Proof of Theorem SLTLT

We simply check the defining properties of a linear transformation (Definition LT). This is a good place to consistently ask yourself which objects are being combined with which operations. \begin{align*} \lt{(T+S)}{\vect{x}+\vect{y}}&= \lt{T}{\vect{x}+\vect{y}}+\lt{S}{\vect{x}+\vect{y}}&&\knowl{./knowls/definition.LTA.knowl}{\text{Definition LTA}}\\ &=\lt{T}{\vect{x}}+\lt{T}{\vect{y}}+\lt{S}{\vect{x}}+\lt{S}{\vect{y}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\lt{T}{\vect{x}}+\lt{S}{\vect{x}}+\lt{T}{\vect{y}}+\lt{S}{\vect{y}}&&\knowl{./knowls/property.C.knowl}{\text{Property C}}\text{ in }V\\ &=\lt{(T+S)}{\vect{x}}+\lt{(T+S)}{\vect{y}}&&\knowl{./knowls/definition.LTA.knowl}{\text{Definition LTA}}\\ \end{align*} and \begin{align*} \lt{(T+S)}{\alpha\vect{x}}&= \lt{T}{\alpha\vect{x}}+\lt{S}{\alpha\vect{x}}&&\knowl{./knowls/definition.LTA.knowl}{\text{Definition LTA}}\\ &=\alpha\lt{T}{\vect{x}}+\alpha\lt{S}{\vect{x}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\alpha\left(\lt{T}{\vect{x}}+\lt{S}{\vect{x}}\right)&&\knowl{./knowls/property.DVA.knowl}{\text{Property DVA}}\text{ in }V\\ &=\alpha\lt{(T+S)}{\vect{x}}&&\knowl{./knowls/definition.LTA.knowl}{\text{Definition LTA}}\\ \end{align*}