# Least Squares Solution - Or is there?

1. Apr 18, 2007

### Mindscrape

I have a problem that says to find the least squares solution to
$$\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)} K x = \colv{2}{2}$$ for
$$K = \left( \begin{array}{cc} 1 & 2\\ 2 & 4 \end{array} \right)$$. 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

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

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. Apr 18, 2007

### Mindscrape

Nevermind, I made a calculation error.