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How to verify that the nullspace is orthogonal to the row space?

  1. Sep 17, 2011 #1

    sharks

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    How to verify that the nullspace is orthogonal to the row space of B?
    I have inserted the screen-shot of the problem below:
    [URL]http://i29.fastpic.ru/big/2011/0918/10/ca341692cc37b831143f5fe32351db10.jpg[/URL]

    2. Relevant equations
    Nullspace and orthogonality.


    3. The attempt at a solution
    I have self-studied that entire chapter of ortho with much difficulty and confusion but i have done the question to the best of my current abilities. Can someone please check on my answers and help with the third part of the question. Please let me know if i have presented my work/answers in the correct/best way. The lecturer is very itchy at corrections if there are slight deviations from the perfect answer/s.

    So, for part (i):
    I did an augmented matrix for Ax =0 and then reduced the augmented matrix to reduced row echelon form. The pivot variables are: x1, x2 and x3
    The free variable is x4.

    The answer for part (i) is x4 (1 -1 1 1 )^T
    or should it be just (1 -1 1 1 )^T? Which would be the special solution? But since it's asking for the nullspace, i suppose i should give the special solution as a multiple of the free variable??

    For part (ii):
    From what i understand, left nullspace (quite a revelation to me) is actually another way of saying that it is the nullspace of B^T
    So, i wrote the matrix B^T by writing the rows as columns.
    Again, i wrote it as augmented matrix, with Ax = 0
    Now, the reduced matrix in reduced row echelon form that i obtained is a bit unusual so i'm not sure of what i got, but it has to be an upper triangular matrix, right?
    Here it is below. Did i do the right thing for the last row?

    1 1 -2
    0 2 1
    0 0 -1
    0 0 0

    The pivot variables are x1, x2 and x3
    The free variable is x4.

    The answer is (not sure if it's correct?)

    x4(0 0 0 1)^T


    For part (iii):

    If i'm right, then,
    The row space of B = column space of B^T

    So, i write the transpose matrix of B. Then i proceed to find the general solutions of Ax = 0 in terms of free variables.

    I write the augmented matrix and then reduce the augmented matrix to reduced row echelon form, and this is what i get:

    1 1 -2 0
    0 2 1 0
    0 0 -1 0
    0 0 0 0

    So, the pivot variables are x1, x2 and x3
    The free variable is x4
    The pivots are 1, 2 and -1

    To verify, i write:

    The row space of B = column space of B^T is a subspace of R^m (where m = 3)
    i think m denotes the number of rows containing pivots, right? and n would then denote the number of pivot columns? I'm sorry for asking so many questions but i am doing self-study and this chapter is quite difficult to grasp.

    Since the null space of B and the row space of B are both subspaces of R^3, therefore, the nullspace of B is orthogonal to the row space B. (verified? is the wording right?)

    There is something mentioned about dimensions/ranks but i don't know if that's the way or alternative way of verifying this last part of the question??
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Sep 18, 2011 #2

    lanedance

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    Homework Helper

    so for i) you can see the matrix rows are linearly independent, and hopefully this was obvious in your row reduction. This means the dimension of the row space is 3, so we expect the dim of the nullspace to be 1.

    Check the vector you found, we see
    [tex] \begin{pmatrix}
    1 & 0 & -2 & 1 \\
    1 & 2 & -2 & 3 \\
    -2 & 1 & 3 & 0 \\
    \end{pmatrix}
    \begin{pmatrix}
    1 \\
    -1 \\
    1 \\
    1 \\
    \end{pmatrix}
    =
    \begin{pmatrix}
    0 \\
    0 \\
    0 \\
    0 \\
    \end{pmatrix}
    [/tex]

    So the null space is given by (1,1,-1,1)^T, also note that this check has in fact answered iii) as well, as performing the matrix multiplication is essentially performing a dot product between the vector (1,1,-1,1)^T and each row vector of the matrix

    so for part ii) as you mention it is actually solving the problem
    [tex] x^TB = 0^T[/tex]

    taking the transpose gives
    [tex] B^Tx = 0[/tex]

    so look like you're heading in the right direction
     
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