Pm6 - Invariant points and lines | OCR A-Level Further Mathematics B (MEI)

A transformation can leave a point fixed or leave a whole line in the same place while moving its points. You learn to distinguish these two meanings of invariance and use coordinate equations to find every invariant point and line of a plane transformation. Checking vertical lines will be just as important as solving for a gradient.

Finding points that stay fixed

Can an entire line stay in the same place even though none of its points stays put?

The shear $(x,y)\mapsto(x+2y,y)$ gives a clue. It moves $(0,1)$ to $(2,1)$ , then $(2,1)$ to $(4,1)$ . Both images are on $y=1$ , but neither point stays fixed. We will find exactly what remains unchanged by the end of the lesson.

You already know

A transformation with matrix $\mathbf{M}$ sends $\begin{pmatrix}x\\y\end{pmatrix}$ to $\begin{pmatrix}x'\\y'\end{pmatrix}=\mathbf{M}\begin{pmatrix}x\\y\end{pmatrix}$ . A shear's fixed axis consists of points that do not move at all.

An invariant point must return to its own coordinates, not merely to the same axis or the same distance from the origin.

Invariant point

A point is invariant under a transformation if its image is the point itself. Thus $(x,y)$ is invariant precisely when $x'=x$ and $y'=y$ . For example, every point on a reflection's mirror line is invariant. A point sent to a different point on that mirror line would not be invariant.

The matrix equation expresses both coordinate conditions at once. Subtracting the original vector gives two homogeneous linear equations.

Invariant-point equations

For $\mathbf{M}=\begin{pmatrix}p&q\\r&s\end{pmatrix}$ , invariant points satisfy

\begin{aligned} \mathbf{M}\begin{pmatrix}x\\y\end{pmatrix}&=\begin{pmatrix}x\\y\end{pmatrix}\\ (p-1)x+qy&=0\\ rx+(s-1)y&=0 \end{aligned}

Both equations must hold. The origin is always invariant under a linear transformation.

Not in the formula booklet: learn it.

Do not assume that the two equations determine a unique point. They may be the same condition written twice, or one may reduce to $0=0$ .

Worked example 1: a line of invariant points

Question. Find all invariant points of the transformation with matrix $\mathbf{H}=\begin{pmatrix}1&2\\0&1\end{pmatrix}$ .

Plan. Equate each image coordinate to its original coordinate and solve both conditions.

\begin{aligned} \mathbf{H}\begin{pmatrix}x\\y\end{pmatrix}&=\begin{pmatrix}x+2y\\y\end{pmatrix}&&\text{image coordinates}\\ x+2y&=x,\qquad y=y&&\text{point must be unchanged}\\ 2y&=0&&\text{the only restriction}\\ y&=0,\qquad x\text{ arbitrary}&&\text{every point on the }x\text{-axis} \end{aligned}

Why this step? The second equation is true for every $y$ ; it does not force $y$ to be any particular number. The first equation supplies the condition $y=0$ .

Answer. All points on the line $y=0$ are invariant.

Check. $\mathbf{H}\begin{pmatrix}t\\0\end{pmatrix}=\begin{pmatrix}t\\0\end{pmatrix}$ for every real $t$ , proving the whole line is fixed.

The next example changes one matrix entry, from $1$ to $2$ in the upper-left corner. The fixed points now lie on a different line.

Worked example 2: a tilted fixed line

Question. Find all invariant points of the transformation with matrix $\mathbf{M}=\begin{pmatrix}2&2\\0&1\end{pmatrix}$ .

Plan. Write both fixed-coordinate equations and retain any free variable.

\begin{aligned} \mathbf{M}\begin{pmatrix}x\\y\end{pmatrix}&=\begin{pmatrix}2x+2y\\y\end{pmatrix}&&\text{multiply the vector}\\ 2x+2y&=x,\qquad y=y&&\text{invariant-point conditions}\\ x+2y&=0&&\text{rearrange the first condition}\\ x&=-2y&&\text{one free coordinate remains} \end{aligned}

Why this step? There is only one independent condition on two coordinates, so the answer is a line rather than a single point.

