Further Maths/Lessons/Linear Transformations/Invariant Points & Lines
Core Pure · Linear Transformations · Year 1

Invariant Points and Lines

Some points and lines are left exactly where they started by a transformation — spotting them, and knowing the difference between a point staying still and a line staying put as a whole, is a favourite exam idea.

14 min read 3 worked examples

Invariant points

Definition 32.1 — invariant point

A point $P$ is an invariant point of a transformation represented by matrix $M$ if $P$ maps to itself: $M\mathbf{p}=\mathbf{p}$, where $\mathbf{p}$ is the position vector of $P$.

The origin is always invariant

For any matrix $M$, $M\binom{0}{0}=\binom{0}{0}$, since multiplying any matrix by the zero vector gives the zero vector. So the origin is an invariant point of every linear transformation — the interesting question is always whether there are any other invariant points.

Example 1 · Finding invariant points other than the origin
Find the invariant point(s), other than the origin, of the transformation represented by $M=\begin{pmatrix}3&1\\2&4\end{pmatrix}$.

Under $M$, the point $(x,y)$ maps to $(x^\prime,y^\prime)$ where $x^\prime=3x+y$ and $y^\prime=2x+4y$.

An invariant point satisfies $x^\prime=x$ and $y^\prime=y$:

$$3x+y=x \;\Rightarrow\; 2x+y=0, \qquad 2x+4y=y \;\Rightarrow\; 2x+3y=0$$

Solving simultaneously: subtracting gives $2y=0 \Rightarrow y=0$, and then $x=0$. So $(0,0)$ — the origin — is the only invariant point.

Why the origin is sometimes the only invariant point

Setting $x^\prime=x$ and $y^\prime=y$ always gives a pair of simultaneous equations of the form $(M-I)\binom{x}{y}=\binom{0}{0}$. If $\det(M-I)\neq0$, this system has only the trivial solution $x=y=0$ — the origin is the only invariant point. If $\det(M-I)=0$, the two equations become multiples of each other, and every point on a whole line through the origin is invariant.

Example 2 · A whole line of invariant points
Show that every point on the line $3x-2y=0$ is invariant under the transformation represented by $M=\begin{pmatrix}4&-2\\3&-1\end{pmatrix}$.

Under $M$, $x^\prime=4x-2y$ and $y^\prime=3x-y$. Setting $x^\prime=x$ and $y^\prime=y$:

$$4x-2y=x \;\Rightarrow\; 3x-2y=0, \qquad 3x-y=y \;\Rightarrow\; 3x-2y=0$$

Both conditions reduce to the same equation, $3x-2y=0$. So instead of forcing $x=y=0$, every point satisfying $3x-2y=0$ is invariant — a whole line of invariant points through the origin, not just the origin itself.

Invariant lines

Definition 32.2 — invariant line

A line is an invariant line of a transformation if every point on the line maps to a point that is also on the line — the line maps to itself as a whole set, even though individual points on it are generally free to slide to a different position along it.

Common mistake — invariant line vs. line of invariant points

A line of invariant points is a stronger condition: every single point on it stays exactly where it is. An invariant line only requires the line to map onto itself as a set — individual points on it can (and usually do) move, just always to another point on the same line. Every line of invariant points is an invariant line, but most invariant lines are not lines of invariant points.

