Matrices as Functions

You have used matrices to move the plane. Now treat each one as a function: its inputs are vectors and its outputs are vectors. Everything you know about functions, composition and inverses carries over, with matrix multiplication doing the work.

Matrices as functions

A $2 \times 2$ matrix $A$ defines the function $\mathbf{v} \to A\mathbf{v}$. Its domain is every vector in the plane, and its outputs are the vectors $A\mathbf{v}$. Different matrices are different functions; the identity $I$ is the function that changes nothing.

Composition: one matrix for two moves

Apply $B$ to $\mathbf{v}$, then apply $A$ to the result. The output is $A(B\mathbf{v})$, and because matrix multiplication can be grouped either way, this equals $(AB)\mathbf{v}$. So the composition of linear transformations "$B$, then $A$" is the single matrix $AB$. The move done first is written on the right, closest to $\mathbf{v}$, just as $g$ is applied first in $f(g(x))$.

Composing rotations adds their angles: rotating by $\frac{\pi}{6}$ three times is the same as rotating by $\frac{\pi}{2}$ once. Areas multiply too: $\det(AB) = \det A \cdot \det B$, since $B$ scales areas by $\left| \det B \right|$ and then $A$ scales them again.

Inverses: undoing a move

If $\det A \ne 0$, the inverse matrix $A^{-1}$ is the inverse function: $A^{-1}(A\mathbf{v}) = \mathbf{v}$ for every $\mathbf{v}$. To find which input gives a known output $\mathbf{w}$, solve $A\mathbf{v} = \mathbf{w}$ by computing $\mathbf{v} = A^{-1}\mathbf{w}$.

Some inverses are easy to see. A rotation by $\theta$ is undone by a rotation by $-\theta$, and a reflection is undone by itself.

When there is no inverse

If $\det A = 0$, the matrix squashes the plane onto a line (or a point), and different inputs land on the same output. For $P = \left[ \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right]$, $P\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] = \left[ \begin{matrix} 2 \\ 0 \end{matrix} \right]$ and $P\left[ \begin{matrix} 2 \\ -1 \end{matrix} \right] = \left[ \begin{matrix} 2 \\ 0 \end{matrix} \right]$. Given only the output $\left\langle 2, 0 \right\rangle$, there is no way to know which input it came from, so $P$ has no inverse function.

Repeating a move

For $M = \left[ \begin{matrix} 3 & 1 \\ 5 & 2 \end{matrix} \right]$, $M^2 = \left[ \begin{matrix} 14 & 5 \\ 25 & 9 \end{matrix} \right]$, so applying $M$ twice sends $\left\langle 1, 0 \right\rangle$ to the first column, $\left\langle 14, 25 \right\rangle$.