4.4 Composition of Transformations
Introduction
“First rotate 90 degrees to the right, then stretch by a factor of 2 in the x-direction.” “First stretch by a factor of 2 in the x-direction, then rotate 90 degrees to the right.”
Do these two sequences produce the same result? No, they do not.
Matrix multiplication is not commutative – swapping the order generally changes the outcome. This is not a bug; it is a fundamental property of how transformations work. When you program a robot arm or design keyframes in a 3D animation, getting the order wrong means getting a completely different motion. Understanding composition of transformations is one of the core skills of linear algebra.
Definition of Composition
Given transformations $T_1 : \mathbb{R}^n \to \mathbb{R}^m$ and $T_2 : \mathbb{R}^m \to \mathbb{R}^k$, the composite transformation applies them in sequence:
\[T_2 \circ T_1 : \mathbb{R}^n \to \mathbb{R}^k, \quad (T_2 \circ T_1)(\vec{x}) = T_2(T_1(\vec{x}))\]$T_1$ is applied first, and $T_2$ is applied second.
Matrix Representation
If $T_1(\vec{x}) = A\vec{x}$ and $T_2(\vec{x}) = B\vec{x}$, then
\[(T_2 \circ T_1)(\vec{x}) = B(A\vec{x}) = (BA)\vec{x}\]The matrix of the composite transformation is $BA$ – the matrix of the transformation applied last goes on the left.
This convention may feel backwards at first, but it follows naturally from how matrix-vector multiplication works: you read from right to left.
Order Matters: $AB \neq BA$
Example: Let $R$ be the 45-degree rotation matrix and $M$ the reflection across the x-axis.
\[R = R_{45°} = \begin{pmatrix} \tfrac{\sqrt{2}}{2} & -\tfrac{\sqrt{2}}{2} \\ \tfrac{\sqrt{2}}{2} & \tfrac{\sqrt{2}}{2} \end{pmatrix}, \quad M = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\]Case 1 – Rotate first, then reflect: The composite matrix is $MR$.
\[MR = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \tfrac{\sqrt{2}}{2} & -\tfrac{\sqrt{2}}{2} \\ \tfrac{\sqrt{2}}{2} & \tfrac{\sqrt{2}}{2} \end{pmatrix} = \begin{pmatrix} \tfrac{\sqrt{2}}{2} & -\tfrac{\sqrt{2}}{2} \\ -\tfrac{\sqrt{2}}{2} & -\tfrac{\sqrt{2}}{2} \end{pmatrix}\]Case 2 – Reflect first, then rotate: The composite matrix is $RM$.
\[RM = \begin{pmatrix} \tfrac{\sqrt{2}}{2} & -\tfrac{\sqrt{2}}{2} \\ \tfrac{\sqrt{2}}{2} & \tfrac{\sqrt{2}}{2} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} \tfrac{\sqrt{2}}{2} & \tfrac{\sqrt{2}}{2} \\ \tfrac{\sqrt{2}}{2} & -\tfrac{\sqrt{2}}{2} \end{pmatrix}\]$MR \neq RM$ – different order, different result.
Inverse Transformations
The inverse of a linear transformation $T(\vec{x}) = A\vec{x}$ is
\[T^{-1}(\vec{y}) = A^{-1}\vec{y}\]The matrix of the inverse transformation is simply the inverse of the original matrix.
For the inverse to exist, $A$ must be invertible: $\det(A) \neq 0$.
Inverse of a composition:
\[(T_2 \circ T_1)^{-1} = T_1^{-1} \circ T_2^{-1}\]In matrix form:
\[(BA)^{-1} = A^{-1}B^{-1}\]The order reverses – just like socks and shoes: you put on socks first and shoes second, but you take off shoes first and socks second.
Worked Example: 45-Degree Rotation Followed by x-Axis Reflection
Apply both transformations to the vector $\vec{x} = (1, 0)^\top$.
Step 1: Rotate by 45 degrees.
\[R_{45°}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}\tfrac{\sqrt{2}}{2}\\\tfrac{\sqrt{2}}{2}\end{pmatrix}\]Step 2: Reflect across the x-axis.
\[M_x\begin{pmatrix}\tfrac{\sqrt{2}}{2}\\\tfrac{\sqrt{2}}{2}\end{pmatrix} = \begin{pmatrix}\tfrac{\sqrt{2}}{2}\\-\tfrac{\sqrt{2}}{2}\end{pmatrix}\]Computing with the composite matrix $M_x R_{45°}$ in one step produces the same result.
Key Takeaways
| Concept | Description / Formula |
|---|---|
| Composition | $(T_2 \circ T_1)(\vec{x}) = T_2(T_1(\vec{x}))$ |
| Composite matrix | $T_1 \to A$, $T_2 \to B$ gives composite matrix $BA$ |
| Order convention | The matrix of the later transformation goes on the left |
| Non-commutativity | In general, $AB \neq BA$ |
| Inverse transformation | Matrix of $T^{-1}$ is $A^{-1}$ |
| Inverse of a composition | $(BA)^{-1} = A^{-1}B^{-1}$ (order reverses) |
| Existence of inverse | Requires $\det(A) \neq 0$ |
In the next post, we explore vector spaces – subspaces and span.