Matrices that move the plane.

Step 1 of 7

The unit vectors

In the trigonometry Unit a vector was an arrow with components. Now we move the whole plane at once. A linear map of the plane sends every point $(x, y)$⁠ to a new point so that straight lines stay straight and parallel, and the origin stays put. The glowing grid on the board is the plane; the F, mint on top and violet underneath, is a shape with no symmetry, so you can see any flip.

Two arrows are enough to describe such a map: $\hat\imath = (1, 0)$⁠, the pink one, and $\hat\jmath = (0, 1)$⁠, the blue one. Every point is $(x, y) = x\hat\imath + y\hat\jmath$$(x, y) =$$x\hat\imath + y\hat\jmath$⁠, so it is enough to know where they go. Here the map is a turn of $90^\circ$⁠. Predict where $\hat\imath$⁠ will land, then watch.

Step 2 of 7

The rotation matrix

The new places of $\hat\imath$⁠ and $\hat\jmath$⁠ are written as the columns of a $2 \times 2$⁠ matrix:

$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \qquad M\hat\imath = \begin{pmatrix} a \\ c \end{pmatrix}, \quad M\hat\jmath = \begin{pmatrix} b \\ d \end{pmatrix}$$M =$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}, \qquad M\hat\imath =$$\begin{pmatrix} a \\ c \end{pmatrix}, \quad M\hat\jmath =$$\begin{pmatrix} b \\ d \end{pmatrix}$⁠

In a turn by $\theta$⁠, $\hat\imath$⁠ lands on $(\cos\theta, \sin\theta)$⁠, like a point on the unit circle. $\hat\jmath$⁠ starts a quarter turn ahead of it, so it lands on $(-\sin\theta, \cos\theta)$⁠. Here Turn is $30^\circ$⁠. The card in the corner shows $M$⁠ as you drag Apply the transformation: before the map it shows a matrix that moves nothing.

Step 3 of 7

Where any point goes

Once $M$⁠ is known, every point goes to the same mix of its columns:

$M\begin{pmatrix} x \\ y \end{pmatrix} = x\begin{pmatrix} a \\ c \end{pmatrix} + y\begin{pmatrix} b \\ d \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}$$M\begin{pmatrix} x \\ y \end{pmatrix} =$$x\begin{pmatrix} a \\ c \end{pmatrix} + y\begin{pmatrix} b \\ d \end{pmatrix} =$$\begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}$⁠

Here Stretch $2$⁠ stretches $\hat\imath$⁠ to $(2, 0)$⁠, and Shear $0.5$⁠ leans $\hat\jmath$⁠ over to $(0.5, 1)$⁠. The unit square, whose far corner is $(1, 1)$⁠, becomes a parallelogram. Work out where that corner lands.

Step 4 of 7

Shear

A shear slides every row of the plane sideways, more the higher it is, like a deck of cards pushed into a slant. Here Shear is $1$⁠: $\hat\imath$⁠ stays put, and $\hat\jmath$⁠ goes to $(1, 1)$⁠. Predict the area of the parallelogram the unit square becomes.

Step 5 of 7

Build a matrix

You can also build a matrix from what you want it to do. Stretch, Shear and Turn set $M$⁠, and the card shows it. Here Shear is locked at $0$⁠; build a map that sends $\hat\imath$⁠ to a given place.

Step 6 of 7

$\hat\jmath$⁠ after a turn

The second column of $M$⁠ is where the blue arrow, $\hat\jmath$⁠, lands.

Step 7 of 7

The diagonal

Every point is a mix of $\hat\imath$⁠ and $\hat\jmath$⁠, so it lands on the same mix of the columns.