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Homework Help: Matrix Kernel

  1. Jun 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Let A=[{1,3,2,2},{1,1,0,-2},{0,1,1,2}]
    i) Find the rank
    ii) Viewing A as a linear map from M4x1 to M3x1, find a basis for the kernel of A and verify directly that these basis vectors are indeed linearly independent.

    2. Relevant equations

    3. The attempt at a solution
    i) is easy enough. Reduce rows to get: A=[{1,3,2,2},{0,1,1,2},{0,0,0,0}] so rank is 2.
    ii) I'm not exactly sure of the question here. At first, I thought it was just find the kernel of the matrix and I had some trouble with that. Using the reduced matrix:
    x1 + 3x2 + 2x3 + 2x4 = 0
    x2 + x3 + 2x4 = 0
    but how do I solve this for 4 variables with only 2 equations :/ Any help is appreciated!
  2. jcsd
  3. Jun 2, 2010 #2
    Multiply the vector by the original matrix, not the reduced row form. You get a pretty comfy vector.
  4. Jun 2, 2010 #3
    How so?
  5. Jun 2, 2010 #4
    If your vector is {x,y,z,w} then you should get the equations
    z= -x-2y
    So thats actually two vectors that you will get. If you don't know how to turn those equations into vectors i recomened you go here http://tutorial.math.lamar.edu/Classes/LinAlg/LinAlg.aspx and read up a bit.
  6. Jun 2, 2010 #5


    Staff: Mentor

    Work with your matrix to get it in reduced row echelon form. In this form all entries above or below a leading 1 entry are zero. Since your matrix looks like this:
    1 3 2 2
    0 1 1 2

    it's not in reduced row echelon form.

    After getting to this form use the first row to write an equation that has x1 in terms of x3 and x4. Use the second row to write an equation that has x2 in terms of x3 and x4. The last two equations are simply x3 = x3 and x4 = x4. This will show two vectors that span the nullspace.
  7. Jun 2, 2010 #6
    Ah of course. So you'd get:

    1 0 -1 -4
    0 1 1 2

    so you end with the two vectors: {(1,0,-1,-4),(0,1,1,2)} which are the basis for the kernel. Sound good?
  8. Jun 2, 2010 #7


    Staff: Mentor

    No, that's not it. The rows don't make a basis for the kernel.

    The matrix you have represents the equation Ax = 0, where the first two rows of A are as you show.

    Solve for x1 in the first row (equation) and for x2 in the second row (equation) to get the following.
    x1 = x3 + x4
    x2 = -x3 - 2x4

    If you add equations for x3 and x4, you get the following system.

    x1 = x3 + x4
    x2 = -x3 - 2x4
    x3 = x3
    x4 = ........................ x4

    Every vector in the nullspace is a linear combination of two vectors that are lurking in the system above.
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