4.03g - Invariant points and lines

You already know

A matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ maps $(x,y)$ to $(ax+by,cx+dy)$. A linear transformation fixes the origin. A line may be written $y=mx+k$, except that vertical lines need the form $x=h$.

Invariant point and invariant line

An invariant point maps to itself. A line of invariant points has every point fixed. An invariant line maps onto itself as a whole; individual points may move to other positions on it. For example, under reflection in the $x$-axis, $y=0$ is a line of invariant points and every vertical line is an invariant line. A line mapped to a different parallel line is not invariant.

Key idea

Fixed point: the same coordinates. Invariant line: the same line. A line of invariant points satisfies both conditions.

Invariant-point equations

For $\mathbf{M}=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, invariant points satisfy

Both equations must hold. Not in the formula booklet: learn it.

Worked example 1: a line of invariant points

Question. Find all invariant points under $\mathbf{A}=\begin{pmatrix}2&1\\1&2\end{pmatrix}$.

Plan. Equate both image coordinates to the original coordinates, then solve their common condition.

Why this step? The two equations are the same condition, so choosing one point would miss infinitely many others.

Answer. Every point on $y=-x$, and no other point, is invariant.

Check. $\mathbf{A}\begin{pmatrix}t\\-t\end{pmatrix}=\begin{pmatrix}2t-t\\t-2t\end{pmatrix}=\begin{pmatrix}t\\-t\end{pmatrix}$ for every real $t$.

Worked example 2: only the origin

Question. Find all invariant points under $\mathbf{B}=\begin{pmatrix}2&1\\1&3\end{pmatrix}$.

Plan. Form the same two fixed-point equations and solve them simultaneously.

Why this step? Satisfying just the first equation gives a candidate line; the second equation removes every non-zero point on it.

Answer. The origin $(0,0)$ is the only invariant point.

Check. $(1,-1)$, which was fixed in the first example, now maps to $(1,-2)$, so it no longer works.

Watch out

Wrong: solving $2x+y=x$ and concluding that all points on $y=-x$ are fixed without checking the second coordinate. Right: impose both $x'=x$ and $y'=y$.

Testing a line through the origin

A non-vertical line $y=mx$ is invariant under $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ if

The non-zero condition ensures that the whole line maps onto a line, rather than collapsing to the origin. Test the vertical line $x=0$ separately: it is invariant if $b=0$ and $d\ne0$. Not in the formula booklet: learn it.

Worked example 3: different stretches along the axes

Question. Find all invariant lines through $O$ for $\mathbf{D}=\begin{pmatrix}2&0\\0&3\end{pmatrix}$.

Plan. Transform $(x,mx)$, compare with $y'=mx'$, then check the vertical line.

Why this step? The form $y=mx$ never includes the $y$-axis, so omitting a separate vertical check would lose a solution.

Answer. The two coordinate axes, $y=0$ and $x=0$.

Check. Points on each axis move along it; neither factor $2$ nor $3$ is zero, so its whole image is that axis.

Worked example 4: a tilted invariant line

Question. Find all invariant lines through $O$ for $\mathbf{E}=\begin{pmatrix}2&1\\0&3\end{pmatrix}$.

Plan. Transform a general point on $y=mx$ and compare coefficients, then test the vertical direction.

Why this step? The condition compares whole lines. For $m=1$, points are multiplied by $3$ along the line, so they move but stay on it.

Answer. $y=0$ and $y=x$, with no other lines through $O$.

Check. $(t,0)\mapsto(2t,0)$, and $(t,t)\mapsto(3t,3t)$; both give their full lines as $t$ varies.

Watch out

Wrong: choosing the origin as the only test point on a proposed invariant line. Every linear transformation fixes $O$, so this says nothing about its direction. Use a general point or a non-zero direction vector, then check that the line does not collapse.

Testing all straight lines

For a non-vertical line $y=mx+k$, transform a general point and impose

For a vertical line $x=h$, require $x'=h$ for every $y$. In either case ensure that the image covers the whole line. These tests are not in the formula booklet: learn them.

Worked example 5: a family of parallel invariant lines

Question. Find all invariant lines under $\mathbf{S}=\begin{pmatrix}2&0\\0&1\end{pmatrix}$.

Plan. Test $y=mx+k$, then test vertical lines separately.

Why this step? The intercept cancels here, so every horizontal line survives, not just the $x$-axis.

Answer. Every horizontal line $y=k$, where $k$ is real, and the $y$-axis $x=0$.

Check. $(t,k)\mapsto(2t,k)$ covers the same horizontal line. Every $(0,t)$ is fixed, so $x=0$ is also a line of invariant points.

Worked example 6: the intercept must be zero

Question. Find all invariant lines under $\mathbf{D}=\begin{pmatrix}2&0\\0&3\end{pmatrix}$.

Plan. Use the same general line and compare both coefficients and constants.

Why this step? Matching the slope alone would admit all horizontal lines; matching the constant term removes every horizontal line except the axis.

Answer. Only $y=0$ and $x=0$.

Check. $y=1$ maps to $y=3$, so it is not invariant despite keeping its direction. Both coordinate axes map onto themselves.

Watch out

Wrong: finding invariant slopes and stopping when the question asks for all invariant lines. A slope specifies a direction, not an intercept. Test $k$, and include vertical lines $x=h$.