Proof of Theorem CLTLT

We simply check the defining properties of a linear transformation (Definition LT). \begin{align*} \lt{(\compose{S}{T})}{\vect{x}+\vect{y}} &=\lt{S}{\lt{T}{\vect{x}+\vect{y}}}&&\knowl{./knowls/definition.LTC.knowl}{\text{Definition LTC}}\\ &=\lt{S}{\lt{T}{\vect{x}}+\lt{T}{\vect{y}}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\text{ for }T\\ &=\lt{S}{\lt{T}{\vect{x}}}+\lt{S}{\lt{T}{\vect{y}}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\text{ for }S\\ &=\lt{(\compose{S}{T})}{\vect{x}}+\lt{(\compose{S}{T})}{\vect{y}}&&\knowl{./knowls/definition.LTC.knowl}{\text{Definition LTC}} \end{align*} and \begin{align*} \lt{(\compose{S}{T})}{\alpha\vect{x}} &=\lt{S}{\lt{T}{\alpha\vect{x}}}&&\knowl{./knowls/definition.LTC.knowl}{\text{Definition LTC}}\\ &=\lt{S}{\alpha\lt{T}{\vect{x}}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\text{ for }T\\ &=\alpha\lt{S}{\lt{T}{\vect{x}}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\text{ for }S\\ &=\alpha\lt{(\compose{S}{T})}{\vect{x}}&&\knowl{./knowls/definition.LTC.knowl}{\text{Definition LTC}} \end{align*}