Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Belonging to row space

  1. Feb 5, 2005 #1
    In my hw problem, I am supposed to find out whether an element belongs to the rowspace of a matrix. So, what I did is to determine the (row)basis of the matrix, dimension of it being one row less of the rows of the original matrix. So, instead of the linearly-dependent row I put the element and if the system turns out to be inconsistent I assume that it does not belong to the row space.
    Is it correct to assume that? If I use dependency equation for the rows and the new row element would that give the same result? When I used it I got there are no solutions at all, I am not sure what it means because for being lin. indep. there has to be one solution: 0.

    Thanks in advance.

    P.S. I decided to post the problem itself afterall:
    2 1 3 1
    1 1 3 0
    0 1 2 1
    3 3 8 2
    and I need to determine whether X = [4, 1, 2, 5] and Y = [1, 2, 3, 4] belong to row space of the matrix.
    The answer is X does, but Y does not.
    Last edited: Feb 5, 2005
  2. jcsd
  3. Feb 6, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    So you need to find a linear combination of the row vector which yields X or Y.
    Putting the row vectors (or a basis for the row space) as columns in A, you're asked to solve:
    Ax=X and Ax=Y, (lousy notation, but A is a matrix, x is the unknown vector and X,Y are given)

    How would you normally solve such an equation?
  4. Feb 6, 2005 #3
    If I were asked to find out whether X belonged to the column space of a matrix, would I do the same thing, i.e. insert that column at the end (in augmented matrix) and see whether it is solvable?
    Thanks a lot, my previous problem worked out!
    And a follow-up question:
    how are basis of rowspace and basis of column space related, besides the fact that dimensions are equal?
    Last edited: Feb 6, 2005
  5. Feb 7, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    Yes. The way I look at Ax in this case is a linear combination of the column vectors of A. So Ax=b (for some vector b) has a solution if and only if b lies in the column space of A.

    Can't think of anything now.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?