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Polynomials vector space

  1. Feb 4, 2010 #1
    1. The problem statement, all variables and given/known data
    find the rank and nullity of the linear transformation T:U -> V and find the basis of the kernel and the image of T


    2. Relevant equations
    U=R[x]<=5 V=R[x]<=5 (polynomials of degree at most 5 over R), T(f)=f'''' (4th derivative)


    3. The attempt at a solution
    Rank = 2
    Nullity = 4

    basis of kernel = {1,x,x^2,x^3} ?
    since a kernel is mapped to V, then the image is the zero vectors? and the basis of the image of T is the empty set?
     
  2. jcsd
  3. Feb 4, 2010 #2

    radou

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    The basis for Im(T) can be {1, x}.
     
  4. Feb 4, 2010 #3
    another QUESTION

    T : U -> V

    and the kernel(T) is the zero vectors,
    then what is the basis? it's not the empty set?
     
  5. Feb 4, 2010 #4

    radou

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    If it happens, for some linear transformation T, that ker(T) = {0} holds, then yes, the basis for ket(T) is the empty set.
     
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