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Finding rank and nullity of a linear map.

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data


    let a be the vector [2,3,1] in R3 and let T:R3-->R3 be the map given by T(x) =(ax)a

    State with reasons, the rank and nullity of T


    2. Relevant equations



    3. The attempt at a solution

    Im having trouble understanding this... I know how to do this with a matrix ie row reduce and no. of leading cols = rank ,, and then no. of non leading cols = nullity.

    But im stuck on how to go about it with this eqn.?

    And that the rank = dim(image)

    but how would I finad that...

    or alternatively the dim(kernal)?

    Thanks.
     
  2. jcsd
  3. Oct 21, 2012 #2

    LCKurtz

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    Isn't the image 1 dimensional, being multiples of a?
     
  4. Oct 21, 2012 #3
    but does the image include zero values?
     
  5. Oct 21, 2012 #4

    LCKurtz

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    Do you mean the zero vector? If ##x = \theta##, the zero vector, what is ##T(\theta)## by your formula?
     
  6. Oct 22, 2012 #5
    still the zero vector
     
  7. Oct 22, 2012 #6

    LCKurtz

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    Does that answer your question?
     
  8. Oct 23, 2012 #7
    yes, so the image cointains the zero value... but I thought the image was the set of all function values except 0, I thought that was the kernal

    or is it that the kernal is a proper subset of the image?
     
  9. Oct 23, 2012 #8

    LCKurtz

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    At this point, I suggest you look up the definition of the kernel of a linear transformation in your text.
     
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