Written by Colin+ in ask uncle colin, matrices.

*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 colin@flyingcoloursmaths.co.uk 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

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 ${\mathbf {M \cdot x}}= \mathbf X$, 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 $x$ (*).

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.

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 $-\frac{18}{5}x = 0$, 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

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

## Tom

If I’m not mistaken, the part where you return to the equation * after finding m=-1/5 , the equation should reduce to (-18/5)x=0 , not -4x=0

## Colin

Good spot! Thanks, I’ve fixed that.

## Igor

Hi, I’m a student currently studying further maths and we’re currently on linear transformations. This was very helpful, thanks!

## Colin

Glad it helped, Igor!

## Hassan Madani

why does a transformation that maps any point onto a straight line ax + by = c not a linear transformation?

## Colin

You’d think that, wouldn’t you? However, linear (in this context) doesn’t mean ‘relating to a line’, it means the transformation has certain properties (which can be summarised, I think, as $T(mA + nB) = mT(A) + nT(B)$.)

In particular, the origin *must* map to the origin, since $O = 1A + (-1)A$ for any $A$, and the origin isn’t necessarily on your line. Even if the origin *is* on the line (i.e., $c=0$), the transformation must still obey the rules – it’s possible to come up with mappings to a line through the origin that aren’t linear transformations.

Hope that makes some sense!