1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Staff Emeritus
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question about nullity and rank
  1. Nullity Question (Replies: 3)

  2. Rank and nullity (Replies: 7)

  3. Rank and nullity (Replies: 3)

Loading...