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Null Space Confusion

  1. Oct 29, 2005 #1
    The question is:
    For each of the following subspaces, find the dimension and a basis:

    {(x,y,z) are elements of R^3: 7x - 3y + z = 0}

    I had actually posted about this before, but I'm confused as to what the Null space is here.

    So, z = -7x + 3y, so there is one dependent variable and two free variables. However, since the equation is equal to 0, the solutions are in the Null Space (or something like that)??? So where:

    -7 3
    1 0
    0 1

    Is in the Null space, it also provides two independent solutions to the equation. Basically, what I'm trying to say is, how many dimensions does the Null Space have, and is the basis of the Null Space also a basis for the subspace defined by the above equation (if that makes sense)?

    Furthermore, written as a matrix:

    (7 -3 1)

    It goes from R^3 -> R^1 (I can say that right?). So if the dimension of the Null Space is 2, then the dimension of the image is 1, but what is the basis of the image, is that needed in the solution to my question, and graphically what is the image?

    I'm also having trouble with Row Reduction of a Matrix, or at least interpreting it. The original matrix is:

    2 5 1 0 8 -4
    0 3 4 3 2 2
    0 6 2 1 3 4

    I then reduced it to get:

    1 0 0 (-5/36) (101/36) (-11/3)
    0 1 0 (-1/9) (4/9) (2/3)
    0 0 1 (5/6) (1/6) 0

    Now, I need to find a basis for and the dimension of the column Space, the row space and the null space.

    So the dimension of the row and column space are = 3 and the dimension of the Null Space = 3.

    The basis for the row space is (The above reduced matrix)
    The basis for the column Space is:
    1 0 0
    0 1 0
    0 0 1 (???)

    Now, the basis for the Null Space is:
    -5/36 101/36 -11/3
    -1/9 4/9 2/3
    5/6 1/6 0
    1 0 0
    0 1 0
    0 0 1

    So when I plug in these values into the original equation, I should get 0 right?
    So (refer back to the original matrix):

    2*(-5/36) + (-1/9)*5 + (5/6)*1 + 1*0 = 0 (or at least it should)

    However, these values do not give me 0 when plugged back into the equation, which leads me to believe I'm checking the basis of the Null Space incorrectly, can anyone show me how to check it correctly?

    If something came across as unclear feel free to ask me to clarify.

    Thanks for the help.
  2. jcsd
  3. Oct 29, 2005 #2
    I have another question, its related to what we did earlier:

    Question: For the linear map of L, find the dimension of and a basis for the null space of L and the Image of L.

    L: P ^ 2 -> R^1 where L(P) = the integral from 1 to 3 of (p(x) + derivative(p(x+1)) dx)

    I started with a basis of P:

    x^2, x, and 1.

    I then performed the transformation:

    x^2 -> (x^3)/3 + x^2 + x + 1
    x -> (x^2)/2 + x + 1
    1 -> (x)

    Evaluated each from 1 to 3 and obtained the values:

    x^2 - > 62/3
    x -> 6
    1 -> 2

    The dimension of the image is 1 and the dimension of the Null Space is 2.
    And a basis for the image is:


    And a basis for the Null Space is???

    x - 3 = 0 (6 - 3(2) = 0)?

    Just so you know, 6 - 3(2), the 6 and the 2 are values being used from the basis, what I'm trying to do is find a vector in the Null Space (0 vector?), so I used a combination of values from the basis of the image to find the vector in the Null Space, and the vector equation was, x - 3 = 0. Can I do this?

    And if so I have to do the same thing again for 62/3 and 2? Then after this I would have answered the question completely, if my methodalogy is corret (which I'm not sure it is). If anyone can fill me in I'd appreciate it, thanks.
  4. Oct 29, 2005 #3


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    Staff Emeritus
    Science Advisor
    Gold Member

    The term "Null Space" does not make sense by itself -- it is only defined in relation to a linear transformation. i.e. if T is a linear transformation, then "The Null Space of T" makes sense.

    So, I think part of your confusion is that you're trying to use the term "Null Space" without specifying a linear transformation.

    Incidentally, you did eventually come up with a linear transformation T (= left multiplication by [7 -3 1]) for which your subspace happens to be the null space of T. (There are many other linear transformations that would work too) This is good if you wanted to apply theorems you know about null spaces.
  5. Oct 29, 2005 #4
    Thanks for the reply.

    So what your saying is if I had to give the number of dimensions of the subspace defined by 7x -3y 1z = 0, that the dimension would be 2, with one dependant variable (z), and that the basis of this subspace is:

    -7 3
    1 0
    0 1

  6. Oct 30, 2005 #5


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    Science Advisor

    You have three variables and one equation connecting them: given any two you could solve for the third so, yes, this is a two dimensional subspace of R3. However, I'm not sure if you intend each column to be a vector- with z first? If you let x= 1, y= 0, you get z= -7= -4. One basis vector is <1, 0, -4>. If you let x= 0, y= 1, you get z= 3. Another basis vector is <0, 1, 3>.
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