Answer. All points on $x+2y=0$ are invariant.

Check. A general point is $(-2t,t)$ , and $\mathbf{M}\begin{pmatrix}-2t\\t\end{pmatrix}=\begin{pmatrix}-4t+2t\\t\end{pmatrix}=\begin{pmatrix}-2t\\t\end{pmatrix}$ .

Other matrices have different fixed sets. For $\operatorname{diag}(2,3)$ , the equations $2x=x$ and $3y=y$ leave only the origin. For the identity matrix, both equations are identities and every point is fixed. These conclusions follow from the equations, rather than from the appearance of a few plotted points.

Watch out

Wrong: from $y=y$ , conclude $y=0$ . This equation imposes no restriction. Right: solve the actual non-trivial equations and state which coordinate is free.

The next matrix changes the same upper-left entry again. Check the whole fixed set, not just one point on it.

Lines that keep their direction

An invariant line stays the same line as a set, although its individual points may move along it.

Reflection in $y=x$ fixes every point on its mirror. It also leaves the line $y=-x$ invariant: $(2,-2)$ moves to $(-2,2)$ , on the same line. Apart from the origin, those points move. “Invariant line” and “line of invariant points” therefore describe different things.

[DIAGRAM: asset_name: An invariant line need not be fixed point by point; asset_slug: invariant_line_vs_points; caption: Reflection in y = x fixes its mirror point by point but exchanges points along y = -x; description: An equal-scale coordinate diagram shows the mirror y equals x in sage, with point A at 1.5,1.5 labelled A maps to A. The line y equals minus x is blue, with P at 2,minus2 and its clay image P prime at minus2,2. Opposing curved arrows indicate that reflection swaps these two points. A side panel distinguishes an invariant point, a line of invariant points and an invariant line.]
Reflection in y = x fixes its mirror point by point but exchanges points along y = -x

Invariant line

An invariant line is a line whose image is the same line. Its points can move to other positions on the line. A line of invariant points is stronger: every point on it is individually fixed. For example, the mirror line of a reflection has invariant points; a perpendicular line can be invariant while its points exchange places.

For a non-vertical line through the origin, write a general point as $(x,mx)$ . The image must satisfy the same equation $y'=mx'$ for every $x$ . It is the relation between the image coordinates that matters, not equality of each old and new coordinate.

Gradient condition for an invariant line

For $\mathbf{M}=\begin{pmatrix}p&q\\r&s\end{pmatrix}$ , a point on $y=mx$ has image

\begin{pmatrix}x'\\y'\end{pmatrix} =\begin{pmatrix}(p+qm)x\\(r+sm)x\end{pmatrix}

Requiring $y'=mx'$ for every $x$ gives

r+sm=m(p+qm)

Use the coordinate relation directly rather than dividing by $x'$ . Check the vertical line $x=0$ separately, using $(0,y)$ .

Not in the formula booklet: learn it.

This condition says that the image stays on the candidate line. In the examples below the coordinate along that line is multiplied by a non-zero factor, so its image is indeed the whole original line.

Worked example 3: two invariant directions

Question. Find all invariant lines through the origin for $\mathbf{A}=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ .

Plan. Apply the matrix to $(x,mx)$ , impose the same gradient, then check the missing vertical line.

\begin{aligned} \mathbf{A}\begin{pmatrix}x\\mx\end{pmatrix}&=\begin{pmatrix}(2+m)x\\(1+2m)x\end{pmatrix}&&\text{general point on the line}\\ 1+2m&=m(2+m)&&y'=mx'\text{ for all }x\\ m^2&=1&&\text{cancel }2m\\ m&=1\text{ or }-1&&\text{real directions} \end{aligned}

For the vertical line,

\mathbf{A}\begin{pmatrix}0\\y\end{pmatrix}=\begin{pmatrix}y\\2y\end{pmatrix}

The image does not generally have first coordinate zero, so $x=0$ is not invariant.

