Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question regarding linear transformation

  1. May 9, 2005 #1
    Hi

    I have a linear transformation T which maps [tex]\mathbb{R}^3 \rightarrow \mathbb{R}^2[/tex] a A is the standard matrix for the linear transformation.

    I'm suppose to determain that T maps [tex]\mathbb{R}^3 \rightarrow \mathbb{R}^2[/tex]

    I was told by my professor about the following theorem.

    If [tex]T:\mathbb{R}^n \rightarrow \mathbb{R}^m[/tex] is a linear transformation and A is the standard matrix for T. Then

    a/ T maps [tex]\mathbb{R}^n \rightarrow \mathbb{R}^m [/tex] if and only if A [tex] \mathrm{span} \{ \mathbb{R}^m \}[/tex]

    b/ T is one-to-one if and only if columns of A are linearly independent.

    If I then apply (a) from to my problem:

    [tex]A = \left[ \begin{array}{ccc} 1 & 0 & 4 \\ 2 & 1 & 6\\ \end{array} \right ][/tex]

    A being the standard matrix of the linear transformation.

    A can also be written:

    [tex]\left[ \begin{array}{cc} 1 \\2 \end{array} \right ] \mathrm{x} + \left[ \begin{array}{cc} 0 \\1 \end{array} \right ] \mathrm{y} + \left[ \begin{array}{cc} 4 \\6 \end{array} \right ] \mathrm{z} = \left[ \begin{array}{cc} 0 \\0 \end{array} \right ][/tex]

    This is equal to:


    [tex]\begin{equation}\nonumber
    x + 4z = 0
    \end{equation}[/tex]

    [tex]\begin{equation}\nonumber
    2x + y + 6z = 0
    \end{equation}[/tex]

    x in equation 1 can be written as [tex]x = -4z[/tex]

    If I insert that x into equation 2 then I get

    [tex]-2z + y = 0[/tex]

    But what do I then do to prove point (a) in the theorem ???

    Sincerley and many thanks in advance

    Fred
     
  2. jcsd
  3. May 11, 2005 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Actually this: "I have a linear transformation T which maps [tex]\mathbb{R}^3 \rightarrow \mathbb{R}^2[/tex]
    a A is the standard matrix for the linear transformation.

    I'm suppose to determain that T maps [tex]\mathbb{R}^3 \rightarrow \mathbb{R}^2[/tex]" doesn't quite make sense! I assume you mean simply that T is from[tex]\mathbb{R}^3 [/tex]to [tex]\mathbb{R}^2[/tex] and you want to prove this is an "onto" mapping.
    Okay, you are given A= [tex]A = \left[ \begin{array}{ccc} 1 & 0 & 4 \\ 2 & 1 & 6\\ \end{array} \right ][/tex]

    I don't understand why you then set Ax= 0: that would prove that the columns are linearly independent which is your professor's (b), not (a) (and this is clearly not one to one). To show that A is "onto", you need to show that for all y in [tex]\mathbb{R}^2[/tex],there exist x in [tex]\mathbb{R}^3[/tex] so that Ax= y. The simplest way to do that is to row-reduce A. If the second row does not reduce to all 0s, it's true.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question regarding linear transformation
Loading...