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Linear Algebra - Transformation / operator

  1. Aug 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Let T:V→V be a linear operator on a vector space V over C:

    (a) Give an example of an operator T:C^2→C^2 such that R(T)∩N(T)={0} but T is not a projection

    (b) Find a formula for a linear operator T:C^3→C^3 over C such that T is a projection with R(T)=span{(1,1,1)} and N(T)=span{(1,1,0);(0,1,1)}

    2. Relevant equations

    If T is a projection, then T^2=T
    R(T) is the range of T (I think it is also the image), N(T) is the null space or kernel of T.

    3. The attempt at a solution

    I don't know how to create one where R(T)∩N(T)={0}. I sort of figured out by using [2 0 , 0 1/2]
    As for (b), I try setting T(1,1,0)=(0,0,0) and T(0,1,1)=(0,0,0) but I'm not sure if this is correct or what to do afterwards.
     
  2. jcsd
  3. Aug 14, 2015 #2

    RUber

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    For part (a), the R(T) ∩ N(T) = {0} implies that only the zero vector maps to zero. So, your idea to use
    ##\begin{pmatrix} 2 & 0 \\ 0 & 1/2 \end{pmatrix}## was valid. The identity matrix satisfies R(T) ∩ N(T) = {0}, but it is a projection, so using non-ones on the diagonal does the trick.
    For part (b) you are right to set the conditions for the null space. The fact that the range is the span of (1,1,1) means that the output in all three columns should be the same, i.e. T( x, y, z) = (f(x,y,z), f(x,y,z), f(x,y,z)).
    Unless I am misinterpreting what you wrote, and you meant that R(T) = span{ (1,0,0), (0,1,0), (0,0,1) } which would be the entire space.
     
  4. Aug 14, 2015 #3

    Orodruin

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    These are finite vector spqces, just represent the transformationa with 2x2 / 3x3 matrices and see what constraints you obtain from the requirements.
     
  5. Aug 14, 2015 #4
    You are correct in saying that the range is the span of (1,1,1), but I am not sure what to do after this.
     
  6. Aug 14, 2015 #5

    RUber

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    As Orodruin mentioned, your goal is to make a 3x3 matrix for the transformation.
    ##\begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3}\\a_{2,1} & a_{2,2}& a_{2,3}\\a_{3,1} & a_{3,2}& a_{3,3} \end{pmatrix}
    \begin{pmatrix} x \\ y \\z\end{pmatrix} = \begin{pmatrix} b_1\\b_2\\b_3\end{pmatrix}##
    Your conditions are that b_1 = b_2 = b_3 to be in the span of (1,1,1). What does this mean about your three rows of the T matrix?
    Add that to your null space requirements, you should be able to provide a general form in terms of some constant value.
     
  7. Aug 14, 2015 #6
    Would I then be correct in saying that all three rows must be identical? If so, using the null space constraints should then give a matrix like :
    \begin{pmatrix} 1 & -1 & 1 \\ 1 & -1 & 1 \\ 1 & -1 & 1 \end{pmatrix}
    If I am not mistaken?
     
  8. Aug 14, 2015 #7

    RUber

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    That's right. And notice that you can multiply that matrix by any constant and it will still satisfy the requirements in your problem.
     
  9. Aug 14, 2015 #8
    Thanks to you both! I appreciate your input and I also appreciate that you helped me through rather than just answering the question. I definitley gained a better understanding
     
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