Three unit arrows
Space has three unit arrows: i = (1, 0, 0) along the x-axis, j = (0, 1, 0) along the y-axis and k = (0, 0, 1) along the z-axis. Every point is made of them: (2, 1, 3) is 2i + j + 3k, two steps along x, one along y and three up z.
A 3 × 3 matrix moves a point when the point, written as a column, is multiplied by it. The reason the columns matter is the same as in the plane. Multiply any matrix by the column for i and the answer is its first column; j gives the second column, and k gives the third. So the three columns of a 3 × 3 matrix are where i, j and k land, and every other point follows them: (x, y, z) lands on x times column 1, plus y times column 2, plus z times column 3.
The identity matrix I = (1 0 0; 0 1 0; 0 0 1) has columns i, j and k themselves, so it moves nothing.
The point (2, 1, 3) is the far corner of a box that runs 2 along the x-axis, 1 along the y-axis and 3 up the z-axis: the arrow from the origin is 2i + j + 3k.
A turn about the z-axis
Take R = (0 −1 0; 1 0 0; 0 0 1). Its first column is (0, 1, 0), so it sends i to j: a quarter turn, just as in the plane. Its second column is (−1, 0, 0), so it sends j to −i. Its third column is (0, 0, 1), which is k itself, so k does not move.
Multiply a point to see the whole effect. For the column (2; 1; 3), row 1 of R gives 0 × 2 + (−1) × 1 + 0 × 3 = −1, row 2 gives 1 × 2 + 0 × 1 + 0 × 3 = 2, and row 3 gives 0 × 2 + 0 × 1 + 1 × 3 = 3, so R sends (2, 1, 3) to (−1, 2, 3). In general R sends (x, y, z) to (−y, x, z). The first two coordinates are turned exactly as the plane's quarter turn (0 −1; 1 0) turns them, which is the block in the top-left corner of R, and the height z is left alone.
So R turns space a quarter turn about the z-axis, counterclockwise as seen from above. The points of the z-axis itself, (0, 0, z), are the only invariant points: they form the axle the rest of space turns around.
Looking down the z-axis from above, R turns the floor plan of a box a quarter turn about the origin, where the z-axis passes through. The corner (3, 0) goes to (0, 3); every height stays the same.
Turns about the other axes
A turn about the x-axis leaves i alone and turns j and k. The quarter turn that sends j to k sends k to −j, so its columns are (1, 0, 0), (0, 0, 1) and (0, −1, 0), and its matrix is (1 0 0; 0 0 −1; 0 1 0). The column that is a unit arrow left in place tells you the axis.
A reflection in a plane
The matrix M = (1 0 0; 0 1 0; 0 0 −1) leaves i and j where they are and sends k to −k, its opposite. It sends (x, y, z) to (x, y, −z): every point keeps its position across the floor and moves to the same distance on the other side of it. That is the reflection in the plane z = 0.
Its invariant points satisfy −z = z, so z = 0. The whole plane z = 0 is a plane of invariant points, as the mirror line was in the plane.
In the same way, (−1 0 0; 0 1 0; 0 0 1) reflects space in the plane x = 0, and (1 0 0; 0 −1 0; 0 0 1) reflects it in the plane y = 0.
Seen from the side, with x across the page and z up the page, M reflects a box standing on the plane z = 0 to hang below it. The base stays where it is, and the top at height 2 goes to height −2.
An enlargement of space
The matrix E = (2 0 0; 0 2 0; 0 0 2) has columns 2i, 2j and 2k, so it sends all three unit arrows twice as far. It sends (x, y, z) to (2x, 2y, 2z), the point twice as far from the origin in the same direction: an enlargement of space with scale factor 2, centered on the origin.
It sends the unit cube to a cube with edges 2, which holds 2 × 2 × 2 = 8 unit cubes. A matrix with different numbers down the diagonal, such as (3 0 0; 0 2 0; 0 0 2), stretches each direction by its own factor.
E doubles every edge, so the unit cube becomes a 2 by 2 by 2 cube made of eight unit cubes.
One transformation after another
Reflecting in z = 0 first and then turning about the z-axis is the product R M: the first transformation stands next to the point, and the product reads from the right, as in the plane.
