Home / A-Level Further Maths 9231 / Matrices / Invariant points and lines
Cambridge International AS & A Level · Topic 1.4
A-Level Further Maths 9231 / Topic 1.4 / AS & A Level

What a transformation leaves alone.

Most points move when a matrix acts on the plane, but some stay exactly where they are, and some whole lines are carried onto themselves. Find them with one equation, \(\mathbf{M}\mathbf{x} = \mathbf{x}\) for points and \(Y = mX\) for lines, and the geometry of the transformation falls out.

Topic 1.4All matrices lessons →
The key idea

A point is invariant when \(\mathbf{M}\mathbf{x} = \mathbf{x}\): it does not move. The origin is always invariant; if \(\det(\mathbf{M} - \mathbf{I}) = 0\) (and \(\mathbf{M} \ne \mathbf{I}\)) there is a whole line of invariant points through the origin. A line is an invariant line when every point on it maps to a point on the same line, not necessarily to itself. For \(y = mx\), map \((x, mx)\) and demand that the image \((X, Y)\) satisfies \(Y = mX\); this gives a quadratic in \(m\). For \(y = mx + c\), demand \(Y = mX + c\) for every \(x\) and compare coefficients. A line of invariant points is always an invariant line, but an invariant line need not be a line of invariant points.

x y O 2 4 6 2 −2 P P′ Q = Q′ line of invariant points: y = −x invariant line: y = x/2
Fig. 1: the matrix M with rows (3, 2) and (1, 2), which maps (x, y) to (3x + 2y, x + 2y). Every point on y = -x stays put, such as Q(-1.5, 1.5). The line y = x/2 is invariant but its points move along it: P(1, 0.5) maps to P'(4, 2)
Section 01

Points that do not move.

A point with position vector \(\mathbf{x}\) is an invariant point of the transformation with matrix \(\mathbf{M}\) when \(\mathbf{M}\mathbf{x} = \mathbf{x}\). Write \(\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}\), multiply out and set each component equal to the original. You get two linear equations through the origin, so the origin is always invariant, and the only question is whether anything else is.

For \(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), the equation \(\mathbf{M}\mathbf{x} = \mathbf{x}\) is the same as \((\mathbf{M} - \mathbf{I})\mathbf{x} = \mathbf{0}\):

\[(a - 1)x + by = 0, \qquad cx + (d - 1)y = 0.\]
ConditionInvariant points
\(\det(\mathbf{M} - \mathbf{I}) \ne 0\)only the origin (the two lines through \(O\) cross only there)
\(\det(\mathbf{M} - \mathbf{I}) = 0\), \(\mathbf{M} \ne \mathbf{I}\)a whole line through \(O\): both equations describe the same line
\(\mathbf{M} = \mathbf{I}\)every point of the plane
Worked example A line of invariant points
Find the invariant points of the transformation with matrix \(\mathbf{M} = \begin{pmatrix} 3 & 4 \\ 1 & 3 \end{pmatrix}\).
  1. \(\begin{pmatrix} 3 & 4 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\) gives \(3x + 4y = x\) and \(x + 3y = y\).
  2. Simplify: \(2x + 4y = 0\) and \(x + 2y = 0\).
  3. The first is twice the second, so they are the same line: \(x + 2y = 0\).
  4. Check with \((-2, 1)\): \(\mathbf{M}\begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} -6 + 4 \\ -2 + 3 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}\).
Answer: Every point on the line \(x + 2y = 0\), that is \(y = -\tfrac{1}{2}x\), is invariant
Worked example Choosing a constant
\(\mathbf{M} = \begin{pmatrix} k & 2 \\ 3 & 4 \end{pmatrix}\). Find the value of \(k\) for which \(\mathbf{M}\) has a line of invariant points, and find that line.
  1. A line of invariant points needs \(\det(\mathbf{M} - \mathbf{I}) = 0\), with \(\mathbf{M} - \mathbf{I} = \begin{pmatrix} k - 1 & 2 \\ 3 & 3 \end{pmatrix}\).
  2. \(3(k - 1) - 6 = 0\), so \(k = 3\).
  3. With \(k = 3\): \(3x + 2y = x\) gives \(2x + 2y = 0\), and \(3x + 4y = y\) gives \(3x + 3y = 0\).
  4. Both reduce to \(x + y = 0\).
Answer: \(k = 3\); the line of invariant points is \(y = -x\)
Your turn
Find the invariant points of the transformation with matrix \(\begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}\).
Examiner trap