MethodFinding invariant lines of a transformation, starting from $y=mx+k$
1
For $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, consider the general line $y=mx+k$ (using $k$ for the intercept, so it isn't confused with the matrix entry $c$), and substitute a general point $(x,mx+k)$ on it into the transformation.
Starting from the general line, rather than assuming it passes through the origin, makes sure no invariant lines are missed.
2
The image is $x^\prime=ax+b(mx+k)=(a+bm)x+bk$ and $y^\prime=cx+d(mx+k)=(c+dm)x+dk$.
Just $M$ applied to the vector $\binom{x}{mx+k}$.
3
For the line to be invariant, every image point must satisfy $y^\prime=mx^\prime+k$. Since this has to hold for every $x$, match the coefficients of $x$ and the constant terms separately:
A relationship that holds for all $x$ forces the $x$-terms and the constant terms to balance independently.
4
Coefficient of $x$: $c+dm=m(a+bm)$, giving $bm^2+(a-d)m-c=0$ — exactly the same quadratic as before, so this still determines the gradient(s) $m$.
This condition doesn't involve $k$ at all, so it's unaffected by whether the line passes through the origin.
5
Constant term: $dk=m(bk)+k$, i.e. $k(d-bm-1)=0$. So either $k=0$, or $bm=d-1$.
The second option only coincides with a genuine root of the quadratic above in special cases (when $\lambda=1$ is an eigenvalue of $M$) — otherwise it forces $k=0$.
6
For a typical transformation, this confirms $k=0$: every invariant line passes through the origin, so it's valid to look for invariant lines directly in the form $y=mx$, and the two approaches give the same gradients $m$. Check the vertical line $x=0$ separately, since it isn't of the form $y=mx$.
$x=0$ maps to itself only when $b=0$, so it needs its own check regardless.
Example 3 · Finding invariant lines through the origin
Find the equations of the invariant lines through the origin of the transformation represented by $M=\begin{pmatrix}5&4\\2&3\end{pmatrix}$.

By the method above, any invariant line here passes through the origin, so let the line be $y=mx$. Substituting $(x,mx)$ into $M$: $x^\prime=5x+4mx=(5+4m)x$ and $y^\prime=2x+3mx=(2+3m)x$.

For the line to be invariant, $y^\prime=mx^\prime$:

$$2+3m=m(5+4m) \;\Rightarrow\; 4m^2+2m-2=0 \;\Rightarrow\; 2m^2+m-1=0$$

Factorising: $(2m-1)(m+1)=0$, so $m=\tfrac12$ or $m=-1$.

The invariant lines through the origin are $y=\tfrac12x$ and $y=-x$.

Exam-style questions

Question 1 4 marks

Show that the transformation represented by $M=\begin{pmatrix}2&3\\1&4\end{pmatrix}$ has a whole line of invariant points, and find its equation.

$x^\prime=2x+3y$, $y^\prime=x+4y$. Setting $x^\prime=x$ and $y^\prime=y$ (M1)

$2x+3y=x\Rightarrow x+3y=0$, and $x+4y=y\Rightarrow x+3y=0$ (A1)

Both equations are the same, so every point on the line is invariant, not just the origin (B1)

The line of invariant points is $x+3y=0$ (A1)

Question 2 4 marks

Show that the origin is the only invariant point of the transformation represented by $M=\begin{pmatrix}5&-1\\1&3\end{pmatrix}$.

$x^\prime=5x-y$, $y^\prime=x+3y$. Setting $x^\prime=x$ and $y^\prime=y$: $4x-y=0$ and $x+2y=0$ (M1, A1)

From the first equation $y=4x$; substituting into the second gives $x+8x=9x=0\Rightarrow x=0$, so $y=0$ (M1)

The only solution is $(0,0)$, so the origin is the only invariant point (A1)

Question 3 5 marks

Find the equations of the two invariant lines through the origin of the transformation represented by $M=\begin{pmatrix}6&-3\\-2&5\end{pmatrix}$.

Let the line be $y=mx$. Substituting: $x^\prime=6x-3mx=(6-3m)x$, $y^\prime=-2x+5mx=(-2+5m)x$ (M1, A1)

For invariance, $y^\prime=mx^\prime$: $-2+5m=m(6-3m)$ (M1)

$3m^2-m-2=0 \Rightarrow (3m+2)(m-1)=0 \Rightarrow m=1$ or $m=-\tfrac23$ (A1)

The invariant lines are $y=x$ and $y=-\tfrac23x$ (A1)

← Previous
The Inverse of a Linear Transformation