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Least squares proof

  1. Feb 27, 2009 #1
    1. The problem statement, all variables and given/known data

    In the least squares method the vector x* that is the best approximation to b statisfies the Least squares equation:

    [tex]A^T A x^*= A^T b [/tex]

    Prove that there's always a solution to this equation.

    2. Relevant equations
    -

    3. The attempt at a solution
    I distinct 2 situations [tex]A^T A [/tex] is invertible and it isn't invertible. If it's invertible then there's no problem [tex]x^*= (A^T A)^{-1} A^T b [/tex]

    But how I prove that it works in the non-invertible case?
     
  2. jcsd
  3. Feb 27, 2009 #2
    If it is not invertible you need another criteria to get a unique solution. For instance you could require that x has minimum norm. In which case use could use the pseudo inverse which is based on singular value decomposition. However, you do not required a unique solution in the above question. So perhaps you could try showing that the columns of A^T form the same column space as the columns of A^TA.
     
  4. Feb 28, 2009 #3
    Don't you mean the columns of [tex]A^T b[/tex] are in the span of the columns of [tex]A^T A[/tex]? If so I don't understand how to prove such a thing.
     
  5. Mar 1, 2009 #4
    I've thought about it and I seriously don't know how to prove that the A^Tb is in the column space of A^TA. Can someone help me?
     
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