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Splitting a vector into a rowspace component and a nullspace component

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Find a basis for the orthogonal complement of the row space of A:
    A =
    [1 0 2
    1 1 4]

    Split x = (3,3,3) into a row space component xr and a nullspace component xn.

    3. The attempt at a solution
    For the first part of the problem I took A to RREF
    R =
    [1 0 2
    0 1 2]
    and then solved to find a basis for the nullspace, x * (-2,-2,1), which should be the basis for the orthogonal complement of the row space.

    I THINK that's right but maybe it's not because I'm stuck on the second part, splitting x = (3,3,3). Do I try to find a vector in Row(A) and a vector in Nul(A) that, when added, produce the vector (3,3,3)? How do I solve for that?

    Help is very much appreciated!

    Also, what's the best way to represent a matrix on these message boards?
  2. jcsd
  3. Nov 1, 2011 #2
    nevermind, I figured it out
  4. Nov 1, 2011 #3


    User Avatar
    Science Advisor

    an observation:

    x(-2,-2,1) isn't the basis for Null(A), it's the entire space. "a" basis, is any particular vector, using a non-zero value for x. x = 1 works well, so one basis is:


    to find xn and xr, we're looking for a,b, and c with:

    (3,3,3) = a(-2,-2,1) + (b(1,0,2) + c(1,1,4))

    the first term is in Null(A) and the second sum of two terms in in Row(A).

    this isn't that hard, we have:

    (3,3,3) = (b-2a+c,c-2a,a+2b+4c), or, if you prefer:

    [tex]\begin{bmatrix}-2&1&1\\-2&0&1\\1&2&4\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}3\\3\\3\end{bmatrix}[/tex]

    solve this for a,b and c.


    the string i entered to input the matrix equation was:

    [tex]\text{[tex]\begin{bmatrix}-2&1&1\\-2&0&1\\1&2&4\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}3\\3\\3\end{bmatrix}\[/tex]}[/tex]
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