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Determining linear transformation

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data
    T4 : R3 -> R4 is defined by T4(x1, x2, x3) = (0, x1, -3 + |x1|, x1 + x2)



    3. The attempt at a solution

    I know that T4(γ1x1 + γ2x2 + γ3x3) is a linear transformation IFF
    γ1.T4(x1) + γ2.T4(x2) + γ3.T4(x3)

    T4(λ10 + λ2x1 + λ3(-3+|x1|) = λ1.T4(0) + λ2.T4(x1) + λ3.T4(-3 + |x1|)

    T4(λ2x1 + λ3(-3 + |x1|) = λ2.T4(x1) + λ3.T4(-3 + |x1|)

    λ2.T4(x1) + λ3.T4(-3 + |x1|) = λ2.T4(x1) + λ3.T4(-3 + |x1|)

    I get a feeling I might be comitting a circular argument with the above proof.
     
  2. jcsd
  3. Mar 25, 2014 #2

    HallsofIvy

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    That's not much of a problem statement! You define T4 but do not say what you are to do with it! From what you do below, it looks like the problem is to determine whether or not T4 is a linear transformation.


    This makes no sense because you cannot apply T4 to numbers and you have already said that x1, x2, and x3 are numbers, not vectors in R3.

    What you want to show is that T4(au+ bv)= T4(a(x1, x2, x3)+ b(y1, y2, y3)= T4(ax1+ by1, ax2+ by2, ax3+ by3)= aT(x1, x2, x3)+ bT(y1, y2, y3).

    You are completely confused as to how T4 is defined!

    What is T4(ax1+ by1, ax2+ by2, ax3+ by3)?

    What are aT4(x1, x2, x3) and bT4(y1, y2, y3)?
     
  4. Mar 25, 2014 #3
    Terrible. I didnt realise they were numbers. Let me rework.
     
  5. Mar 25, 2014 #4

    HallsofIvy

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    You said T4 was applied to R3 and then gave a formula for T4(x1, x2, x3). If (x1, x2, x3) is in R3 the each of x1, x2, and x3 must be in R- a number.
     
  6. Mar 25, 2014 #5

    Mark44

    Staff: Mentor

    T4 is the transformation. As HallsOfIvy already said, T4(γ1x1 + γ2x2 + γ3x3) doesn't make any sense, nor do T4(x1), T4(x2), or T4(x3). The usual formulation that I remember is that T is a linear transformation if T(cx) = cT(x) and T(x + y) = T(x) + T(y), where x and y are vectors in the domain space, and c is a scalar.

    Also, it would be helpful to us if you learned how to make subscripts, using either LaTeX or the commands that are part of the advanced menu.

    In LaTex you can do this:
    Code (Text):
    ##c_1x_1##
    which renders like this: ##c_1x_1##

    From the advanced menu (click the Go Advanced button to open it), click the X2 button. This button puts [ sub ] and [ /sub ] tags (without the extra spaces) around the exponent.
    which renders like this
    Try to be more careful in what you type. As I read the line above, it took me awhile to figure out that λ10 and T4(0) were really λ1x0 and T4(x0). If you can't be bothered to check what you've written, why should we go out of our way to figure out what you really meant?
     
    Last edited: Mar 28, 2014
  7. Mar 28, 2014 #6
    I apologize for all the nitty gritty mistakes. I'm trying to learn as much as I can and in the process (lots of anxieties), I have a greater disposition to overlook the nitty gritty stuffs.

    In any case, I've solve the problem.
     
    Last edited by a moderator: Mar 28, 2014
  8. Mar 28, 2014 #7
    And then it conveniently slip my mind. It's so easy to overlook the information and take the real number as vectors in the midst of rushing to solve the problem.
    I'll be more careful.
     
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