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Homework Help: Matrix of Linear Transformations

  1. Apr 25, 2010 #1
    L: R^2=>R^2 is defined by L(x,y)= (x+2y), (2x-y)

    let S be the natural basis for R^2 and T=(-1,2), (2,0). T is another basis for R^2.

    Find the matrix representing L with respect to

    A) S
    B) S and T
    C) T and S
    D) T
    E) Compute L
    L(1,2)
    using the definition of L and also the matrices obtained in a,b,c,d.


    I am pretty lost here. I plugged in 1,0 0,1 for the natural basis, solving for part A. plugged in T=(-1,2), (2,0) to solve for part C. Is this correct? Any ideas on the rest?
     
  2. jcsd
  3. Apr 25, 2010 #2

    HallsofIvy

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    Science Advisor

    I'm not sure what you mean by "plugging in 1, 0, 0, 1" but you can get the matrix representing a linear transformation from a vector space with a given basis, to a vector space with another basis, by applying the linear transformation to each basis vector in the domain space in turn and writing the result in terms of the basis for the range space. The coefficients will be a column in the matrix.

    For example, you are told that L(x,y)= (x+2y), (2x-y) so L(1, 0)= ((1+ 0),(2- 0)= (1, 2) and L(0,1)= (0+2(1), 2(0)- 1)= (2, -1). Since (1, 2)= 1(1, 0)+ 2(0,1) and (2, -1)= 2(1, 0)+ (-1)(0, 1) those are the columns of the matrix for (A).

    (B) asks for the matrix with respect to "S and T". Again, we have L(1, 0)= (1, 2) and L(0,1)= (2, -1). Now, we need to write those results in terms of basis T: find a and b so that a(-1, 2)+ b(2, 0)= (1, 2) and c and d so that c(-1, 2)+ d(2, 0)= (2, -1). a and b will be the values for the first column and c and d for the second column.

    (C) asks for the reverse of that. L(-1, 2)= (-1+ 2(2), 2(-1)- 2)= (-3, -4) and L(2, 0)= (2+ 2(0),2(2)- 0)= (2, 4) and those are already written in terms of basis S.

    (D) Knowing that L(-1, 2)= (-3, -4) and L(2, 0)= (2, 4), write them in terms of basis T. Find a and b so that a(-1, 2)+ b(2, 0)= (-3, -4) and c and d so that c(-1, 2)+ b(2, 0)= (2, 4).
     
  4. Apr 25, 2010 #3
    Thanks Hall, that was very clear. By pluggin in I meant using the vectors in the transformations.

    Just to clarify, in part C "L(-1, 2)= (-3, -4)' did you mean (3,-4)?

    Thanks
     
  5. Apr 25, 2010 #4
    Can you shed some light on how to evaluate part E?
     
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