Ask Uncle Colin: Invariant Lines

Dear Uncle Colin,

I’ve got a matrix, and I’m not afraid to use it. It’s $\left(\begin{array}{cc}3& -5\\ -4& 2\end{array}\right)$

Apparently, it has invariant lines. Those, I’m afraid of.

How do I find them?

-- Terrors About Rank, Safely Knowing Inverses

Hi, TARSKI!

An invariant line of a transformation is one where every point on the line is mapped to a point on the line – possibly the same point.

We can write that algebraically as $\mathbf{M}\cdot \mathbf{x}=\mathbf{X}$, where $\mathbf{x}=\left(\begin{array}{c}x\\ mx+c\end{array}\right)$ and $\mathbf{X}=\left(\begin{array}{c}X\\ mX+c\end{array}\right)$. Our job is to find the possible values of $m$ and $c$.

So, for this example, we have:

$\left(\begin{array}{cc}3& -5\\ -4& 2\end{array}\right)\left(\begin{array}{c}x\\ mx+c\end{array}\right)=\left(\begin{array}{c}X\\ mX+c\end{array}\right)$

And now it gets messy. We have two equations which hold for any value of $x$:

$3x-5(mx+c)=X$

$-4x+2(mx+c)=mX+c$

Substituting for $X$ in the second equation, we have:

$-4x+2(mx+c)=m(3x-5(mx+c))+c$

… which tidies up to …

$(2m-4)x+2c=(-5m^2+3m)x+(-5m+1)c$

… or …

$(5m^2-m-4)x+(5m+1)c=0$, for all $x$ (*).

Aside, on the difference between variables and constants

There are three letters in that equation, $m$, $c$ and $x$. For a long while, I thought “letters are letters, right? bits of algebraic furniture you can move around.” This isn’t true. In fact, there are two different flavours of letter here.

The $m$ and the $c$ are constants: numbers with specific values that don’t change. The $x$, on the other hand, is a variable, a letter that can mean anything we happen to find convenient.

Back to work

Considering $x=0$, this can only be true if either $5m+1=0$ or $c=0$, so let’s treat those two cases separately.

If $m=-\frac{1}{5}$, then equation (*) becomes $-\frac{18}{5}x=0$, which is not true for all $x$; $m=-\frac{1}{5}$ is therefore not a solution.

Instead, if $c=0$, the equation becomes $(5m^2-m-4)x=0$, which is true if $x=0$ (which it doesn’t, generally), or if $(5m^2-m-4)=0$, which it can; it factorises as $(5m+4)(m-1)=0$, so $m=-\frac{4}{5}$ and $m=1$ are both possible answers when $c=0$.

So the two equations of invariant lines are $y=-\frac{4}{5}x$ and $y=x$.

Just to check: if we multiply $\mathbf{M}$ by $(5,-4)$, we get $(35,-28)$, which is also on the line $y=-\frac{4}{5}x$. Similarly, if we apply the matrix to $(1,1)$, we get $(-2,-2)$ – again, it lies on the given line.

(It turns out that these invariant lines are related in this case to the eigenvectors of the matrix, but sh. Let’s not scare anyone off.)

-- Uncle Colin

* Edited 2019-06-08 to fix an arithmetic error. Thanks to Tom for finding it!