Invariant points solve \(\mathbf{M}\mathbf{x} = \mathbf{x}\), not \(\mathbf{M}\mathbf{x} = \mathbf{0}\). The test is \(\det(\mathbf{M} - \mathbf{I}) = 0\), not \(\det \mathbf{M} = 0\): subtract 1 from the diagonal entries before you take the determinant. And give a line of invariant points as an equation, \(x + 2y = 0\), not as one point.

Section 02

Lines through the origin that map to themselves.

A line is invariant when the image of every point on it lies on the same line. The points may slide along the line, as \(P\) does in Fig. 1; they need not stay still. To find the invariant lines \(y = mx\), take a general point \((x, mx)\), find its image \((X, Y)\), and require \(Y = mX\).

For \(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\): \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} (a + bm)x \\ (c + dm)x \end{pmatrix}\). Requiring \(Y = mX\) for every \(x\):

\[c + dm = m(a + bm) \quad\Longrightarrow\quad bm^{2} + (a - d)m - c = 0.\]
Worked example The matrix in Fig. 1
Find the invariant lines through the origin for \(\mathbf{M} = \begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix}\), and decide whether either is a line of invariant points.
  1. \(\begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} (3 + 2m)x \\ (1 + 2m)x \end{pmatrix}\).
  2. Require \(Y = mX\): \(1 + 2m = m(3 + 2m)\), so \(2m^{2} + m - 1 = 0\).
  3. \((2m - 1)(m + 1) = 0\), so \(m = \tfrac{1}{2}\) or \(m = -1\). Since \(b = 2 \ne 0\), \(x = 0\) is not invariant.
  4. \(y = -x\): \((1, -1) \mapsto (3 - 2, 1 - 2) = (1, -1)\), so every point stays put. \(y = \tfrac{1}{2}x\): \((2, 1) \mapsto (8, 4)\), so points move along the line, four times as far from \(O\).
Answer: \(y = \tfrac{1}{2}x\) (an invariant line) and \(y = -x\) (a line of invariant points)
Worked example Points that flip through the origin
Find the invariant lines through the origin for \(\mathbf{M} = \begin{pmatrix} 1 & 4 \\ 1 & 1 \end{pmatrix}\).
  1. \((x, mx) \mapsto \big((1 + 4m)x,\ (1 + m)x\big)\).
  2. Require \(1 + m = m(1 + 4m)\): \(4m^{2} - 1 = 0\), so \(m = \pm\tfrac{1}{2}\).
  3. \(y = \tfrac{1}{2}x\): \((2, 1) \mapsto (6, 3)\), three times as far out.
  4. \(y = -\tfrac{1}{2}x\): \((2, -1) \mapsto (-2, 1)\). The point jumps to the other side of \(O\), but it is still on the line.
Answer: \(y = \tfrac{1}{2}x\) and \(y = -\tfrac{1}{2}x\); neither is a line of invariant points
Your turn
Find all the invariant lines through the origin for \(\begin{pmatrix} 2 & 0 \\ 3 & -1 \end{pmatrix}\).
Examiner trap

Do not mix up an invariant line with a line of invariant points: in Example 1 the line \(y = \tfrac{1}{2}x\) is invariant although \(P\) moves from \((1, 0.5)\) to \((4, 2)\). And watch the order of terms in \(bm^{2} + (a - d)m - c = 0\); rederive it from \(Y = mX\) each time rather than trusting a half-remembered formula.

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A point with position vector \(\mathbf{x}\) is an invariant point of the transformation with matrix \(\mathbf{M}\) when

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Find the invariant points of \(\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\).

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Which statement is true?

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For \(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), the line \(y = mx\) is invariant when

Section 03

Lines that miss the origin, and the geometry.

