How to find lines of invariant points? Every time I try a question on this topic I get it wrong.
My textbook says:

Invariant points satisfy $B\begin{pmatrix}u\\ v\end{pmatrix}=\begin{pmatrix}u\\ v\end{pmatrix}$
Re-write this as a system of equations.
Check whether both equations are in fact the same.
If so, they give a line of invariant points.

So I tried this question using that method (and the method needed for finding invariant lines):

For [this] matrix, find any lines of invariant points and any other invariant lines through the origin.
$\begin{pmatrix}3&-2\\ 4&-3\end{pmatrix}$

What I did was:
$\begin{pmatrix}3&-2\\ 4&-3\end{pmatrix}\begin{pmatrix}u\\ mu\end{pmatrix}=\begin{pmatrix}u\\ v\end{pmatrix}=\begin{pmatrix}3u-2mu\\ 4u-3mu\end{pmatrix}$
These equations are not the same, so no lines of invariant points.
Then, to find invariant lines:
$\begin{pmatrix}3&-2\\ 4&-3\end{pmatrix}\begin{pmatrix}u\\ mu\end{pmatrix}=\begin{pmatrix}u'\\ v'\end{pmatrix}=\begin{pmatrix}3u-2mu\\ 4u-3mu\end{pmatrix}$
$4u-3mu=m\left(3u-2mu\right)$
$2m^2-6m+4\:=0\:\Rightarrow \:m\:=1\: $or$ \:m=2$
So the invariant lines should be $y=2x$ or $y=x$. As far as I know this is sort of right, except $y=x$ is apparently also a line of invariant points. What did I do wrong?
 A: Let
$$\mathbf{B} = \left [ \begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix} \right ], \quad \mathbf{p} = \left [ \begin{matrix} u \\ v \end{matrix} \right ]$$
If $\mathbf{p}$ is an invariant point with respect to $\mathbf{B}$, then
$$\mathbf{B} \mathbf{p} = \mathbf{p} \tag{1}\label{NA1}$$
which is equivalent to
$$\left\lbrace \begin{aligned}
b_{11} u + b_{12} v &= u \\
b_{21} u + b_{22} v &= v \end{aligned} \right .$$
and
$$\left\lbrace \begin{aligned}
(b_{11} - 1) u + b_{12} v &= 0 \\
b_{21} u + (b_{22} - 1) v &= 0 \end{aligned} \right . \tag{2}\label{NA2}$$
To find invariant points, you solve $\eqref{NA2}$ for $u$ and $v$.  In some cases, the solution is not a single point, but a line. If the matrix is all zeros, then all points are invariant.

Let's examine $\mathbf{B} = \left [ \begin{matrix} 1 & 4 \\ 2 & -1 \end{matrix} \right ]$. Substituting to $\eqref{NA2}$ we get
$$\left\lbrace\begin{aligned}
(1 - 1)u + 4 v &= 0 \\
2 u + (-1 - 1) v &= 0 \\
\end{aligned} \right . \quad \iff \quad \left\lbrace\begin{aligned}
4 v &= 0 \\
2 u - 2 v &= 0 \\
\end{aligned}\right .$$
The only solution to this is $u = v = 0$. So, this matrix has an invariant point at origin.

Let's examine $\mathbf{B} = \left [ \begin{matrix} 3 & -2 \\ 4 & -3 \end{matrix} \right ]$. Substituting to $\eqref{NA2}$ we get
$$\left\lbrace\begin{aligned}
2 u - 2 v &= 0 \\
4 u - 4 v &= 0 \\
\end{aligned}\right .$$
If we divide the upper one by 2, or the lower one by 4, we get the same equation: $u - v = 0$. Thus, this matrix has an invariant line $u = v$.
