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The kernel exists in both?

  1. May 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove for every subspace B of vector space C, there is at least 1 linear operator L: C→C with ker (L) = B and there's at least 1 linear operator L':C→C with L'(C) = B.

    2. Relevant equations



    3. The attempt at a solution

    The first operator with Ker(L) = B would be anything mapping something to 0. So this would map to 0. So by the definition of a vector space, there exists a 0 element. And since the definition of a subspace is something that has the same conditions as the vector space. Therefore, the kernel exists in both?

    And for the second operator, this would be the identity element. So would the same argument as above work?
     
  2. jcsd
  3. May 8, 2013 #2

    Dick

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    You define what a linear operator is by defining what it does to a basis. So pick a basis for B, say ##v_1,v_2,...,v_k## and extend it to a basis for C, ##{v_1,v_2,...,v_k,v_{k+1},...,v_n}##. Now define your operators by saying what they do to each of those basis vectors.
     
  4. May 8, 2013 #3
    Okay so B = (v1, v2, ...., vK) is a basis and C = (v1, v2, ..., vk, vk+1, ....., vn) is also a basis. I need an operator where the kernel of the linear operator is equal to the subspace B of C.
     
  5. May 8, 2013 #4

    Dick

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    Let f be your operator. What should f(v1) be?
     
  6. May 8, 2013 #5

    Mark44

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    You're going to confuse yourself by using the same names for subspaces as you do for basis sets. I don't think you need to give names to the basis sets.

    Let {v1, v2, ... , vk} be a basis for B, and
    let {v1, v2, ... , vk, vk+1, ... , vn} be a basis for C.
     
  7. May 8, 2013 #6
    f(v1) = 0, because the kernel is all the vectors that map to 0
     
  8. May 8, 2013 #7

    Dick

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    Good so far. Now take a crack at defining the rest. You have to make sure all of the vectors in B map to 0 and none of the vectors in C that aren't in B map to 0.
     
    Last edited: May 8, 2013
  9. May 9, 2013 #8
    f(v1)=0
    f(v2)=0
    .......
    f(vk)=0
    f(vk+1) = a1
    .........
    f(vn) = an

    where a1,....,an are natural numbers
     
  10. May 9, 2013 #9

    Dick

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    f is supposed to map C to C. Not C to the natural numbers.
     
  11. May 9, 2013 #10
    f(v1)=0
    f(v2)=0
    .......
    f(vk)=0
    f(vk+1) = c1
    .........
    f(vn) = cn

    where c is contained in C?
     
  12. May 9, 2013 #11

    Dick

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    You need to be a little more definite than that. You have to be able to show that if an element of C, call it v, is not an element of B, then f(v) is not zero. If v=a1*v1+...+ak*vk+...+an*vn, where the v's are your basis vectors and the a's are scalars, how can you tell if v is in B?
     
  13. May 9, 2013 #12
    Let v be an element of C. Need to show that if v is not an element of B, then f(v) is not zero.
    Let v = a1*v1+...+ak*vk+...+an*vn. v is an element of B, if every vector maps to 0.
     
  14. May 9, 2013 #13

    Dick

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    No, how would you tell v is an element of C that's not in B by looking at the a's?
     
  15. May 9, 2013 #14
    If the a is not equal to zero. For instance L[(a1, a2, ...., an)] = [a1, a2,0,.....0]. Here, a1=a2=0, but a3,....,an can have any values
     
  16. May 9, 2013 #15

    Dick

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    I don't see what that answer has to do with my question.
     
  17. May 9, 2013 #16
    B/c if the a is not 0, then it would not map to 0. Therefore, it would not be apart of the kernel and it would be in C not B.
     
  18. May 9, 2013 #17

    Mark44

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    Not true. A vector doesn't have to be the zero vector to be in the kernel.
     
  19. May 9, 2013 #18

    Dick

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    The idea here is to pick values for the ##f(v_i)## so that you can prove that to be true. Suppose ##a_{k+1}## is not zero. Is v in B? Why or why not?
     
  20. May 9, 2013 #19
    v is in B if the sum of a1v1 + .... + akvk+ ..... + anvn = 0
    So, if ak+1 is not zero, v is not in B.
     
  21. May 9, 2013 #20

    Dick

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    Right answer. Completely wrong reason. v1,...,vk is supposed to be a basis for B. What does that mean? Define 'basis'.
     
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