Points that do not move
A point is invariant under a transformation when the transformation sends it back to exactly where it was. For a matrix M, the invariant points are the points (x, y) with M (x; y) = (x; y).
Multiplying out gives two equations in x and y, and the invariant points are their solutions. The origin is always one of them, because a matrix never moves the origin. The question is which other points stay.
The reflection in the x-axis
The reflection in the x-axis has matrix M = (1 0; 0 −1), and it sends (x, y) to (x, −y). So M (x; y) = (x; y) says (x, −y) = (x, y).
Compare the coordinates one at a time. The first coordinates give x = x, which is always true, so x can be any number. The second coordinates give −y = y, so 2y = 0 and y = 0.
The invariant points are all the points (x, 0): every point on the x-axis, the mirror line itself, stays where it is.
The triangle with base from (−2, 0) to (2, 0) and top corner (0, 2) is reflected in the x-axis. The base lies on the mirror line and does not move; the top corner goes to (0, −2).
Every point off the mirror line is thrown across it to the other side, the same distance away. A point at height 2 lands at height −2, so it is not invariant. Only the points of the mirror line are left untouched.
Invariant lines
A line is invariant when the transformation sends every point of the line to a point of the same line. The points may move, as long as they stay on the line.
Under the reflection in the x-axis, a point (3, y) of the vertical line x = 3 goes to (3, −y), which is on the line x = 3 again. So x = 3 is an invariant line. Its points slide along it: (3, 2) goes to (3, −2). The only point of it that stays put is (3, 0), where it crosses the mirror.
This gives two different things to look for. A line of invariant points is a line on which every point stays where it is, like the x-axis here. An invariant line only has to land on itself, like every vertical line here.
A matrix whose answer is not obvious
Take N = (3 −2; 1 0). It sends (x, y) to (3x − 2y, x). For an invariant point, 3x − 2y = x and x = y. The first equation gives 2x = 2y, which is y = x again, so the two equations agree, and every point on the line y = x is invariant. Check one: N sends (1, 1) to (3 − 2, 1) = (1, 1).
Now look for invariant lines through the origin, y = mx. The point (x, mx) goes to (3x − 2mx, x). It lands on the line y = mx when x = m(3x − 2mx). Divide by x: , so , which factors as (2m − 1)(m − 1) = 0. So m = 1 or .
The line y = x is invariant, as it must be: it is the line of invariant points. On the line , the point (2, 1) goes to (3 × 2 − 2 × 1, 2) = (4, 2), which is on the same line, twice as far from the origin. That line is invariant, but its points move out along it.
The two invariant lines of (3 −2; 1 0): y = x, the steeper one, and . The dot at (1, 1) on y = x does not move. The point P at (2, 1) on moves along its line to P' at (4, 2).
Rotations
The quarter turn R = (0 −1; 1 0) sends (x, y) to (−y, x). For an invariant point, −y = x and x = y. Put the second into the first: −x = x, so x = 0, and then y = 0. Only the origin is invariant.
Every rotation about the origin through an angle that is not a whole number of turns fixes the origin and nothing else, because every other point swings round the center to somewhere new.
Lines behave differently for different angles. The quarter turn sends every line through the origin to the line at right angles to it, so it has no invariant lines at all. The half turn, (−1 0; 0 −1), sends (x, y) to (−x, −y), so every line through the origin is invariant: each point jumps to the other side of the origin along its own line.
A quarter turn about the origin. The corner at the origin stays where it is; the corner (2, 0) goes to (0, 2), and (2, 1) goes to (−1, 2).
The usual mistakes
Solving M x = 0. That finds the points sent to the origin. Invariant points solve M x = x.
Calling an invariant line a line of invariant points. Under the reflection in the x-axis, the line x = 3 is invariant, but (3, 2) moves to (3, −2).
Solving for m and keeping only one root. For (3 −2; 1 0), has two roots, and both lines y = x and are invariant.
Thinking a rotation fixes the points on an axis. The quarter turn sends the x-axis onto the y-axis, and only the origin stays.
A stack of cards and a piece of fabric
In the applications below, a stack of cards is pushed sideways by a shear. The table line is the line of invariant points, and every horizontal line is an invariant line whose points slide along it. Then a square of stretchy fabric is pinned along one diagonal, which is its line of invariant points, and pulled along the other, which is an invariant line found from y = mx.
Worked example: A Stack of Cards Pushed Sideways, and the Table Line That Does Not Move
Question Seen from the side, a neat stack of cards is a rectangle with corners A(1, 0), B(5, 0), C(5, 3) and D(1, 3), in centimeters, with the table along the x-axis. A hand pushes the stack so that each card slides sideways by its height above the table, which is the shear S = 1101. (a) Find the corners of the pushed stack, and its area as seen from the side. (b) Find the invariant points of S, and show that every line y = c is an invariant line.
1.Sxy = x + yy: each point slides sideways by its height y and keeps its height.
