Proof of Theorem ILTLT

We work through verifying Definition LT for $\ltinverse{T}$, using the fact that $T$ is a linear transformation to obtain the second equality in each half of the proof. To this end, suppose $\vect{x},\,\vect{y}\in V$ and $\alpha\in\complex{\null}$. \begin{align*} \lt{\ltinverse{T}}{\vect{x}+\vect{y}}&= \lt{\ltinverse{T}}{\lt{T}{\lt{\ltinverse{T}}{\vect{x}}}+\lt{T}{\lt{\ltinverse{T}}{\vect{y}}}} &&\knowl{./knowls/definition.IVLT.knowl}{\text{Definition IVLT}}\\ &=\lt{\ltinverse{T}}{\lt{T}{\lt{\ltinverse{T}}{\vect{x}}+\lt{\ltinverse{T}}{\vect{y}}}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\lt{\ltinverse{T}}{\vect{x}}+\lt{\ltinverse{T}}{\vect{y}} &&\knowl{./knowls/definition.IVLT.knowl}{\text{Definition IVLT}} \end{align*}

Now check the second defining property of a linear transformation for $\ltinverse{T}$, \begin{align*} \lt{\ltinverse{T}}{\alpha\vect{x}}&= \lt{\ltinverse{T}}{\alpha\lt{T}{\lt{\ltinverse{T}}{\vect{x}}}} &&\knowl{./knowls/definition.IVLT.knowl}{\text{Definition IVLT}}\\ &=\lt{\ltinverse{T}}{\lt{T}{\alpha\lt{\ltinverse{T}}{\vect{x}}}} &&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\alpha\lt{\ltinverse{T}}{\vect{x}} &&\knowl{./knowls/definition.IVLT.knowl}{\text{Definition IVLT}} \end{align*}

So $\ltinverse{T}$ fulfills the requirements of Definition LT and is therefore a linear transformation.