Suppose that $\ltdefn{T}{\complex{2}}{\complex{4}}$ and $\ltdefn{S}{\complex{4}}{\complex{3}}$ are defined by \begin{align*} \lt{T}{\colvector{x_1\\x_2}}=\colvector{x_1+2x_2\\3x_1-4x_2\\5x_1+2x_2\\6x_1-3x_2} && \lt{S}{\colvector{x_1\\x_2\\x_3\\x_4}}= \colvector{2x_1-x_2+x_3-x_4\\5x_1-3x_2+8x_3-2x_4\\-4x_1+3x_2-4x_3+5x_4} \end{align*}
Then by Definition LTC \begin{align*} \lt{(\compose{S}{T})}{\colvector{x_1\\x_2}}&= \lt{S}{\lt{T}{\colvector{x_1\\x_2}}} =\lt{S}{\colvector{x_1+2x_2\\3x_1-4x_2\\5x_1+2x_2\\6x_1-3x_2}}\\ &=\colvector{ 2(x_1+2x_2)-(3x_1-4x_2)+(5x_1+2x_2)-(6x_1-3x_2)\\ 5(x_1+2x_2)-3(3x_1-4x_2)+8(5x_1+2x_2)-2(6x_1-3x_2)\\ -4(x_1+2x_2)+3(3x_1-4x_2)-4(5x_1+2x_2)+5(6x_1-3x_2) }\\ &=\colvector{ -2x_1+13x_2\\ 24x_1+44x_2\\ 15x_1-43x_2 } \end{align*} and by Theorem CLTLT $\compose{S}{T}$ is a linear transformation from $\complex{2}$ to $\complex{3}$.