

    I am a little unclear of what the question is asking.
    ## [T]_b = \begin{bmatrix} 1&0\\-3&1 \end{bmatrix} ##
    And in stacked matrix form:
    ## B = \begin{bmatrix} -1&0\\1&1 \end{bmatrix} ##
    So, you are looking for a matrix A such that AB = [T]_b. Is that the correct understanding of the question?
     
  4. Sep 23, 2015 #3
    Sorry, I just want to know if the following statement is true: True or False: T(x,y) = (x, 3x+y) for all x,y∈ℝ?
     
  5. Sep 23, 2015 #4

    RUber

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    ##[T]_b## refers to the matrix operation conducted relative to the basis set B.
    So if you start with (x,y) in the standard basis, what is (x,y) in basis B? ( y-x, y).
    Then apply T_b to that and you will get T(x,y).
     
  6. Sep 23, 2015 #5
    I did what you said, but I get that T(x,y) = (-x+y, 3x-2y)
    How I convert it to standard basis again?
    I tried to multiply T(x,y).B , but my result is (4x -3y, 3x-2y)
    I don't know what I am doing wrong..
     
  7. Sep 23, 2015 #6

    RUber

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    This looks right.
    You don't -- x and y are representative of the standard basis.
     
  8. Sep 23, 2015 #7
    But the answer says that the statement T(x,y) = (x, 3x+y) for all x,y∈ℝ? is True.

    Also, [T]b.(x,y)b return the answer in Basis B, not?
     
  9. Sep 23, 2015 #8

    RUber

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    Hmmm. Okay. So perhaps we did the operation backward to get from (x,y) in the standard basis to (a,b) in the basis B.
    To point to ( x,y )_{x,y} you need a = -x, and b = x+y, or (-x, x+y)_B.
    Hit that with T_b.
    You get ( -x, 3x+x + y ) = (-x, 4x+y) in basis B.
    Putting that back into the canonical basis, gives ( x , -x+4x+y).
    Now I can see that the answer should be true.

    The problem was that we were using the definition for basis vectors of B, as the operation to move from (x,y)_{x,y} to (a,b)_B, where really it is B * (a,b)_B = (x,y).
    This means that:
    (a,b)_B = B^{-1} (x,y).
    T_b (a,b)_B = T_b B^{-1} (x,y), which as you said was in basis B.
    So B T_b (a,b)_B =B T_b B^{-1} (x,y) will return the answer in the standard basis.

    Sorry that took so long.
     
  10. Sep 24, 2015 #9

    HallsofIvy

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    Since you are told what T is "in relation to basis B = {(-1, 1), (0, 1)}", write (x, y) in that basis:
    (x, y)= a(-1, 1)+ b(0, 1)= (-a, a+ b) so we have -a= x and a+ b= y. a= -x so a+ b= -x+ b= y, b= x+ y.

    (x, y)= -x(-1, 1)+ (x+ y)(0, 1). T(x, y)= -xT(-1, 1)+ (x+ y)T(0, 1).
     
  11. Sep 24, 2015 #10
    Thank yu both, guys!!!!!!!
    It took long to understand but thanks to you all I managed to grasp it. And I think it'll stick now!
     
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