# Compositions of Linear Transformations

Suppose that $T_A: \mathbb{R}^n \to \mathbb{R}^k$ and $T_B: \mathbb{R}^k \to \mathbb{R}^m$. If we wanted to take a vector $\vec{x} \in \mathbb{R}^n$, transform it under $T_A$ and then take the image and transform it under $T_B$, our result would be the combined effect of first applying $T_A$ and then $T_B$. We say call this combined transformation the composition of $T_B$ with $T_A$ and mathematically we denote it as $T_B(T_A(x))$ or $(T_B \circ T_A)(x)$. Furthermore:

(1)Thus we say this transformation is *multiplication by the product $BA$*, and the standard matrix for this composition is $BA$. We thus obtain the following relationship between standard matrices:

Furthermore, we are not restricting to only composing 2 transformations together. In fact, we can compose as many transformations as we like, and for $n$ transformations composed together, that is $(T_n \circ T_{n-1} \circ ... \circ T_2 \circ T_1)(x) = T_n(T_{n-1}(...(T_2(T_1(x)))))$, the same rules apply regarding standard matrices, that is:

(3)## Example 1

**Find the standard matrix for the linear operator $T: \mathbb{R}^3 \to \mathbb{R}^3$ that first reflects a vector $\vec{x}$ about the $xy$-plane and then rotates it counterclockwise around the positive $z$-axis by $30^\circ$.**

Let $T_A: \mathbb{R}^3 \to \mathbb{R}^3$ be the linear operator that reflects a vector about the $xy$-plane. The standard matrix for this transformation is given as $A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1 \end{bmatrix}$.

Also let $T_B: \mathbb{R}^3 \to \mathbb{R}^3$ be the linear operator $T: \mathbb{R}^3 \to \mathbb{R}^3$ that rotates a vector counterclockwise around the positive $z$-axis by $30^\circ$. The standard matrix for this transformation is given as $B = \begin{bmatrix} \cos 30^\circ& -\sin 30^\circ & 0\\ \sin 30^\circ & \cos 30^\circ & 0\\ 0 & 0 & 1 \end{bmatrix}$.

We now want to take the composition $(T_B \circ T_A)(x)$, and thus the standard matrix for this composition is given as [[ [T_B \circ T_A] = [T_B][T_A] $]], and thus:

(4)