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One-to-One Linear Transformations

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Give a thorough explanation as to why a linear transformation:
    gif.gif with a standard matrix A CANNOT be one to one.

    2. Relevant equations

    3. The attempt at a solution
    I think I have figured this one out, but I was hoping somebody could confirm whether this example is sufficient:
    gif.gif

    gif.gif

    gif.gif

    gif.gif

    gif.gif

    gif.gif

    gif.gif

    There are two equations with three unknowns. Therefore, this system cannot have a unique solution, instead it will have an infinite number of solutions. Therefore, there cannot be a one-to-one mapping from gif.gif
     
  2. jcsd
  3. Mar 1, 2015 #2

    Stephen Tashi

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    Science Advisor


    If you are using "[itex] a,b,c [/itex]" to represent scalars, it doesn't make sense to put arrows over them.

    It's possible for some systems of two equations in three unknowns to have a unique solution (or no solutions) so you would have to give more specifics to argue along those lines. You need to show explicitly that there are two different elements in [itex] \mathbb{R}^3 [/itex] that are mapped to the same element in [itex] \mathbb{R}^2 [/itex].

    It's hard for me to guess what approach the problem expects you take, since I don't know what material has been covered prior to its being asked.
     
  4. Mar 1, 2015 #3
    The instructor gave a hint that we are supposed to use pivots to prove the statement is true, but I'm not really seeing how to do that...
     
  5. Mar 1, 2015 #4

    Mark44

    Staff: Mentor

    In addition to what Stephen Tashi said, your matrix A is incorrect. If T is a map from R3 to R2, any matrix representation will have to be 2 X 3, not 3 X 2.

    IOW, like this:
    $$ A = \begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\end{bmatrix}$$

    I would use ##a_{ij}## for the components of A, ##x_1, x_2, x_3## for the components of x, and ##y_1, y_2## for the components of y.
     
  6. Mar 1, 2015 #5
    Thank you...so basically I should go back to the drawing board right? :D
     
  7. Mar 1, 2015 #6

    Mark44

    Staff: Mentor

    Back to the drawing board, but using the definition of "one-to-one transformation," which is something that Stephen alluded to.
     
  8. Mar 2, 2015 #7
    I looked at this problem again this morning and I think my main problem was that I was writing the matrix as 3x2 instead of 2x3. With that in mind, I took a more simple approach (the approach I think the instructor intended us to take) and said:
    A 2x3 matrix can have at most one pivot in each of its two rows. However, 3>2, so the same matrix cannot have a pivot in each of its three columns. Therefore, the given transformation can never be one-to-one.
     
  9. Mar 2, 2015 #8

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I can't speak for your teacher's hint, but I would do it this way: Apply A to the standard basis for R3, u= (1, 0, 0), v= (0, 1, 0), and w= (0, 0, 1). The three vectors, Au, Av, and Aw can't be independent because they are in R2 which has dimension 2. Therefore, there exist numbers x, y, and z such that xAu+ yAv+ zAw= 0. From that, zAw= -xAu- yAv so A(zw)= A(-xu- yv). Now show that zw is NOT equal to -xu- yv.
     
  10. Mar 2, 2015 #9
    Great, thank you for the hint!
     
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