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Least Squares Solution - Or is there?

  • Thread starter Mindscrape
  • Start date
  • #1
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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(
\begin{array}{cc}
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
\left(
\begin{array}{cc}
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:

Answers and Replies

  • #2
1,860
0
Nevermind, I made a calculation error.
 

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