Let $F: U\longrightarrow V$ and $G: V\longrightarrow W$ be two maps. The composite map $G\circ F: U\longrightarrow W$ is defined by
$$G\circ F(v)=G(F(v))$$
for each $v\in U$.

Example. Let $A$ be an $m\times n$ matrix and let $B$ be a $q\times m$ matrix. Let $L_A: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ be the linear map such that $L_AX=AX$ for each $X\in\mathbb{R}^n$, and let $L_B:\mathbb{R}^m\longrightarrow\mathbb{R}^m$ such that $L_BY=Y$ for each $Y\in\mathbb{R}^m$. Then
\begin{align*}
L_B\circ L_A(X)&=L_B(L_A(X))\\
&=L_B(L_A(X))\\
&=B(AX)\\
&=BA(X)\\
&=L_{BA}(X)
\end{align*}
for each $X\in\mathbb{R}^n$. Therefore, $L_B\circ L_A=L_{BA}$.

Example. Let $A$ be an $m\times n$ matrix, and let $L_A:\mathbb{R}^n\longrightarrow\mathbb{R}^m$ be the linear map such that $L_A(X)=AX$ for each $X\in\mathbb{R}^n$. Let $C$ be a vector in $\mathbb{R}^m$ and let $T_C:\mathbb{R}^m\longrightarrow\mathbb{R}^m$ be the translation by $C$
$$T_C(Y)=Y+C$$
for each $Y\in\mathbb{R}^m$. Then for each $X\in\mathbb{R}^n$,
\begin{align*}
T_C\circ L_A(X)&=T_C(AX)\\
&=AX+C.
\end{align*}

Example. Let $V$ be a vector space, and let $w$ be an element of $V$. Let $T_w: V\longrightarrow V$ be the translation by $w$ i.e.
$$T_w(v)=v+w$$
for each $v\in V$. Then
\begin{align*}
T_{w_1}\circ T_{w_2}(v)&=T_{w_1}(T_{w_2}(v))\\
&=T_{w_1}(v+w_2)\\
&=v+w_2+w_1
\end{align*}
for each $v\in V$. Therefore, $T_{w_1}\circ T_{w_2}=T_{w_1+w_2}$ i.e. the composite of translations is again a translation.

Remark. Note that translations are not linear. One easy way to see this is that $T_w(O)=O+w=w$. So $T_w$ is not linear unless $w=O$ in which case $T_w$ is the identity map.

Example. [Rotations] Let $\theta$ be a number, and $A(\theta)$ the matrix
$$A(\theta)=\begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix}.$$
Let $R_\theta:\mathbb{R}^2\longrightarrow\mathbb{R}^2$ be the rotation by angle $\theta$ i.e.
$$R_\theta(X)=A(\theta)X$$
for each $X\in\mathbb{R}^2$. Clearly, rotations are linear.
Now,
\begin{align*}
R_{\theta_1}\circ R_{\theta_2}(X)&=R_{\theta_1}(R_{\theta_2}(X))\\
&=R_{\theta_1}(A(\theta_2)X)\\
&=A(\theta_1)A(\theta_2)X\\
&=A(\theta_1+\theta_2)X\\
&=R_{\theta_1+\theta_2}.
\end{align*}

Theorem. Let $U,V,W$ be vector spaces. Let $F:U\longrightarrow V$ and $G:V\longrightarrow W$ be linear maps. Then the composite map $G\circ F:U\longrightarrow W$ is also linear.

Proof. The proof is straightforward and is left for an exercise.