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Vector Space, dimensions and kernal rank

  1. Dec 1, 2008 #1
    Please could someone help me with this question, thank you.

    Find dim[Ker(D^2 -D: P_3(F_3) ==>P_3(F_3))]

    Where dim is dimension, Ker is kernal

    D is the matrix

    D^2 is the derivative of D is it equals


    And F_3 is the field subscript3

    so D^2 -D

    Should equal


    But where do I go from here? I have tried reducing this matrix to row echelon form however this doesn't seem logical, do you have any ideas on finding the dimension of the kernal?

  2. jcsd
  3. Dec 1, 2008 #2
    I will write T = D2 - D.

    Well, the dimension of the image space of T is the number of independent columns in a matrix of T (indeed, the image space is spanned by the columns of the matrix); from what you have found, clearly dim(im T) = 2 (you can be sure of this by reducing the matrix to column echelon form). But since the domain of T, P3(F3), has dimension 4, we have 4 = dim(im T) + dim(ker T).

    Equivalently, dim(ker T) is the number of zero columns in a column echelon form of a matrix of T.
    Last edited: Dec 1, 2008
  4. Dec 2, 2008 #3


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    Don't think this is right. Recheck your working again.

    One way to do it, as adrian suggested, would be to find dim(R(T)) and then use the dimension theorem to find dim(ker(T)) or nullity. The other way would be to let a vector v = (w x y z)^T. Then solve the homogenous system of linear equations Tv = 0, to find a basis for nullspace(T). Then simply count the number of vectors in that basis for the answer.
  5. Dec 2, 2008 #4
    It is correct. Remember that we're working in the finite field F3.
  6. Dec 2, 2008 #5


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    Homework Helper

    Yeah, you're right. Silly me.
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