FP1.6 - Transformations using matrices

A 2 by 2 matrix represents a linear transformation by multiplying each position column vector. The two columns of the matrix are the images of the coordinate basis vectors.

In particular, $\binom{1}{0}\mapsto\binom{a}{c}$ and $\binom{0}{1}\mapsto\binom{b}{d}$. Knowing those two images therefore determines the whole matrix.

If $A=\begin{pmatrix}2&1\\-1&3\end{pmatrix}$, then $$\begin{pmatrix}4\\-2\end{pmatrix}\mapsto\begin{pmatrix}2(4)+1(-2)\\-1(4)+3(-2)\end{pmatrix}=\begin{pmatrix}6\\-10\end{pmatrix}.$$

If $B$ acts first and $A$ acts second, then a vector follows $$\mathbf v\mapsto B\mathbf v\mapsto A(B\mathbf v)=(AB)\mathbf v.$$ Thus $AB$ represents $B$ followed by $A$: the right-hand factor acts first.

A standard transformation matrix is identified by what it does to coordinates, or equivalently by the images of the basis vectors in its columns. All transformations here are centred at the origin where a centre is needed.

A rotation of $210^\circ$ anticlockwise has $\cos210^\circ=-\sqrt3/2$ and $\sin210^\circ=-1/2$, so its matrix is $$\begin{pmatrix}-\frac{\sqrt3}{2}&\frac12\\-\frac12&-\frac{\sqrt3}{2}\end{pmatrix}.$$ Exact trigonometric values keep the transformation exact.

To identify an unfamiliar matrix, test its effect on $(1,0)$ and $(0,1)$ and compare both images with the table. A full description includes the transformation type, line or centre where relevant, direction and angle for a rotation, and scale factor for a stretch or enlargement.

A sequence of linear transformations is represented by a matrix product in reverse reading order: the matrix for the last transformation is placed on the left.

Let $R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ rotate $90^\circ$ anticlockwise, and let $S=\begin{pmatrix}2&0\\0&1\end{pmatrix}$ stretch parallel to the $x$-axis by factor $2$. Rotation followed by stretch has matrix $$SR=\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}0&-2\\1&0\end{pmatrix}.$$

For $\mathbf v=\binom{1}{2}$, the separate transformations give $\binom{1}{2}\mapsto\binom{-2}{1}\mapsto\binom{-4}{1}$. The product gives $SR\binom{1}{2}=\binom{-4}{1}$, confirming the order.

Repeating the same transformation $A$ exactly $n$ times gives $A^n$. For example, four successive quarter-turns give $R^4=I$, returning every vector to its original position.

A non-singular transformation can be undone by its inverse matrix. Its determinant also controls area: every finite region has its area multiplied by the absolute value of the determinant.

For $M=\begin{pmatrix}2&1\\0&3\end{pmatrix}$, $\det M=6$. Hence $$M^{-1}=\frac16\begin{pmatrix}3&-1\\0&2\end{pmatrix},$$ and a region of area $5$ is transformed to a region of area $|6|\times5=30$. Conversely, divide the image area by $6$ to recover the original area.

For invertible transformations, $(AB)^{-1}=B^{-1}A^{-1}$: the inverse first undoes the action performed last. This is the transformation meaning of the reverse-order inverse rule.