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R2 to R3 transformation

  1. Sep 23, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the transformation T from ℝ2 to ℝ3 given by,

    $$T\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = x_1\begin{bmatrix} 1 \\ 2 \\ 3\end{bmatrix} + x_2\begin{bmatrix} 4 \\ 5 \\ 6\end{bmatrix}$$

    Is this transformation linear? If so, find its matrix

    2. Relevant equations
    A transformation is not linear unless:
    a. T(v+w) = T(v) + T(w)
    b. T(kv) = kT(v)
    for all vectors v and w and scalars k in R^m



    3. The attempt at a solution
    All I know right now are the two relevant equations that I posted, but I can't seem to find how to apply them here. I'm not sure how to work with all of these column vectors. Any help is appreciated
     
    Last edited by a moderator: Sep 24, 2013
  2. jcsd
  3. Sep 23, 2013 #2

    Mark44

    Staff: Mentor

    For your problem, the vectors v and w are in R2.
    Start with a couple of arbitrary vectors v = <v1, v2> and w = <w1, w2>, and determine whether T(v + w) = T(v) + T(w). If these expressions aren't equal, you're done - T is not a linear transformation.
    If they are equal, determine whether T(kv) = kT(v). If these aren't equal, T is not a linear transformation. If they are equal, then T is a linear transformation.

    BTW, it's not too hard to use LaTeX to render vectors. Here's your defining equation for the transformation. Right-click anywhere in the expression to see the code that produced it.
    $$ T\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = x_1\begin{bmatrix} 1 \\ 2 \\ 3\end{bmatrix} + x_2\begin{bmatrix} 4 \\ 5 \\ 6\end{bmatrix} $$
     
    Last edited: Sep 23, 2013
  4. Sep 23, 2013 #3
    So, if my vector v is <1,1> and w is <2,2>, my matrix should look like the below?

    $$T\begin{bmatrix} 1 \ 2 \\ 1 \ 2 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3\end{bmatrix} + \begin{bmatrix} 1 \ 2 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \\ 6\end{bmatrix}$$

    I'm still not sure even on the format. If this is right, does solving this simply yield T(v+w)= 1(1)+1(2)+1(3)+2(1)+2(2)+2(3) for the first part, and etc. ?
     
    Last edited by a moderator: Sep 24, 2013
  5. Sep 23, 2013 #4

    Mark44

    Staff: Mentor

    No, that's not it. T operates on vectors in R2, not matrices. The output is a vector in R3.

    $$T\left(\begin{bmatrix} 1 \\ 1 \end{bmatrix} + \begin{bmatrix} 2 \\ 2 \end{bmatrix}\right)= T\left(\begin{bmatrix} 3 \\ 3 \end{bmatrix}\right)$$

    What do you get for the above?
    How does it compare to T(<1, 1>) + T(<2, 2>)?
     
  6. Sep 23, 2013 #5
    I just don't understand how to multiply column vectors... and it doesn't seem right to me to multiply a column vector with 2 rows to one of 3. But I suppose you would get 1+2+3+1+2+3=12 for the first part, 2+4+6+2+4+6=24 for the second part?

    EDIT: Okay, I think I'm starting to see what you mean. If I plug in T(3,3), then I get 3 times the column vector 1,2,3 plus 3 times the column vector 4,5,6 right? So 18+45?
     
    Last edited: Sep 23, 2013
  7. Sep 24, 2013 #6

    Mark44

    Staff: Mentor

    You're partly right, but 18 + 45 is a number, and you should be getting a vector in R3.

    You get this:
    $$ 3\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} +3\begin{bmatrix}4 \\ 5 \\ 6 \end{bmatrix} $$

    When you multiply each column vector by 3, and then add the two vectors together, what do you get?
     
  8. Sep 24, 2013 #7
    Ahh, so the vector would be <3+12,6+15,9+18> = <15,21,27> ? So then I plug in T(1,1) and T(2,2) to get two vectors, add them together, and see if they are equal to <15,21,27>?
     
  9. Sep 24, 2013 #8

    Mark44

    Staff: Mentor

    Yes to both.
     
  10. Sep 24, 2013 #9
    T(1,1) = <5,7,9> T(2,2) = <10,14,18> and T(1,1)+T(2,2) = <15,21,27> = T(3,3). Okay, so now I have verified that the first equation is true. Now for the second, let's say I use the same vector v = <1,1>, and I'll let k = 4. T(4v) is now <20,28,36> and 4T(v) = <20,28,36>.

    So the equation is infact linear?
     
  11. Sep 24, 2013 #10

    Mark44

    Staff: Mentor

    Well, maybe. You can't use specific vectors such as <1, 1> to show that the transformation is linear. The relationships have to hold for any choices of x = <x1, x2>T and y = <y1, y2>T, and any scalar k. (The T exponent means the transpose of the vectors, to indicate that they are column vectors rather than row vectors.)

