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Finding the null space, matrix fun! wee!

  1. Nov 16, 2005 #1
    Hello everyone i'm confused on finding the null space on this problem:
    the matrix is:
    0 2 0 -5
    0 1 4 0
    0 0 1 0
    0 0 0 1

    null(A) =
    2b - 5d = 0
    b + 4c = 0;
    c = 0
    d = 0;
    b = 0;
    a = ?
    You don't know what a is, so i'm quite confused. Any help?
     
  2. jcsd
  3. Nov 16, 2005 #2
    Just a thought..your matrix row reduces to:

    0 0 0 0
    0 1 0 0
    0 0 1 0
    0 0 0 1

    Somebody else may have to check this.. but it suggests to me that the only values for b c and d are for the trivial solution, and a could be anything.
     
    Last edited: Nov 16, 2005
  4. Nov 16, 2005 #3
    Ok, so here is your matrix:
    [tex] A = \left( \begin{array}{cccc} 0 & 2 & 0 & -5 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) [/tex]
    You want to know what the nullspace is of the matrix [itex] A [/itex]. So this means you have an equation of the form.
    [tex] A\vec x = \vec 0 [/tex]
    which will look like (sorry LaTeX wasn't playing nice for me here, I'll just write it out):
    [0 2 0 -5][x1] = [0]
    [0 1 4 0][x2] = [0]
    [0 0 1 0][x3] = [0]
    [0 0 0 1][x4] = [0]



    So to solve this you could:
    Use gaussian-jordon elimination and reduce it to:
    [tex] \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) [/tex]
    as Hammie pointed out.

    So this leaves you with:
    [tex] x_1 = \alpha [/tex]
    [tex] x_2 = 0 [/tex]
    [tex] x_3 = 0 [/tex]
    [tex] x_4 = 0 [/tex]

    Which is a matrix of the form:
    [tex] \left( \alpha,0,0,0 \right) ^T [/tex]

    which typically you want to factor out the variables to yield:
    [tex] \vec x = \alpha \left( 1,0,0,0 \right) ^T [/tex]


    Now thinking of this in the algebraic sense, the solution to the system is:
    (going back to the [itex] a, b, c, d [/itex] notation:
    [tex] a = \alpha [/tex]
    [tex] b = 0 [/tex]
    [tex] c = 0 [/tex]
    [tex] d = 0 [/tex]

    Which, just as Hammie said. b,c,d are for the trivial solution while a could be anything.

    I think an easy general method for solving for the null space is to:
    1) reduce it as much as possible:
    ex) (1 0 0, 0 1 0, 0 0 0)^T type of matrix

    2) any row that is all 0's then set it equal to some variable.
    3) write down a column vector describing the solutions, and factor out the variables.
     
    Last edited: Nov 16, 2005
  5. Nov 16, 2005 #4
    Wow excellent explanation!! thank you very much, both of you! :)
     
  6. Nov 16, 2005 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that's precisely correct. You can't solve for a because a doesn't appear in any of the equations. In particular if b= c= d= 0, No matter what a is, the linear transformation is will take (a, 0, 0, 0) to (0, 0, 0, 0).
    What does that tell you about one basis vector for the null space?

    You also have b+ 4c= 0 so that c= -(1/4)b as well as 2b- 5d= 0 so that
    d= (2/5)b. That seems to say that if you take b to be anything, c= -(1/4)b, d= (2/5)b and a to be anything, all the equations will be satisfied. What does that tell you about the dimension of the kernel? What is a basis for the kerne?
     
  7. Nov 16, 2005 #6
    Quick question, I'm now trying to find the imagine space of this matrix, but I took the determinant of the matrix and it is equal to 0, doesn't that mean there is no image space? Because there is a row of 0's it can't span R^4, so there can't be a unique solution for all varaibles right?
     
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