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Existent solution to the linear system Ax=b

  1. Jun 21, 2012 #1
    Suppose that the linear system Ax=b is given for some symmetric A, and it is known that vector c spans the null-space of A.

    How could one formally show that if b is not orthogonal to c, the solution to the system Ax=b does not exist?
    To remind you, the null-space of A contains all vectors u for which Au=0.
  2. jcsd
  3. Jun 21, 2012 #2
    Think about

    [tex](Ax)^T c[/tex]

    where x is an arbitrary vector and c spans the null-space.
  4. Jun 22, 2012 #3
    That would imply x^T Ac=x0=0. How do you relate this to the above problem?
  5. Jun 22, 2012 #4


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    Hey onako.

    I'm assuming A is nxn (since you said it is symmetric). From this if c spans the null-space it must be an n-dimensional column vector.

    From this you can use the decomposition argument that a basis can be broken into something and its perpendicular element (some books write it as v_perp + v = basis). Your zero vector c is perpendicular to b if you wish to have full rank.

    If this is not the case, then you can show that you don't have full rank and that a solution should not exist. For specifics you should look at rank nullity, and for the v_perp + v = basis thingy, this is just a result of core linear algebra with spanning, dimension, and orthogonality.
  6. Jun 22, 2012 #5
    Doesn't this imply that Ax is perpendicular to c?
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