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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


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    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


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    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.
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