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Homework Help: Question about nullity and rank

  1. Apr 19, 2006 #1
    So i have the following matrix:
    A= [2,0,-1; 4,0,-2;0,0,0]
    I do r-r-e
    I get
    [1,0,-1/2;0,0,0;0,0,0]
    So my rank for A is 1, because I only have 1 leading one.
    Now for my nullity, i get the following
    x_1 - 1/2 X_3 = 0
    -->
    x_1=1/2 x_3
    therefore

    [x_1,X_2,X_3] =[1/2;0;1]t
    Which would imply that nullity = 1
    which then wouldn't satisfy the Dimension theorem for matrices..
    which states
    rank(a)+nullity(a) = n, where n is the # of columns.
    ..
    so my question is.. what am I doing wrong?
     
  2. jcsd
  3. Apr 20, 2006 #2
    How did you come up with x_2 = 0?
     
  4. Apr 20, 2006 #3

    HallsofIvy

    User Avatar
    Science Advisor

    The equation x1- (1/2)x3= 0 tells you that x3= 2x1 alright but tells you nothing about x2. A basis for the null space is [1, 0, 2] (equivalent to your [1/2, 0, 1]) and [0, 1, 0]. The nullity is 2.
     
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