Problem source

The matrix A is given by
$$\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.$$
\begin{questionparts}
\item You are given that the transformation represented by A has a line $L_1$ of invariant points (so that each point on $L_1$ is transformed to itself). Let $(x, y)$ be a point on $L_1$. Show that $((a - 1)(d - 1) - bc)xy = 0$.
Show further that $(a - 1)(d - 1) = bc$.
What can be said about A if $L_1$ does not pass through the origin?
\item By considering the cases $b \neq 0$ and $b = 0$ separately, show that if $(a - 1)(d - 1) = bc$ then the transformation represented by A has a line of invariant points. You should identify the line in the different cases that arise.
\item You are given instead that the transformation represented by A has an invariant line $L_2$ (so that each point on $L_2$ is transformed to a point on $L_2$) and that $L_2$ does not pass through the origin. If $L_2$ has the form $y = mx + k$, show that $(a - 1)(d - 1) = bc$.
\end{questionparts}

Solution source

\begin{questionparts}
\item Suppose $(x,y)$ is on the line of invariant points, then

\begin{align*}
&&\begin{pmatrix} x \\ y \end{pmatrix}  &= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\
&&&= \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix} \\
\Rightarrow && \begin{cases} (a-1)x + by = 0 \\ (cx + (d-1)y = 0 \end{cases} \tag{*}
\end{align*}
Therefore either $x  = 0, y = 0$ or $(a-1)(d-1)-bc = 0$ $\Rightarrow ((a-1)(d-1)-bc)xy = 0$. We also know this is true for all values $x,y$ on the line of invariant points. If there is one where both $x \neq 0, y \neq 0$ we are done, otherwise the line of invariant points must be one of the axes. ie but then one of $\begin{pmatrix} a \\ c \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ or $\begin{pmatrix} b \\ d \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ is true and we'd also be done.

If the line doesn't go through the origin then there are points on every line, not equal to the origin which are fixed. But then every point on those lines is fixed (since $\mathbf{A}$ is a linear operator) and so every point is fixed. ie $\mathbf{A} = \mathbf{I}$.

\item Suppose $(a-1)(d-1) -bc = 0$ and $b \neq 0$ then I claim that $y = \frac{1-a}{b}x$ is a line of invariant points. It's clear that the first equation will be satisfied in $(*)$ so it suffices to check the second, but the first condition is equivalent to the equations being linearly dependent, ie both equations are satisfied.

If $b = 0$ then $(a-1)(d-1) = 0$, so our matrix must look like $\begin{pmatrix} 1 & 0 \\ c & d\end{pmatrix}$ (if $d \neq 1$)or $\begin{pmatrix} * & 0 \\ * & 1\end{pmatrix}$. In the first case, the line $y = \frac{c}{1-d}x$ and in the second $x = 0$ is an invariant line.

\item Suppose the invariant line is $y = mx+k$ then we must have that 
\begin{align*}
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ mx + k \end{pmatrix} &= \begin{pmatrix} (a + mb)x + bk \\
(c+dm)x + dk
 \end{pmatrix}
\end{align*}

and $(c+dm)x + dk = m((a + mb)x + bk) +k \Rightarrow k(d-mb-1) = x(-c+(a-d)m+m^2b)$

Since this equation must be true for all values of $x$, and $k \neq 0$ we can say that $mb = d-1$ and $-c+(a-d)m+m^2b = 0$, ie $-c + (a-d)m + m(d-1) = 0 \Rightarrow (a-1)m-c = 0$ if $m \neq 0$ then $(a-1)\frac{(d-1)}{b} - c = 0$ ie our desired relation is true. If $m = 0$ then we must have that $y = k$ is an invariant line, ie $d-1=0$ and $c=0$ which also satisfies our relation.

\end{questionparts}