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Linear Transformations

  1. Nov 22, 2005 #1
    I have an assignment problem and I dont even know where to start... I'm taking the course through correspondence so i have no notes or prof to talk to... i've read my text and course manual over and over again but I just can't figure it out
    Let T: P2->P2 be a linear transformation defined by T(a+bx+cx^2) = (a+b+c)+2(b+c)x+3cx^2. Let B = {1,1+x,3+4x+2x^2} and B' = {1,x,x^2} be two ordered bases of P2. Find the following:
    a) (T)B'B'
    b) P the transition matrix from B to B'
    c) P-1 the transition matrix from B' to B
    d) (T)BB
    i somewhat understand part b and c... i know there is explanation of those in my text that i understand... but i dont know how to even approach part a and d... there is nothing even remotely similar in my text examples
     
  2. jcsd
  3. Nov 22, 2005 #2

    Physics Monkey

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    To calculate the matrix of a linear transformation in a given basis, you must calculate the result of the transformation on each of your basis vectors and then decompose that result in terms of the basis vectors. The coeffecients of that decomposition are the matrix elements.

    For example, consider the transformation T that maps P1 to P1 and is given by T(a + bx) = (a+b) + (a-b)x. To write this in the basis consisting of 1+x and 1-x, I follow the procedure outlined above. First, T(1+x) = 2, and I decompose this in terms of the basis vectors as 2 = 1*(1+x) + 1*(1-x). This tells me that the first column of my matrix has 1 in the first row (from the number in front of 1+x) and 1 in the second row (from the number in front of 1-x). Now I do the same for 1-x to find T(1-x) = 2x = 1*(1+x) + (-1)*(1-x). This means the second column of my matrix has 1 in the first row and -1 in the second row. Therefore in this basis, the first basis vector 1+x is represented by
    [tex]
    \left[ \begin{array}{c} 1 \\ 0 \end{array} \right],
    [/tex]
    my second basis vector 1-x is represented by
    [tex]
    \left[ \begin{array}{c} 0 \\ 1 \end{array} \right],
    [/tex]
    and the linear transformation is represented by the matrix
    [tex]
    \left[ \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right].
    [/tex]

    Hope this helps.

    Edit: I got something backwards in the original post, fixed now.
     
    Last edited: Nov 22, 2005
  4. Nov 22, 2005 #3
    how did you get T(1+x) = 2 and T(1-x) = 2x ?
     
  5. Nov 23, 2005 #4

    HallsofIvy

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    "For example, consider the transformation T that maps P1 to P1 and is given by T(a + bx) = (a+b) + (a-b)x."

    He's not doing your T, he made up one of his own so you would have the experience of apply the concept to yours.

    In 1+ x, a= 1, b= 1 so (a+b)= (1+1)= 2 and (a-b)= (1-1)= 0.
    T(1+x)= 2+ 0x= 2.
    In 1- x, a= 1, b= -1 so (a+b)= (1+(-1))= 0 and (a-b)= (1-(-1))= 2.
    T(1-x)= 0+ 2x= 2x.

    Now for your problem: You said "T(a+bx+cx^2) = (a+b+c)+2(b+c)x+3cx^2" and B = {1,1+x,3+4x+2x^2}
    Write 1 as 1+ 0x+ 0x^2 so a= 1, b= 0 ,c= 0. What is T(1)?
    Write 1+ x as 1+ 1x+ 0x^2 so z= 1, b= 1, c= 0. What is T(1+x)?
    For 3+ 4x+ 2x^2, a= 3, b= 4, c= 2. What is T(3+4x+2x^2)?

    Now, write each of those answers as a linear combination of {1,1+x,3+4x+2x^2}. The coefficients will be the columns of the matrix (T)BB. The other way, (T)B'B' is the same idea but much easier!
    For the transition matrices, apply T to each basis vector but write the result in terms of the other basis.
     
  6. Nov 23, 2005 #5
    okay, so am I doing this right?:
    T(1) = 1
    T(1+x) = 2
    T(3+4x+2x^2) = 9
    then for the matrix TBB i would get

    |1 0 0|
    |1 1 0|
    |3 4 2|

    is this right?

    why is it written TBB... why are there two B's?
     
  7. Nov 23, 2005 #6

    HallsofIvy

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    No, that's not at all right. You were given that
    T(a+bx+cx^2) = (a+b+c)+2(b+c)x+3cx^2
    if a+ bx+ cx^2= 1, then a= 1, b= 0, c= 0 so T(1)= 1 as you have
    but if a+ bx+ cx^2= 1+ x, then a= 1, b= 1, c= 0 so T(1+x)= (1+1)+ 2(1+0)x+ 3(0)x^2= 2+ 2x, not just "2".
    If a+ bx+ cx^2= 3+ 4x+ 2x^2, then a= 3, b= 4, c= 2 so T(3+4x+2x^2)= (3+4+2)+ 2(4+2)x+ 3(2)x^2= 9+ 12x+ 6x^2.
    Now write each of those as a linear combination of 1, 1+x, 3+4x+2x^2.
    Clearly 1= 1(1)+ 0(1+x)+ 0(3+4x+2x^2 so the first column of the matrix is [1 0 0].
    2+ 2x= 0(1)+ 2(1+x)+ 0(3+4x+2x^2) so the second column is [0 2 0].
    9+ 12x+ 6x^2 is harder. 9+ 12x+ 6x^2= a(1)+ b(1+x)+ c(3+4x+2x^2)=
    (a+b+3c)+ (b+ 2c)x+ 2cx^2 so we must have a+ b+ 3c= 9, b+ 2c= 12, and 2c= 2. From the last, c= 1. Then b+ 2= 12 so b= 10. Finally, a+ 10+ 3= 9 so a= -4. The final column is [-4 10 1].
    [/quote]then for the matrix TBB i would get
    |1 0 0|
    |1 1 0|
    |3 4 2|
    is this right?
    why is it written TBB... why are there two B's?[/QUOTE]
    Every linear transformation has a "domain" space and a "range" space which could have different bases. Here you are using the same base (1, 1+x, 3+4x+2x^2) for both the domain and the range.
     
  8. Nov 24, 2005 #7
    okay... i think i get it, but for the last one... why is it b + 2c = 2 and 2c = 2... shouldn't it be b+4c=12 and 2c=6?
     
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