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Finding the Kernel of a transformation

  1. Aug 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Let T:[R[/3]→[R[/3] so that when u=[R][/3] and v=(1,2,1), then T(u)=u×v

    a) Show that T is a linear transformation.
    b) Find T((3,0,2))
    c) Find a basis for Ker( T ). Give a geometric description of Ker( T ).
    2. Relevant equations
    Properties of a linear transformation:
    i) T(u+v)= T(u) + T(v)
    ii) T(ku)= KT(u)

    3. The attempt at a solution
    a) I have proven T is a linear transformation by closing it under both addition and scalar multiplication and am confident in my calculations for this part.
    b) I am not so confident in this part. Here are my calculations;

    T(u)= u×v
    T((3,0,2)) = (3,0,2)×(1,2,1)
    =i(0⋅1-2⋅2)-j(3⋅1-2⋅1)+k(3⋅2-0⋅1)
    =i(0-4)-j(3-2)+k(6-0)
    =(-4, -1, 6)
    ∴T((3,0,2))=(-4, -1, 6)

    I had to make an educated guess at how to approach this question as we haven't covered it in our coursework so far. Please let me know if this is correct or correct me if I am on the completely wrong path!

    c) I don't know how to approach this part. I know how to find a kernel, given a matrix but I don't have a matrix. Should I form a matrix with vectors (3,0,2), (1,2,1) and (-4,-1,6)? I attempted this but found consisted only of the zero vector and subsequently has no basis. Is the because T:[R[/3]→[R[/3] is an orthogonal projection and so the points T maps onto 0=(0,0,0)? If this is the case, should I still include my calculations of the matrix formed by the three vectors or could I just state this information and move on?

    Any help would be greatly appreciated. I am finding this topic quite difficult to get my head around as I am studying it through correspondence and am very disappointed by the amount of education and support I am getting from the university.
     
  2. jcsd
  3. Aug 27, 2015 #2

    fzero

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    Your solution to part (b) is fine. Since you say that you are comfortable with finding the kernel of a matrix, I would suggest that you find an expression for ##T## as a matrix. Repeat the calculation you did in part (b) for a general vector ##\mathbf{u} = (u_1,u_2,u_3)##. The result is going to be a vector linear in the components ##u_i##. You can find the matrix that has to act on ##\mathbf{u}## in order to give this expression. Then the remaining calculations will be familiar to you.
     
  4. Aug 27, 2015 #3
    Thank you for your help. So my matrix should consist of the vectors, v=(1,2,1) and ui=(0,1,-1)?
     
  5. Aug 27, 2015 #4
    I am thinking there is a vector missing? As the result should be a 3x3 matrix?
     
  6. Aug 27, 2015 #5

    fzero

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    You will compute
    $$T( (u_1,u_2,u_3) ) = (t_1,t_2,t_3)$$
    where the components ##t_i## are linear in the ##u_i##. In fact, the coefficients in the linear combinations are the matrix elements of ##T## that you are looking for.
     
  7. Aug 27, 2015 #6
    Oh, that makes sense! Thank you for all your help. I have now calculated the coefficients and formed the matrix
    1 2 1
    0 1 -1
    1 1 -3
    Which has a kernel of zero (what I was looking for). I appreciate all of your suggestions. Is there an option on this site to give feedback of users? Or can we only "like" a comment?
     
  8. Aug 27, 2015 #7

    fzero

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    Very good and you're welcome! Liking posts is suggested by the forum admins, but I am not personally worried about them (likes, not the admins!). I am happy enough that you replied to say that you understood the help and had success in your problem solving.
     
  9. Aug 28, 2015 #8
    Sorry to bother you again but were so helpful before.
    I have now come across this part (in the same question)
    e) Using the standard basis B={(1,0,0),(0,1,0),(0,0,1)}, find the
    matrix representation of T.

    I attempted the question by expressing the vector v=(1,2,1) with respect to the basis B. I found;

    1 0 0
    0 2 0
    0 0 1

    But I think I have done this wrong. Any help would be really appreciated.
     
  10. Aug 28, 2015 #9

    Zondrina

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    The matrix representation of ##T## has columns that are images of the standard basis vectors ##e_1, e_2,## and ##e_3##.

    That is, you must find ##T(e_1), T(e_2),## and ##T(e_3)##. Then you must form a matrix from the columns like so:

    $$A = [T(e_1), T(e_2), T(e_3)]$$
     
  11. Aug 28, 2015 #10

    HallsofIvy

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    The simplest way to write a linear transformation as a matrix is to apply it to the basis vectors. The results give the columns of the matrix.
    So what is T(1, 0, 0), T(0, 1, 0), and T(0, 0, 1).

    But, of course, you don't have to work with matrices. If (a, b, c) is any vector then T(a, b, c) is the cross product, (a, b, c)x (1, 2, 3). Write that out for general a, b, c and set each component equal to 0.
     
  12. Aug 28, 2015 #11

    fzero

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    Zondrina and HallsofIvy have already pointed out the correct way to answer that part. I would point out that this matrix form should agree with the one I suggested that you find earlier. However, you seem to have made a mistake when finding the matrix in post #6, that I apologize that I didn't realize earlier. Once you find the matrix using the basis vectors, you should go back and recheck that part.
     
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