Why is the eigenvector of the smallest eigenvalue of this matrix gives the intersection of a set of lines in homogenous coordinates? Consider $L = \{l_1 \cdots l_n\}$ be a set of lines in the plane in homogeneous coordinates. In this definition, a line $ax+by+c = 0$ is given by its direction $l^T = (a/c,b/c,1)^T$ such that it consists of the points where $l^T(x,y,1)^T=0$. Now suppose that I want to find the intersection of lines in $L$, assuming that it exists. Otherwise, I want to find the point given by the least square solution had I written all this as a system of linear equations.
I have found a PDF that claims the eigenvector of the smallest eigenvalue of this matrix is the answer:
$$M = \sum_i l_i l_i^T$$
I wonder why this is true. In case anyone wonders where I have found this, here's the link: click here.
 A: If $v=(x,y,1)^T$ is the intersection of all lines, then $l_i^Tv=0$ for every $i$, i.e. $l_il_i^Tv=\vec 0$ for every $i$. It follows that $Mv=\vec 0$, i.e. $v$ is an eigenvector of $M$ with zero eigenvalue.
If the known coefficient of the lines are only approximations of their real values, then the eigenvalue is not zero, but very near to it (of course we are supposing the approximations to be very good). Hence the claim is true, provided we specify "the smallest eigenvalue in modulus".
Very likely, this is also the point given by the least square solution, in case the lines don't coverge: it shouldn't be difficult to carry out the computation and verify the claim.
EDIT.
I must retract my last sentence above: in general, the eigenvector with smallest eigenvalue does not yield the least square solution. 
Here's a counterexample: lets take three non-coverging lines $l_1=(1,-1,1)$, $l_2=(-1,-1,1)$, $l_3=(0,-2,1)$. If $v=(x,y,1)^T$ and $l_iv=a_i$, then to find the least square $v$ we have to minimize $S=\sum_i a_i^2$. But:
$$
{\partial S\over\partial x}=2\sum_i a_i{\partial a_i\over\partial x}=
2\sum_i v^Tl_i^Tl_i{\partial v\over\partial x}=2v^TM\pmatrix{1\\0\\0}
\quad\text{and}\quad
{\partial S\over\partial y}=2v^TM\pmatrix{0\\1\\0}.
$$
You can verify that imposing $\partial S/\partial x=0$ and 
$\partial S/\partial y=0$ gives as solution $v=(0,2/3,1)$.
On the other hand, the smallest eigenvalue of $M$ is $(9-\sqrt{73})/2$, corresponding to the eigenvector 
$v=(0,(\sqrt{73}-3)/8,1)$, which is different from the solution found above.
