Finding the invariant lines of $\left(\begin{smallmatrix} 3 & -1 \\ 6 & -2 \end{smallmatrix}\right)$ How do I find the invariant lines of:
\begin{pmatrix}
3 & -1 \\
6 & -2
\end{pmatrix}
So my thoughts were to assume the invariant line has the form $y=mx$ as it will pass through the origin, giving any point on the line coordinates $(x,mx)$.
Hence multiplying the above matrix with the vector:
\begin{pmatrix}
x \\
mx
\end{pmatrix}
Gives: 
\begin{pmatrix}
3x-mx\\
6x-2mx
\end{pmatrix}
Substituting the x and y components back into $y=mx$ and rearranging yields:
$m^2x-5mx+6x=0$ which has solutions $m=2$ and $m=3$ hence the invariant lines are $y=2x$ and $y=3x$. However, apparently the only invariant line is $y=2x$ because the original matrix is singular. Could somebody explain why we do not accept $y=3x$ as an invariant line?
Many thanks.

It depends upon how you define invariant line. Is it, given a linear transformation $T$ a line $l$ such that $T(l)\subset l$? Or is it a line $l$ such that $T(l)=l$?
If the first possibility is the correct one, then you are right.
But if it's the second ine, you are not. That's so because if $l$ is the line $y=3x$, then $T(l)=\{0\}\neq l$.
