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


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.


17 comments on “Ask Uncle Colin: Invariant Lines

  • 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!

  • Peter Kempner

    Hi Colin, I’m just looking at an Edexcel Further Maths text book question

    it says if $T$ is a matrix $\mattwotwo{ -3}{ 0}{0}{-3}$ and the line $y=mx$ is an invariant line under $T$, then
    find the possible values of $m$. The answer says a general point $(x,y)$ gets mapped to $(-3x,-3mx)$ – which is fine and dandy but then it just says “The point $(−3x, -3mx)$ lies on the line $y =mx$ for $m = 1$ and $m = −1$, so the line is invariant under $T$ for these two values of $m$”

    I just don’t see how they got this.

    When I use the method you have shown us, I get the 2 equations:
    $-3x = -3X$ and $-3mx = -3mX$ which doesn’t seem to help!


    [Edited for LaTeX/formatting]

    • Colin

      I think there is an error in the question – the matrix there is an enlargement, so any line through the origin is invariant ($m$ could be anything, which is why there are infinitely many solutions using either method).

      I suspect the question meant to ask about $\mattwotwo{0}{-3}{-3}{0}$, but I couldn’t say for sure.

      • Peter Kempner

        Wow! You are right. I was careless and was solving the wrong problem mixing up the diagonals. Thank you for your quick response. Looking on the bright side I now have a deeper understanding of why at GCSE yo draw lines from the centre of enlargement to help find images.

        Thank you again.

        • Colin

          Don’t worry, it’s easily done – and I like your attitude, keep it up!

  • Michelle

    Hi Uncle Colin! Thanks for your help!
    I was wondering how we could find invariant lines in the three dimensional space.
    If a linear transformation is as follows
    T(x,y,z)=x(1,-1,0) + y(-2,0,0) + z(2,1,2)
    how can we find the invariant lines?
    Thanks for your help!

    • Colin

      Hi, Michelle! It turns out, in much the same way. Set it up as

      $\matthreethree{1&}{-2&}{2}{-1&}{0&}{2}{0&}{0&}{2}\colvecthree{t}{a+bt}{c+dt} = \colvecthree{T}{a+bT}{c+dT}$ and solve for $a$, $b$, $c$ and $d$. It’s a bit messy, but it should (hopefully) come out. Let me know if it doesn’t and I’ll have a proper play with it.

  • Linda Manas

    Very helpful.
    I understand this but am collecting invariant line questions to collate for my students.
    Best wishes

  • Josh

    Hi Colin, thanks very much for the useful post. I watched a video on invariant lines ( and in that the guy just looked at a point (x,y) and then set up a simultaneous equation, multiplying (x,y) by the transformation matrix and setting that equal to (x,y). I was just wondering how that method relates to your one above: obviously you don’t end up with a +c, but is it a good method to get an invariant line of a transformation? Thanks!

    • Colin


      This will give you the invariant points rather than invariant lines. The difference is that on an invariant line, points can move to *other* points on the line, while invariant points stay put. Does that make sense?

      • Josh

        Makes sense, thanks Colin!

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