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Matrices and Invertible Linear Transformations

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data

    How do I know if this linear transformation is invertible or not?

    T: [ x ] ---> [ 2y ]
    [ y ] [ x-3y ]

    (I also uploaded a small .bmp file to represent this if this looks too ugly)

    3. The attempt at a solution

    Well, I thought maybe it could be represented by a transformation matrix T
    [ 0 2 ]
    [ 1 -3 ]

    So then I just took the inverse of T, which I got as
    [ 1.5 1 ]
    [ 0.5 0 ]

    So does that mean that in the question, it is invertible? Because if it is, I'm getting the impression that these R2 to R2 linear transformations are invertible. Would this impression be correct?

    Attached Files:

  2. jcsd
  3. Nov 21, 2008 #2


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    Science Advisor

    Yes, a linear transformation is invertible if and only if a matrix representing it in some basis is invertible. And that is the true if and only if the determinant is non-zero so just observing that 0(-3)- (1)(2)= -2 is sufficient.

    Of course, from the basic definition a function is invertible if and only if it is "one-to-one" and "onto" so you could do this:
    Suppose f[(x,y])= f(]x',y']). Then [2y, x- 3y]= [2y', x'- 3y'] so 2y= 2y' and y= y'. Then x- 3y= x' -3y'= x'- 3y so x= x'. The function is "one-to-one".

    If [u, v] is any vector in R2, if f([x,y])= [u, v], then 2y= u and x- 3y= v. From the first equation, y= u/2 so x- 3y= x- (3/2)u= v and x= (3/2)u+ v. Yes, there exist [x, y] such that f([x,y])= [u, v] for any [u,v] so f is "onto".
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