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Proof that the nullspace is closed in addition/multiplication

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    It is number three on the following page.
    http://people.math.carleton.ca/~mezo/A3math1102-11.pdf


    2. Relevant equations

    No idea.

    3. The attempt at a solution

    I have no idea how to incorporate the kj.

    Best I could reason through this is supposing: b1 ∈ N(A) , c1 ∈ N(A)

    Ab1 + Ac1 = 0
    A(b1 + c1) = 0

    => (b1 + c1) ∈ N(A)

    This clearly isn't as rigorous as is desired though.

    Any help would be so appreciated, I'm completely stuck here!

    Thank you
     
  2. jcsd
  3. Oct 11, 2011 #2
    Anyone? Please?
     
  4. Oct 11, 2011 #3

    Mark44

    Staff: Mentor

    akj is nothing more than the entry in row k, column j of the matrix. Each number in the matrix is an element of the field F.
    Probably not, but it's not too far off.
    Let b = <b1, b2, ..., bn> and c = <c1, c2, ..., cn> be vectors in N(A).

    Show that b + c is in N(A).

    And similar for scalar multiplication.
     
  5. Oct 11, 2011 #4


    Thank you so much for your reply!

    b = <b1, b2, ..., bn>
    c = <c1, c2, ..., cn>

    b + c = < (b1 + c1), (b2 + c2), ..., (bn + cn)>

    We know:

    Ab = 0
    s.t. 0 = Ab1, Ab2, ..., Abn

    Ac = 0
    s.t. 0 = Ac1, Ac2, ..., Acn



    => Ab + Ac = <(Ab1 + Ac1, Ab2 + Ac2, Abn + Acn)>
    => A(b+c) = <(A(b1 + c1), A(b2 + c1), A(bn + cn)>

    So, addition is closed in the null space of A


    Does this prove it?
    Other info is given to the nature of the matrix i.e. k ≥ 1, j ≤ n, etc.
    Should I mention this at all? What is the significance?

    Thanks again
     
  6. Oct 11, 2011 #5

    Mark44

    Staff: Mentor

    The notation in the problem means that both j and k are between 1 and n.
    I don't think what you have above is what they're looking for. By telling you the entries of the matrix, I believe that they are expecting you to use them.

    For example, with b = <b1, b2, ..., bn>, you are given that b is in N(A). This means that Ab = <0, 0, ..., 0>.

    I believe that you're supposed to write something that represents the multiplication of A and b. In other words, you should get an n x 1 vector whose i-th entry is the dot product of the i-th row of the matrix and the vector b. Since Ab = 0, each of those dot products is equal to 0.

    Same thing for c. Then use this information to show that b + c is in N(A).
     
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