Follow the unit arrows to find its columns. M leaves i alone and R sends it to j, so column 1 is (0, 1, 0). M leaves j alone and R sends it to −i, so column 2 is (−1, 0, 0). M sends k to −k and R leaves that alone, so column 3 is (0, 0, −1). So R M = (0 −1 0; 1 0 0; 0 0 −1), which is the same as multiplying the rows of R into the columns of M.
Check it on a point. M sends (2, 1, 3) to (2, 1, −3), and R turns that to (−1, 2, −3). R M sends (2, 1, 3) to (−1, 2, −3) in one step.
The product R M: its third column, (0, 0, −1), is where k lands after the reflection and then the turn.
When the order matters
This pair happens to commute: M R is also (0 −1 0; 1 0 0; 0 0 −1), because the turn changes only x and y and the reflection changes only z. Most pairs do not.
Take F = (−1 0 0; 0 1 0; 0 0 1), the reflection in the plane x = 0. Reflecting first and then turning, R F sends i to −i and then to −j, so its first column is (0, −1, 0). Turning first and then reflecting, F R sends i to j, which F leaves alone, so its first column is (0, 1, 0). The two products are different, so the two orders move space differently.
The usual mistakes
Reading a row as the image of an arrow. The first row of R is (0, −1, 0), but i lands on the first column, (0, 1, 0).
Writing the third column as zeros. k is not sent to nothing by a turn about the z-axis; it stays as k, so the third column is (0, 0, 1). With a column of zeros, every point would drop onto the floor z = 0.
Using (0 1 0; −1 0 0; 0 0 1) for the counterclockwise turn. That sends i to −j, which is the clockwise quarter turn.
Writing the product in the order things happen. Reflecting with M and then turning with R is R M, with M next to the point.
A crate on a turntable
In the application below, a crate on a turntable is turned a quarter turn about its upright axle, the z-axis. The matrix is built column by column from where i, j and k go, a corner of the crate is multiplied by it, and the invariant points turn out to be the axle.
Worked example: A Crate on a Turntable Turned Through a Quarter Turn About Its Upright Axle
Question A crate 2 m long, 1 m wide and 1 m high stands on a turntable. It fills the box from (2, 0, 0) to P(4, 1, 1), in meters, and the turntable's axle is the z-axis, pointing straight up. The turntable turns through 90° counterclockwise, seen from above. (a) Write down the matrix R of the turn, and find the image P' of the corner P. (b) Find det R, and say what it tells you about the crate's volume. Find the invariant points of R.
1.Seen from above, the quarter turn sends i to j and sends j to −i, and k points along the axle and does not move. These images are the columns: R = 0−10100001.
i goes to j, j to −i, and k stays: R = 0−10100001. 2.(a) R411 = 0 × 4 − 1 × 1 + 01 × 4 + 0 + 00 + 0 + 1 × 1 = −141, so P' is (−1, 4, 1). The corner has swung round the axle and kept its height of 1 m.
(a) R411 = −141: the corner swings round the axle and keeps its height of 1 m. 3.Expand the determinant along the third row, whose first two entries are 0: det R = 1 × (0 × 0 − (−1) × 1) = 1. So the turn multiplies every volume by 1, and the crate keeps its volume of 2 × 1 × 1 = 2 cubic meters.
Along the third row, det R = 1 × (0 × 0 − (−1) × 1) = 1: the crate keeps its volume of 2 cubic meters. 4.An invariant point satisfies Rxyz = xyz, so −y = x, x = y and z = z. Together the first two give x = −x, so x = 0 and then y = 0, while z can be any number.
Rp = p gives −y = x and x = y, so x = y = 0, and z can be any number. 5.(b) det R = 1, so the crate's volume of 2 cubic meters is unchanged. The invariant points are the points (0, 0, z) of the axle: they are the only points the turn does not move.
(b) det R = 1, so the volume is unchanged, and the invariant points are the points (0, 0, z) of the axle.
Answer: (a) R = 0−10100001 and P'(−1, 4, 1); (b) det R = 1, so the crate keeps its volume of 2 cubic meters, and the invariant points are the points (0, 0, z) on the axle
Common mistakes
- Leaving out the third column, or writing it as a column of zeros. k is not sent to nothing: it stays as k, so the third column is 001, and without it every point would drop to the floor.
- Using 010−100001, which turns the crate clockwise and sends P to (1, −4, 1). The image of i under a counterclockwise turn is j, so the first column is 010.
More matrices as transformations problems, worked step by step →