Why this step? A gradient represents every line through the origin except a vertical one. Checking it separately makes the answer complete.

Answer. Exactly $y=x$ and $y=-x$ .

Check. $(t,t)\mapsto(3t,3t)$ , so $y=x$ is invariant while its non-zero points move. Also $(t,-t)\mapsto(t,-t)$ , so $y=-x$ is a line of invariant points.

The next example changes only the sign of the upper-right entry. A small algebraic change removes every real invariant direction.

Worked example 4: no real invariant line through the origin

Question. Find all invariant lines through the origin for $\mathbf{B}=\begin{pmatrix}2&-1\\1&2\end{pmatrix}$ .

Plan. Repeat the gradient condition, but accept that it may have no real solution.

\begin{aligned} \mathbf{B}\begin{pmatrix}x\\mx\end{pmatrix}&=\begin{pmatrix}(2-m)x\\(1+2m)x\end{pmatrix}&&\text{coordinate images}\\ 1+2m&=m(2-m)&&\text{same line required}\\ 1&=-m^2&&\text{cancel }2m \end{aligned}

No real gradient satisfies this. For a vertical line,

\mathbf{B}\begin{pmatrix}0\\y\end{pmatrix}=\begin{pmatrix}-y\\2y\end{pmatrix}

so the image is not generally on $x=0$ .

Why this step? A line in the real coordinate plane needs a real direction. Complex solutions for $m$ would not describe additional lines in that plane.

Answer. There are no invariant lines through the origin.

Check. The horizontal direction $(1,0)$ becomes $(2,1)$ , already turning away from its original line. The algebra proves that every other real direction turns too, including the separately checked vertical direction.

In contrast, an enlargement centre the origin has every line through the origin invariant: multiplying a position vector by a non-zero scalar keeps it on its line. This is another reason that checking one point is not enough to decide whether an entire line is invariant.

Watch out

Wrong: solve $\mathbf{M}\mathbf{r}=\mathbf{r}$ to find all invariant lines. That finds invariant points only. Right: points may move along an invariant line, so require their image coordinates to satisfy that line's equation.

The next matrix has the same sign pattern as Worked example 4, with a different amount of mixing between coordinates.

Invariant lines away from the origin

To find invariant lines that may miss the origin, keep the intercept in the line equation throughout the calculation.

The origin staying fixed does not mean that every invariant line must pass through it. The opening shear keeps the height of every point unchanged, so every horizontal line is a candidate. Using a general intercept lets you establish the entire family rather than guessing a few members.

General line condition

For the line $y=mx+c$ and matrix $\begin{pmatrix}p&q\\r&s\end{pmatrix}$ :

\begin{aligned} x'&=(p+qm)x+qc\\ y'&=(r+sm)x+sc \end{aligned}

Require $y'=mx'+c$ for every $x$ . Matching the coefficient of $x$ and the constant term gives

r+sm=m(p+qm),\qquad c(s-mq-1)=0

For a vertical line $x=h$ , substitute a general point $(h,y)$ and require $x'=h$ for every $y$ . Derive these conditions from the image coordinates whenever you use them.

Not in the formula booklet: learn it.

The coefficient condition determines possible directions. The constant condition decides which parallel lines in each direction are allowed.

Worked example 5: the opening shear

Question. Find all invariant lines of the transformation with matrix $\mathbf{H}=\begin{pmatrix}1&2\\0&1\end{pmatrix}$ . State which of them consist entirely of invariant points.

