Are approximate least square intersections unique? I seem to be getting a different approximate intersections for the same three lines by multiplying one of the line equations (so that the equation still defines the same line but has different numbers in the equation. I'm trying to figure out if I'm making a mistake or if the method I'm using using to find the approximate intersection gives non-unique answers.
My first set of lines is:
$$
2x-y=2\\
x+2y=1\\
x+y=4
$$
The only difference being in the second line, my second set of lines is:
$$
2x-y=2\\
\frac{1}{2}x+y=\frac{1}{2}\\
x+y=4
$$
With the left hand side of these line equations forming my $A$ matrix and the right hand side forming my $b$ matrix, I setup the equation $A^TAx = A^Tb$, form the augmented matrix of $[A^TA\,\vert\,A^Tb]$, and find the reduced row echelon form of the left side of the augmented matrix. I expect the right side to then be the approximate intersection. However, using these two different line equations produces two different intersections ($\frac{10}{7},\frac{3}{7}$ and $\frac{52}{31},\frac{69}{62}$). Both seem like plausible solutions. I have looked back over the math a few times and I'm not finding any errors I made. So I'm wondering if I can actually get two different answers from the different numbers in the line equations despite the fact that those different numbers defined the same line. Thanks!
 A: Yes, this can happen, and your answer is (answers are) correct.  I will try to give some intuitive (not rigorous) explanation.
Remember that your basic least squares problem is this: find $x,y$ such that
$$x\pmatrix{2\cr1\cr1\cr}+y\pmatrix{-1\cr2\cr1\cr}
  \quad\hbox{is as close as possible to}\quad \pmatrix{2\cr1\cr4\cr}\ ,$$
or in the second version, find $x,y$ such that
$$x\pmatrix{2\cr\frac12\cr1\cr}+y\pmatrix{-1\cr1\cr1\cr}
  \quad\hbox{is as close as possible to}\quad \pmatrix{2\cr\frac12\cr4\cr}
  \ .$$
Now you will assess the "as close as possible" by finding the distance between the LHS and the RHS in each case.  In the second case, the contribution from the second coordinate is half what it was in the first case.  So the second coordinate is now "less important", and if we allow it to increase in such a way that the "more important" first and third coordinates decrease, we may be able to get a better overall result.
I hope this gives some idea as to why the two problems actually are different and why you should expect to get different answers.
A: @David provides a fine explanation and two comments are added.
First, this scaling issue only occurs for the overdetermined case where $m$, the number of rows exceeds $\rho$, the matrix rank, $m>\rho$.
Second, to see the effect explicitly, introduce the scaling parameter $\alpha \ne 0$. The system matrix transforms as
$$
\begin{align} 
  \mathbf{A} &\to \hat{\mathbf{A}} \\
\left[
  \begin{array}{cr}
    2 & -1 \\
    1 &  2 \\
    1 &  1
  \end{array}
\right]
 &\to 
\left[
  \begin{array}{cr}
    2 & -1 \\
    \alpha &  2\alpha \\
    1 &  1
  \end{array}
\right]
\end{align}
$$
while the data vector becomes
$$
\begin{align} 
  b &\to \hat{b} \\
\left[
  \begin{array}{c}
    2 \\
    1 \\
    4
  \end{array}
\right]
 &\to 
\left[
  \begin{array}{cr}
    2 \\
    \alpha \\
    1 
  \end{array}
\right]
\end{align}.
$$
The solution changes accordingly:
$$
\begin{align} 
  x &\to \hat{x} \\
\left[
  \begin{array}{c}
    \frac{10}{7} \\
    \frac{3}{7}  
  \end{array}
\right]
 &\to 
\left[
  \begin{array}{c}
    \frac{18+32\alpha^{2}}{9+26\alpha^{2}} \\
    \frac{18-3\alpha^{2}}{9+26\alpha^{2}}  
  \end{array}
\right]
\end{align}
$$
which established $\hat{x}|_{\alpha=1} = x$ as expected.
