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Homework Help: Least Squares Solution - Or is there?

  1. Apr 18, 2007 #1
    I have a problem that says to find the least squares solution to
    [tex]\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}
    K x = \colv{2}{2}[/tex] for
    [tex] K = \left(
    1 & 2\\
    2 & 4
    \end{array} \right)[/tex]. Then express the solution in the form x = w + z, where w is in the corange, and z is in the kernel.

    Since the determinate of the matrix K will be zero, then there should be infinite least squares solutions since the kernel is non-zero, correct? When I try to find a solution by multiplying each side by K^T, I get no solution rather than infinite solutions. Specifically the sytem

    [tex] \newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2
    5 & 10\\
    0 & 0
    \end{array} \right) x = \colv{0}{10}[/tex]

    How can I find the least square of a matrix with determinate zero? Everything I know is based off the fact that the matrix will be invertible.

    *My last matrix is messed up somehow, it should be the augmented matrix {{5,10 | 5},{0,0 | 10}}.
    Last edited: Apr 18, 2007
  2. jcsd
  3. Apr 18, 2007 #2
    Nevermind, I made a calculation error.
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