An invariant line need not pass through \(O\). For \(y = mx + c\), map the general point \((x, mx + c)\) to \((X, Y)\) and require \(Y = mX + c\) for every value of \(x\). An identity in \(x\) means you compare the coefficients of \(x\) and the constant terms; the \(x\)-terms give the same quadratic in \(m\) as before, and the constants decide which values of \(c\) are allowed.

The standard transformations have invariant points and lines you can predict before calculating:

TransformationInvariant pointsInvariant lines
Reflection in a line \(l\) through \(O\)every point of \(l\)\(l\), and every line perpendicular to \(l\)
Shear with the \(x\)-axis invariantevery point of the \(x\)-axisevery line \(y = c\) (parallel to the invariant axis)
Stretch parallel to the \(x\)-axis, factor \(k \ne 1\)every point of the \(y\)-axisthe \(y\)-axis and every line \(y = c\)
Enlargement, centre \(O\), factor \(k \ne 1\)only \(O\)every line through \(O\)
Rotation about \(O\), not through a multiple of 180°only \(O\)none

For example, the reflection in \(y = x\) maps \((x, -x + c)\) to \((-x + c, x)\), which lies on \(y = -x + c\) again: each point swaps to the other side of the mirror along the same perpendicular line.

Worked example Every line of gradient 1
Show that every line with gradient 1 is invariant under the transformation with matrix \(\begin{pmatrix} 4 & -3 \\ 3 & -2 \end{pmatrix}\).
  1. Take the line \(y = x + c\) and the point \((x, x + c)\) on it.
  2. Image: \(X = 4x - 3(x + c) = x - 3c\) and \(Y = 3x - 2(x + c) = x - 2c\).
  3. Then \(Y - X = (x - 2c) - (x - 3c) = c\), so \(Y = X + c\).
  4. The image lies on \(y = x + c\), the same line, whatever the values of \(x\) and \(c\).
Answer: Every line \(y = x + c\) is invariant (and \(y = x\) itself is a line of invariant points, since \(c = 0\) gives \(X = x,\ Y = x\))
Worked example All the invariant lines of one matrix
Find all the invariant lines of the transformation with matrix \(\mathbf{M} = \begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix}\) (Fig. 1).
  1. \((x, mx + c) \mapsto (X, Y)\) with \(X = (3 + 2m)x + 2c\) and \(Y = (1 + 2m)x + 2c\).
  2. Require \(Y = mX + c\) for all \(x\): \((1 + 2m)x + 2c = m(3 + 2m)x + 2mc + c\).
  3. Coefficients of \(x\): \(2m^{2} + m - 1 = 0\), so \(m = \tfrac{1}{2}\) or \(m = -1\). Constants: \(2c = 2mc + c\), so \(c(1 - 2m) = 0\).
  4. \(m = \tfrac{1}{2}\): \(c(1 - 1) = 0\) holds for every \(c\). \(m = -1\): \(3c = 0\), so \(c = 0\).
Answer: \(y = -x\) (a line of invariant points) and every line \(y = \tfrac{1}{2}x + c\)
Your turn
The matrix \(\begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}\) is a stretch parallel to the \(x\)-axis. Find its invariant points and show that every line \(y = c\) is invariant.
Examiner trap

When you compare coefficients you must use BOTH equations. The \(x\)-terms alone give only the gradients; it is the constant terms that decide \(c\). For the matrix in Fig. 1, \(m = -1\) forces \(c = 0\), so \(y = -x + 5\) is NOT invariant: \((0, 5) \mapsto (10, 10)\), which is not on \(y = -x + 5\).

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Find the invariant lines through the origin for the transformation with matrix \(\begin{pmatrix} 5 & 2 \\ 2 & 2 \end{pmatrix}\).

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\(\mathbf{M} = \begin{pmatrix} k & 3 \\ 1 & 2 \end{pmatrix}\) has a line of invariant points. Find \(k\).

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Which of these lines is invariant under the reflection in the line \(y = x\)?

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Six original Cambridge-style questions
Lines of invariant points, a matrix with only the origin fixed, invariant lines through the origin, an unknown entry, a family of parallel invariant lines, and a proof for a shear.