Ask Uncle Colin: Invariant Lines

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to and Uncle Colin will do what he can.

Dear Uncle Colin,

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

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

How do I find them?

— Terrors About Rank, Safely Knowing Inverses


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 ${\mathbf {M \cdot x}}= \mathbf X$, where $\mathbf x = \begin{pmatrix} x \\ mx + c\end{pmatrix}$ and $\mathbf X = \begin{pmatrix} X \\ mX + c\end{pmatrix}$. Our job is to find the possible values of $m$ and $c$.

So, for this example, we have:

$\begin{pmatrix} 3 & -5 \\ -4 & 2\end{pmatrix}\begin{pmatrix} x \\ mx + c\end{pmatrix} = \begin{pmatrix} X \\ mX + c\end{pmatrix}$

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 $m = – \frac 15$, then equation (*) becomes $-4x = 0$, which is not true for all $x$; $m = -\frac15$ 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 $y = -\frac45x$ and $y = x$.

Just to check: if we multiply $\mathbf{M}$ by $(5, -4)$, we get $(35, -28)$, which is also on the line $y = – \frac 45 x$. 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


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.


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