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Homework Help: Finding the basis of the nullspace, can you see if i'm doing this right?

  1. Dec 6, 2005 #1
    Hello everyone, I think i did this right, but i want to make sure this is what they want. The questions says:
    In each case find a basis for and caclulate the dimension of nullA:
    Here is my work and problem!
    http://img207.imageshack.us/img207/2948/lastscan6ib.jpg [Broken]

    I just found out its wrong...but why? the book got:
    {[-2 4 1 0 0 ]^T, [7 -11 0 2 1]^T}
    dim(nullA) = 2.
    Those kind of resemble my columns, i got
    [2 -4 0 0 0] and [-7 11 2 0 0], hm...
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Dec 6, 2005 #2


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    First of all, do you understand where the null space is?

    If A: U->V is a linear transformation from U to V, then the null space of L is the subspace of U such that if x is in the null space, then Ax= 0.

    Your A is a 5 by 3 matrix: from R5 to R3, but you give {[1,0,0], [0,1,0], [0,0,1]} as the basis for the null space. That's impossible! The nullspace of A a subpace of R5, not R3.

    Now, that's really mystiftying! In your image, your columns contain only 3 numbers, your rows, 5!

    I haven't checked all of your arithmetic (you shift rows around in a way I find confusing) but assuming the last, row-reduced, matrix is correct, the equation AX= 0 becomes (letting X= [x, y, z, u, v]),
    x+ 2z- 7v= 0, y- 4z+ 11v= 0, and u- 2v= 0. That is 3 equations in 5 unkowns so we can solve for 3 of them in terms of the other 2: in this row-reduced form, x= -2z+ 7v, y= 4z- 11v., u= 2v.
    Taking z= 1, v= 0, that gives x= -2, y= 4, u= 0. One vector in the nullspace is [-2, 4, 1, 1, 0].
    Taking z= 0, v= 1 gives x= 7, y= -11, u= 2. Another vector in the nullspace is [7, -11, 0, 2, 1].
    Those 2 vectors constitute a basis for the nullspace.

    The nullspace is 2 dimensional. That's no surprise. In order to "squeeze" a 5 dimensional space into a 3 dimensional space, at least 2 dimensions have to go to 0. Assuming no other dependencies, that would be exactly 2.
  4. Dec 6, 2005 #3
    ohh wow, i did totally screw that one up, thank you so much for the detailed explanation as always :biggrin: !!
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