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Solutions of Ax = b (for singular A)

  1. Aug 16, 2010 #1
    A system of linear equations, Ax = b (with A a square matrix), has a unique solution iff [tex]det(A) \ne 0[/tex]. If b = 0, the system is homogeneous and can be solved using SVD (which gives the null space of A).

    Now, how can the solution set be characterized for singular A and [tex]b \ne 0[/tex]? If a single solution [tex]s[/tex] is known, [tex]s + v[/tex] is also a solution for all [tex]v[/tex] from the null space of A... but how is it possible to determine whether such an [tex]s[/tex] exists at all, and if so, find it?
  2. jcsd
  3. Aug 16, 2010 #2
    Gaussian elimination can be employed to determine whether a solution exists.
  4. Aug 17, 2010 #3


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    A singular matrix, A, will map Rn into a proper subspace of Rn. There will exist x such that Ax= b if and only if b is in that subspace. What that subspace is, and whether or not b is in it, can be determined by row reduction of A augmented by adding b as a last column. Any row that reduces to "all 0s" in the first n columns (and, since A is singuar, there will be at least one such) must also have a "0" in the last column in order that Ax= b have a solution.

    Of course, if a solution exists, it is not unique. Adding any non-zero vector in the kernel of A (which is non-trivial for A singular) to a solution will give another solution.
    Last edited by a moderator: Aug 18, 2010
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