1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Prove whether or not a transformation is one to one.

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Let T: R3 -> M22 by
    a \\
    b \\
    a-b & b-c \\
    a+b & b+c\\

    Is this transformation one-to-one?

    2. Relevant equations

    3. The attempt at a solution
    I am not really certain how to go about this. A nudge in the right direction?
  2. jcsd
  3. Mar 6, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper

    A linear transformation is one-to-one if it's kernel is {0}. Try computing the kernel.
  4. Mar 6, 2013 #3


    User Avatar
    Science Advisor

    First, use "tex" and "/tex" at the beginning and end of an entire equation, not for individual parts:
    [tex]T\begin{bmatrix}a \\ b \\ c \end{bmatrix}= \begin{bmatrix}a- b & b- c \\ a+ b & b+ c\end{bmatrix}[/tex].

    Now, can we assume that you understand what "one to one" means? T is one to one if Tu= Tv only if u= v. Taking
    [tex]u= \begin{bmatrix}a \\ b \\ c\end{bmatrix}[/tex]
    [tex]v=\begin{bmatrix}x \\ y \\ z\end{bmatrix}[/tex]
    then Tu= Tv means that
    [tex]\begin{bmatrix}a- b & b- c \\ a+ b & b+ c\end{bmatrix}= \begin{bmatrix}x- y & y- z \\ x+ y & y+ z\end{bmatrix}[/tex]
    which is the same as a- b= x- y, b- c= y- z, a+ b= x+ y, and b+ c= y+ z.

    Can you prove from those equations that a= x, b= y, and c= z?
    (Dick's method is equivalent to this, taking x= y= z= 0.)
  5. Mar 6, 2013 #4
    So if T(u) = T(v) then 0 = T(u)-T(v) = T(u-v) so u-v is in kerT and u-v=0 so u=v?
    But only if the kerT=0.
    Last edited: Mar 6, 2013
  6. Mar 6, 2013 #5


    User Avatar
    Science Advisor
    Homework Helper

    That's it exactly. So just find ker(T).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted