invariant lines/points under linear transformation

Assume a linear transformation matrix, \[T\]. Then, an invariant point is any point which stays the same under the transformation. Mathematically, it would be written as \[T\vec{x}=\vec{x}\], where \[\vec{x}\] is the position vector of the invariant point. A clearer way to write it is \[T\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x\\y\end{pmatrix}\]. Note that the origin \[(0,0)\] is always an invariant point as this is a property of linear transformations.

A line of invariant points is a straight line where all the points are invariant, which means any points lying on this line will stay the same under the transformation. A line of invariant points must normally pass through the origin, but not always, so we can give it the (2 dimensional) equation \[y=mx\] (as the y-intercept is the origin, thus \[c=0\]).

Substituting \[y\] into our original equation, we get \[T\begin{pmatrix}x\\mx\end{pmatrix}=\begin{pmatrix}x\\mx\end{pmatrix}\].

An invariant line is slightly different than a line of invariant points. Although line stays the same, the points on the line could move along the line. Meaning, the individual points lying on the line can change positions by some factor of \[k\], but will always stay on the line.
Mathematically, assume \[T\] is the transformation matrix, where \[T=\begin{pmatrix}-3&2\\-8&5\end{pmatrix}\].

\begin{align*} \begin{pmatrix} -3 & 2 \\ -8 & 5 \\ \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} X \\ mX \end{pmatrix} \end{align*}

To find the equation of the line,

\begin{align*} -3x+2mx&=X\\ -8x+5mx&=mX\\ \end{align*}

Dividing both equations we get,

\begin{align*} \frac{-3+2m}{-8+5m}&=\frac{1}{m}\\ 2m^{2}-3m&=5m-8\\ 2m^{2}-8m+8&=0\\ (m-2)^{2}&=0\\ \therefore m=2,y=2x \end{align*}

Therefore, the invariant line is \[y=2x\].

Also, if \[T\begin{pmatrix}x\\mx\end{pmatrix}=k\begin{pmatrix}x\\mx\end{pmatrix}\], this proves that the line is an invariant line, as the point on the line has shifted by a factor of \[k\], but still fulfills the equation \[y=mx\].