Do invariant lines of linear transformations contain a fixed point? Suppose $A$ is a $2$-by-$2$ matrix, and $\mathcal{l}$ is an invariant line under $A$, so $(x,mx+c)$ is mapped to $(X,mX+c)$ for some variable $X$ linear in $x$. Then is there a point on the line $\mathcal{l}$ which is a fixed point of $A$, i.e. there is some $x' \in \mathcal{l}$ such that $Ax'=x'$?
The reason I ask is that apparently if $y=mx+c$ is an invariant line under $A$, so is $y=mx$ - the above is equilivant to this by linearity, I believe. Certainly, the case where the invariant line goes through the origin is simple - but unfortunately this isn't the only case, e.g. the matrix
$$ \left(\begin{matrix} 0 & 1 \\ 5 & -4 \end{matrix}\right)$$
has an invariant line $y=-5x+2$ as 
$$ \left(\begin{matrix} 0 & 1 \\ 5 & -4 \end{matrix}\right)\left( \begin{matrix} x\\ -5x + 2 \end{matrix} \right) = \left( \begin{matrix} -5x+2\\ -5(-5x+2) + 2 \end{matrix} \right)$$
However, I can't see this example being very illuminating, as it's quite contrived to force $(0,1)$ to be mapped to itself, and I can't imagine the $y$-intercept always being a fixed point in these circumstances. Does anyone have any ideas?
 A: Obviously if such a fixed point exists then $Au=u$ and $u$ is an eigenvector of $A$ with the eigenvalue $1$. So $1$ has to be an eigenvalue of $A$, as is the case for your matrix. Then consider the direction vector of the line $v$, it can't be parallel to $u$ since the line is not through the origin, and since $A(u+tv)=u+sv$ and $Au=u$ this $v$ also has to be an eigenvector. So $A$ has to have two non-parallel eigenvectors.
Conversely, if $u$ is an eigenvector corresponding to $1$ and $v$ is some other eigenvector not parallel to $u$ then you can get a line not through the origin with a fixed point on it, namely $u+tv$ in parametric form, where $t$ is a real parameter. This is how your example works: $(x,-5x+2)=(0,2)+(1,-5)x$ and both $u=(0,2)$ and $v=(1,-5)$ are eigenvectors with $Au=u$.
It's not quite true that for such a line to exist it's enough that $A$ has $1$ as an eigenvalue, because a matrix like $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ has only one eigenvector up to parallel ones. The correct condition is that $A$ has to be diagonalizable, which for $2\times2$ matrices means having two non-parallel eigenvectors, and have $1$ as an eigenvalue.
A: To directly answer the question asked (although I think Conifold has already done so just fine), consider the transformation
$$
T(x, y) = (x + y, y).
$$
Then every horizontal line (i.e, line with $y = const$) is an invariant line. In particular, the line $y = 1$ is an invariant line. But on that line, we have
$$
T(x, 1) = (x + 1, 1)
$$
which clearly has no fixed points. 
(The transformation itself does have a fixed point --- any point on the $x$-axis is fixed, for instance --- but there isn't one on this particular invariant line, which is what the first paragraph asks, even if the title doesn't exactly.) 
By the way, for this $T$, we have (if you're using column vectors), 
$$
A = \pmatrix{1 & 1 \\ 0 & 1},
$$
which tends to provide counterexamples to lots of conjectures about matrices, and should always be your first "go-to example" when you think something might be true. :) 
