Composition of Transformations

Overview

Composing linear transformations means applying one transformation after another. This corresponds to matrix multiplication, with the order of matrices reversed from the order of application.

Definition

For transformations $S: U \to V$ and $T: V \to W$, the composition is:

$(T \circ S)(\mathbf{u}) = T(S(\mathbf{u}))$

First apply $S$, then apply $T$.

Composition is Linear

If $S$ and $T$ are linear, then $T \circ S$ is linear:

$(T \circ S)(c\mathbf{u} + d\mathbf{v}) = T(S(c\mathbf{u} + d\mathbf{v})) = T(cS(\mathbf{u}) + dS(\mathbf{v}))$

$= cT(S(\mathbf{u})) + dT(S(\mathbf{v})) = c(T \circ S)(\mathbf{u}) + d(T \circ S)(\mathbf{v})$

Matrix Representation

If $S$ has matrix $A$ and $T$ has matrix $B$:

$T \circ S \text{ has matrix } BA$

Note the order: matrices multiply in reverse order of application.

Why Reversed?

$(T \circ S)(\mathbf{x}) = T(S(\mathbf{x})) = B(A\mathbf{x}) = (BA)\mathbf{x}$

Example

$S$ is rotation by 90°, $T$ is scaling by 2:

$A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \quad \text{(rotation)}$

$B = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \quad \text{(scaling)}$

$T \circ S$ (first rotate, then scale):

$BA = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$

$S \circ T$ (first scale, then rotate):

$AB = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$

In this case, $AB = BA$ (scaling and rotation commute).

Non-Commutativity

In general, $T \circ S \neq S \circ T$.

Example

$S$ = reflection across x-axis, $T$ = rotation by 90°:

$A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

$BA = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

$AB = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}$

$BA \neq AB$: order matters!

Properties of Composition

Property	Formula
Associativity	$(T \circ S) \circ R = T \circ (S \circ R)$
Identity	$T \circ I = I \circ T = T$
Inverse	$T \circ T^{-1} = T^{-1} \circ T = I$ (if $T$ invertible)
Not commutative	$T \circ S \neq S \circ T$ in general

Inverse Transformation

If $T$ is invertible (bijective):

$T^{-1} \text{ exists and } T^{-1} \circ T = T \circ T^{-1} = I$

Matrix form:

$A^{-1}A = AA^{-1} = I$

Multiple Compositions

For $T_1, T_2, T_3$ with matrices $A_1, A_2, A_3$:

$T_3 \circ T_2 \circ T_1 \text{ has matrix } A_3 A_2 A_1$

Rightmost matrix corresponds to first transformation.

Application: Composite Transformations

Rotate then Scale

To rotate by $\theta$ then scale by $s$:

$\begin{bmatrix} s & 0 \\ 0 & s \end{bmatrix}\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} = \begin{bmatrix} s\cos\theta & -s\sin\theta \\ s\sin\theta & s\cos\theta \end{bmatrix}$

Reflect then Rotate

Reflect across x-axis then rotate by 90°:

$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

This is reflection across $y = x$.

Decomposing Transformations

Any transformation can often be decomposed:

• $A = QR$ (orthogonal × upper triangular)
• $A = LU$ (lower × upper triangular)
• $A = PDP^{-1}$ (diagonalization)
• $A = U\Sigma V^T$ (singular value decomposition)