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The nonhomogenous system Ax=b

  1. May 28, 2013 #1
    The invertible matrix equation tells us that the following statements are equivalent, for any square matrix A:
    1) A is invertible
    2) Ax=0 has only the trivial solution
    3) Ax=b has a unique solution for any column vector b

    My question:
    Suppose you know that A is a singular matrix. Then can you conclude that for every column vector b, either Ax=b has no solution or Ax=b has infinitely many solutions? Or is there a vector b for which Ax=b has a unique solution, despite the fact that A is singular?

    Also, given that A is singular, how can you tell whether Ax=b has no solutions or has infinitely many solutions?

    BiP
     
  2. jcsd
  3. May 28, 2013 #2

    CompuChip

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    If A is not invertible then it is not true that (Ax = b has a unique solution for any column vector b).
    Negating that statement,
    ##\neg\left( \forall b \in \text{column vectors}, Ax = b \text{ has a unique solution } \right)##
    gives
    ##\exists b \in \text{column vectors}, \neg \left( Ax = b \text{ has a unique solution } \right)##
    i.e.
    for at least one column vector b, Ax = b does not have a unique solution.

    It doesn't say anything about any column vector b, and it also doesn't say whether "not a unique solution" means "no solution at all" or "infinitely many".

    One way to check is to form the augmented matrix
    $$[A | b] = \left[\begin{array}{ccc|c}
    A_{11} & \cdots & A_{1n} & b_1 \\
    \vdots & \ddots & \vdots & b_2 \\
    A_{m1} & \cdots & A_{mn} & b_3
    \end{array}\right].$$

    If you row-reduce it, you will end up with one or more rows of zeroes at the bottom. If the corresponding entries in the right-most column are non-zero, there are no solutions; otherwise there are infinitely many.
     
  4. May 29, 2013 #3

    HallsofIvy

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    If A is not invertible, then its "kernel", the set of all x such that Ax= 0, is non-trivial. It is a subspace of the domain of A with dimension at least 1. Let [itex]x_0[/itex] be a non-zero vector such that [itex]Ax_0= 0[/itex]. If Ax= b has a solution, that is, if there exist [itex]x_1[/itex] such that [itex]Ax_1= b[/itex] then [itex]A(x_1+ x_0)= Ax_1+ Ax_0= b+ 0= b[/itex] so [itex]x= x_1+ x_0[/itex] is another vector such that Ax= b.

    Do you know what a "linear manifold" is? In R2, a subspace is a line through the origin. A linear manifold is a line NOT through the origin. In R3 a two dimensional subpace is a plane through the origin. A two dimensional linear manifold is a plane NOT containing the origin. Given a linear manifold, we can take any vector v in it and show that the set {x- v| x in the linear manifold} is a subspace. We say that the linear manifold is "parallel"
    to the kernel.

    If A is not invertible, then its kernel, {x| Ax= 0}, is a subspace of dimension at least 1. And if Ax= b has at least one solution then {x|Ax= b} is a linear manifold of the same dimension as the kernel and "parallel" to it.

    In particular, if A is not invertible and Ax= b has a solution then there are an infinite number of solutions.
     
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