In class, I was learning about finding the invariant lines of a linear transformation. For example this question:
Find the invariant lines of the transformation $\begin{pmatrix} 5&1\\2&4 \end{pmatrix}$.
I was taught by my teacher this method:
Let $(x, y)$ be a point that gets mapped to the point $(x', y')$.
$$ \begin{pmatrix} 5&1 \\ 2&4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix} $$
Using matrix multiplication,
$$ x'=5x+y\\ y'=2x+4y $$
Then I substitute these expressions into the general equation of a line ($y=mx+c$):
$$ 2x+4y=m(5x+y)+c $$
Rearranging into the form $y=mx+c$
$$ y=\frac{5m-2}{4-m}x+\frac{c}{4-m} $$
Up to this point, me and my friend's methods are the same. I would continue by substituting $y=mx+c$ into the above equation and after a whole ton of rearranging, I would get this:
$$ x(m-1)(m+2)=4c-mc $$
And I would reason that $4c-mc$ won't change no matter what value of $x$ is. Therefore $(m-1)(m+2)$ must be 0 and $m=1$ or $m=-2$. And using the fact that $4c-mc$ is also 0, we can easily find $c=0$, hence finding the 2 invariant lines.
My friend found another easier way though. He notices that this line:
$$ y=\frac{5m-2}{4-m}x+\frac{c}{4-m} $$
has the same gradient as $m$, so
$$ \frac{5m-2}{4-m}=m $$
We can then easily solve for $m$. Then substitute the values back into $mx+c=\frac{5m-2}{4-m}x+\frac{c}{4-m}$, getting $c=0$.
My question is, does my friend's method give the same answers as my method for all matrices? If not, for what kind of matrix will it give the same answers?
The two methods are equivalent. Your friend is equating the coefficients of two equal polynomials: if $Ax+B=Cx+D$ for all values of $x$, then $A=C$ and $B=D$. This sets up a system of two equations in $m$ and $c$, which he then proceeds to solve by back-substitution. You’re solving the same system of equations via a different route: what you’re doing amounts to subtracting one equation of the line from the other and then rearranging.