    The work that you were doing doesn't prove anything, but it has helped you understand how to evaluate T for a specific vector.

    You need to show two things:
    $$ T\left( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} y_1 \\ y_2\end{bmatrix}\right) = T\left( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \right) + T\left( \begin{bmatrix} y_1 \\ y_2\end{bmatrix}\right)$$
    and
    $$ T\left( k\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \right) = k T\left( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \right)$$

    These aren't actually very hard to show.
     
  12. Sep 24, 2013 #11
    Yes, I know this is what I need to show. I just don't know how to multiply <x1,x2> by a column vector...

    Does [itex]T \left (\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} y_1 \\ y_2\end{bmatrix} \right ) = \begin{bmatrix} x_1 \ x_2 \\ 2x1 \ 2x2 \\ 3x1 \ 3x2 \end{bmatrix} + \begin{bmatrix} 4x1 \ 4y2 \\ 5y1 \ 5y2 \\ 6y1 \ 6y2\end{bmatrix} ?[/itex]
     
    Last edited: Sep 24, 2013
  13. Sep 24, 2013 #12

    Mark44

    Staff: Mentor

    You shouldn't have to.

    Start with the expression on the left side, below.
    Add the two vectors - you should get a column vector with two entries. Then take the first entry (upper) and multiply <1, 2, 3>^T by it, as a scalar. Multiply the vector <4, 5, 6>^T by the second entry (lower), as a scalar. Then add the two resulting vectors together.
    The above with corrections:
    $$T \left (\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} y_1 \\ y_2\end{bmatrix} \right ) = T\begin{bmatrix} x_1 + y_1 \\ x_2 + y_2 \end{bmatrix} =? $$
     
  14. Sep 24, 2013 #13
    Is that equal to [itex] \left (\begin{bmatrix} 6x_1+6y_1 \end{bmatrix} + \begin{bmatrix} 15x_2+15y_2 \end{bmatrix}\right ) [/itex] ? and the final vector = <12, 30> ?
     
  15. Sep 24, 2013 #14

    Mark44

    Staff: Mentor

    No, not at all.
    T is a map from R2 to R3. That means that the inputs are vectors in R2 and the outputs are vectors in R3.

    You started to get the idea back in post #5, but it doesn't seem to have stuck. Take a look at the first part of your Edit in that post.

    $$T\begin{bmatrix} x_1 + y_1\\ x_2 + y_2 \end{bmatrix} = ?\begin{bmatrix} 1 \\ 2 \\ 3\end{bmatrix} + ?\begin{bmatrix} 4 \\ 5 \\ 6\end{bmatrix}$$

    What comes out of this is a single vector in R3, after you do the scalar multiplications and the vector addition. You do know how to do those, right?
     
  16. Sep 24, 2013 #15
    No, not in this form. That's what I've been trying to tell you. I don't know if the x1 and x2 both go where the first ? is, or the x1 and y1? And I'm not sure how to multiply a vector of two coordinates by three scalars...
     
  17. Sep 24, 2013 #16

    Mark44

    Staff: Mentor

    The only multiplication going on here is the multiplication of a 3-element vector by a single scalar. There is absolutely no multiplication of vectors by vectors.

    Let's go back to where you weren't confused.

    What are each of these:
    $$T\left( \begin{bmatrix} 1 \\ 2\end{bmatrix}\right)$$

    $$T\left( \begin{bmatrix} 1 + 3 \\ 2 + 4\end{bmatrix}\right)$$

    Note that 1 + 3 is a number (scalar), in exactly the same way as x1 + y1 is a number.
     
  18. Sep 24, 2013 #17
    Oh alright, I think I'm understanding now... So would T(x1+x2) = <5x1+5x2,7x1+7x2,9x1+9x2> ?
     
  19. Sep 24, 2013 #18

    Mark44

    Staff: Mentor

    I don't think you do. T(x1 + x2) doesn't make any sense. T takes a vector in R2 (not a scalar such as x1 + x2) and maps it to a vector in R3.

    See if you can figure out the two that I asked you to do in my previous post, and we can go from there.
     
  20. Sep 25, 2013 #19
    It's too late, this is due today. This is the one problem I couldn't answer and it's probably the easiest of them all. I appreciate your help, but it's a shame you guys can't just post the work and answer to the problem so I can learn from it
     
  21. Sep 25, 2013 #20
    I think I've figured it out anyway, (again). I hope this IS linear or else I'm still really really off in this process. But I just wasn't understanding the notations, I thought you literally had to multiply <x1, x2> by each column vector, not split it into its components..
     
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