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Fundamental subspaces of A?

  1. Sep 27, 2012 #1
    When studying linear algebra when encountering a system Ax=b, I always read of the fundamental subspaces of A: N (the null space, all solutions x of Ax=0), the column or domain space of A: (the space spanned by the columns of A, or in other words, all possible b for Ax=b), the row space (the space spanned by the rows of A. I have a harder time wrapping my head around this one).

    But I had another question. What about all the possible vectors x for a given b, such that Ax=b ? I get that this wouldn't always work because say, some systems are inconsistent and have no solutions so it would be empty in this case. It seems like this wouldn't necessarily be a subspace because it wouldn't necessarily be closed under addition (just because x is a solution to Ax=b doesn't mean A(2x)=b is a solution). But sometimes it would be a subspace. Because for example, the nullspace would just be the special case where b = 0, the 0 vector, and we know the nullspace is a subspace. Is this "solution space" given special importance? If so what's it called? (I keep calling it a solution space and I just don't know what the proper name of it is)
  2. jcsd
  3. Sep 28, 2012 #2


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  4. Sep 28, 2012 #3


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    If Ax= b, then A(x+ v)= Ax+ Av= b for any v in the null space of A.

    The only time a "solution space" (which is, in fact, a standard name for it) is a subspace is when b= 0. In the case that [itex]b\ne 0[/itex], the "solution set" is a "linear manifold"- something like a line or plane in R3 that does NOTcontain the origin. It can be shown that if v is a vector satisfying Ax= b, then any solution of Ax= b can be written as v plus a vector satisfying Ax= 0. Since Ax= 0 is a subspace, it contains a basis so any solution to Ax= 0 can be written as a linear combination of basis vectors. And so any solution of Ax= b can be written as such linear combination of basis vectors plus v.
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