Plan. Transform a general point on $y=mx+c$ , compare coefficients, then check every possible vertical line.

\begin{aligned} \begin{pmatrix}x\\mx+c\end{pmatrix}&\mapsto\begin{pmatrix}(1+2m)x+2c\\mx+c\end{pmatrix}&&\text{apply the shear}\\ mx+c&=m\big((1+2m)x+2c\big)+c&&\text{same line equation}\\ m&=m+2m^2,\qquad c=2mc+c&&\text{match coefficients}\\ m&=0,\qquad c\text{ arbitrary}&&\text{both conditions satisfied} \end{aligned}

For $x=h$ , a general point maps as

(h,y)\mapsto(h+2y,y)

The first coordinate is not always $h$ , so no vertical line is invariant.

Why this step? Choosing $y=mx$ at the start would have lost every horizontal line except the $x$ -axis. The intercept carries the information needed for the whole family.

Answer. Exactly the lines $y=c$ , for every real $c$ , are invariant. Only $y=0$ consists entirely of invariant points: on $y=c$ , every point moves horizontally by $2c$ .

Check. Along any $y=c$ , the rule is $(x,c)\mapsto(x+2c,c)$ . As $x$ ranges over all real values, so does $x+2c$ , so the image is the whole original line. If $c\ne0$ , none of its points is fixed.

The next example changes the transformation while retaining the unchanged $y$ -coordinate. A stretch preserves the same horizontal family and gains one vertical invariant line.

Worked example 6: a horizontal family and a vertical line

Question. Find all invariant lines of the transformation with matrix $\mathbf{S}=\begin{pmatrix}3&0\\0&1\end{pmatrix}$ .

Plan. Substitute a general point on $y=mx+c$ , then treat $x=h$ separately.

\begin{aligned} (x,mx+c)&\mapsto(3x,mx+c)&&\text{only }x\text{ is scaled}\\ mx+c&=m(3x)+c&&\text{require }y'=mx'+c\\ m&=3m&&\text{match the }x\text{ coefficients}\\ m&=0,\qquad c\text{ arbitrary}&&\text{all horizontal lines} \end{aligned}

For vertical lines,

\begin{aligned} (h,y)&\mapsto(3h,y)&&\text{general point on }x=h\\ 3h&=h&&\text{require the same vertical line}\\ h&=0&&\text{only the }y\text{-axis} \end{aligned}

Why this step? The equation $y=mx+c$ cannot represent $x=0$ . Omitting the separate test would give an incomplete answer.

Answer. All horizontal lines $y=c$ , together with the vertical line $x=0$ .

Check. On $y=c$ , the first coordinate ranges through every real value under $x\mapsto3x$ . On $x=0$ , every point is actually fixed.

For an invariant line, test a general point. Two isolated examples can suggest the answer but do not establish it algebraically. Likewise, showing that one family is invariant does not prove that it is the only family: solving for all $m,c$ and checking vertical lines supplies that stronger conclusion.

Exam tip

If asked to show that all lines parallel to the $x$ -axis are invariant, use $(x,c)$ with $c$ arbitrary and show that its image has second coordinate $c$ . Using only $(x,0)$ proves a statement about the $x$ -axis alone.

The next matrix changes the vertical scale from $1$ to $2$ . The constant-term condition now removes the shifted horizontal lines.

Watch out

Wrong: every point on an invariant line must satisfy $(x',y')=(x,y)$ . That condition finds fixed points. For the shear, the correct line condition allows $(x,c)\mapsto(x+2c,c)$ : the point can move while remaining on the same line $y=c$ .

Require membership of the same line, then test vertical lines separately.

Quick Recap

Invariant points satisfy $x'=x$ and $y'=y$ ; solve both equations and retain any free coordinate.
An invariant line maps to the same line, while its points may move along it.
For $y=mx+c$ , require $y'=mx'+c$ for a general point and match both coefficients and constants.
Check vertical lines $x=h$ separately; they have no finite gradient.

Yes: under the opening shear, every line $y=c$ is invariant, but if $c\ne0$ , all its points move by $2c$ . The line stays put even though none of its points does.

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Pm6 - Invariant points and lines

Finding points that stay fixed

Lines that keep their direction

Invariant lines away from the origin

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