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Showing that a basis for the ker(A) is in the kernel

  1. Jan 21, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm new to this and I was wondering if any one could help me out

    given:

    x+z-w=1
    y-z+w=1
    x+y+z=3

    find the coefficient matrix A, the vector of constants B, use Gauss-jordan elimination to solve the system. Find the Rank(A), the Null(A) and a basis for the im(A) and a basis for the ker(A) then verify that the vectors and in the kernel

    2. Relevant equations



    3. The attempt at a solution

    I reduced the matrix, found the rank to be 3 and the null to be 1, the basis for the

    im(A) =( [1;0;1] , [0;1;1] , [1;-1;1] )

    then I get to the kernel, I set the equations from the reduced matrix to zero and found

    x=w
    y=-w
    z=0

    so the basis for the kernel is [1;-1;0;1] right?

    how do I show that that basis is in the kernel?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 21, 2008 #2
    If anything is in the kernel, then it maps to the zero vector. Does this happen?
     
    Last edited: Jan 21, 2008
  4. Jan 21, 2008 #3
    I'm not sure what " maps to the zero vector" means, this is my first course in Linear Algerbra ( two class periods so far ) and we don't have a book to work from.

    After I reduce the matrix A|B I get

    1 0 0 -1 | 0
    0 1 0 1 | 2
    0 0 1 0 | 1

    which gives

    x -w = 0
    y+w = 2
    z=1

    i set these to zero ( not sure if that is right ) and then solve the equations for x and y
    then i have

    X = [x;y;z;w] = [1;-1;0;1]*w

    that column vector then becomes my ker(A) right?
     
  5. Jan 21, 2008 #4
    "maps to the zero vector" means that if you plug the basis vector into your system you get zero.

    So what I was getting at was - if you plug the basis vector that you got for the kernel in your system, does zero come out? If so, it's a basis for the kernel (assuming you did the rest correctly).
     
  6. Jan 21, 2008 #5
    That a vector v "maps to the zero vector" just means that A*v = 0. You can talk about A as a matrix, but you can also talk about it as function from one vector space to another. That's where the "mapping" terminology comes from.
     
  7. Jan 21, 2008 #6
    yeah, if i take my basis for the ker(A) and plug the vaules back into the original system i get all of the equations to equal zero. is that right then?

    thanks for the help by the way :)
     
  8. Jan 21, 2008 #7
    As long as the rest is correct - you have a kernel of dimension 1, you have a nonzero vector in it, so it is a basis. I haven't actually looked at the system itself.
     
  9. Jan 21, 2008 #8
    Thanks!!!!!!
     
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