Sxy = x + yy: each point slides sideways by its height. 2.(a) A and B are on the table, where y = 0, so they stay at (1, 0) and (5, 0). C(5, 3) goes to C'(5 + 3, 3) = (8, 3), and D(1, 3) goes to D'(1 + 3, 3) = (4, 3).
(a) A and B stay on the table; C goes to C'(8, 3) and D to D'(4, 3). 3.The pushed stack is a parallelogram with base AB = 4 cm and height 3 cm, so its area is 4 × 3 = 12 square centimeters, the same as the rectangle's. This agrees with det S = 1 × 1 − 1 × 0 = 1: the shear keeps every area.
The pushed stack has base 4 cm and height 3 cm: area 12 square centimeters, as det S = 1 says. 4.An invariant point satisfies Sxy = xy, so x + y = x and y = y. The first equation gives y = 0, so every point on the table line y = 0 is an invariant point.
x + y = x gives y = 0: every point on the table line is an invariant point. 5.(b) A point (x, c) on the line y = c goes to (x + c, c), which is on the same line, so every line y = c is an invariant line: each card slides along its own level. Only on y = 0 does every point stay where it is.
(b) (x, c) goes to (x + c, c) on the same line: every line y = c is invariant, and only on y = 0 do the points stay put.
Answer: (a) A(1, 0), B(5, 0), C'(8, 3) and D'(4, 3), and the area is still 12 square centimeters; (b) the invariant points are the points of the table line y = 0, and every line y = c is an invariant line, though only y = 0 is a line of invariant points
Common mistakes
- Calling the top line y = 3 a line of invariant points because it is an invariant line. Its points all move 3 cm to the right; the line is invariant, but only the points of y = 0 stay put.
- Using the slanted side AD', of length √32 + 32, as the height of the parallelogram. The height is the distance between the parallel sides, which is still 3 cm, so the area is still 12 square centimeters.
More matrices as transformations problems, worked step by step →
Worked example: A Square of Stretchy Fabric Pinned Along One Diagonal and Pulled Along the Other
Question A square of stretchy fabric has corners P(1, 1), Q(−1, 1), R(−1, −1) and S(1, −1), in decimeters from its center. It is pinned along the diagonal through Q and S and pulled out along the other diagonal, which is the transformation M = 2112. (a) Find the images of the four corners, and find the line of invariant points of M. (b) Find both invariant lines of M through the origin, one of which you have already met in (a), in the form y = mx, and find the area of the stretched fabric.
1.Multiply each corner by M: M11 = 33, M−1−1 = −3−3, M−11 = −2 + 1−1 + 2 = −11 and M1−1 = 1−1. So P' is (3, 3) and R' is (−3, −3), while Q and S do not move.
P(1, 1) goes to P'(3, 3) and R to R'(−3, −3), while Q and S do not move. 2.An invariant point satisfies Mxy = xy, so 2x + y = x and x + 2y = y. Both equations simplify to x + y = 0.
Mp = p gives 2x + y = x and x + 2y = y, which both simplify to x + y = 0. 3.(a) The line of invariant points is y = −x, the pinned diagonal through Q and S.
(a) The line of invariant points is y = −x, the pinned diagonal through Q and S. 4.A point (x, mx) on the line y = mx goes to (2x + mx, x + 2mx). The image is on the same line when x + 2mx = m(2x + mx). Divide by x: 1 + 2m = 2m + m2, so m2 = 1.
A point (x, mx) goes to (2x + mx, x + 2mx), on the line again when 1 + 2m = m(2 + m), so m2 = 1. 5.So m = 1 or m = −1. On y = x each point moves three times as far from the center, as P(1, 1) goes to (3, 3), so the line is invariant but its points move along it. On y = −x no point moves at all.
m = 1 or m = −1: on y = x the points move three times as far out, and on y = −x none moves. 6.(b) The invariant lines through the origin are y = x and y = −x. det M = 2 × 2 − 1 × 1 = 3, so the fabric's area of 2 × 2 = 4 square decimeters becomes 3 × 4 = 12 square decimeters. Check: the diagonals of the stretched fabric are 6√2 and 2√2, and 12 × 6√2 × 2√2 = 12.
(b) The invariant lines through the origin are y = x and y = −x, and det M = 3 takes the area from 4 to 12 square decimeters.
Answer: (a) P'(3, 3), Q'(−1, 1), R'(−3, −3) and S'(1, −1), and the line of invariant points is y = −x; (b) the invariant lines through the origin are y = x and y = −x, and the stretched fabric has an area of 12 square decimeters
Common mistakes
- Stopping at x + y = 0 and giving y = −x as the only invariant line. That is the line of invariant points; y = x is also invariant, because every point on it lands on it again, three times as far out.
- Solving m2 = 1 as m = 1 only. A square number has two square roots, and the root m = −1 is the pinned diagonal itself.
More matrices as transformations problems, worked step by step →