Proof of Theorem IMILT

Choose bases $B=C=\set{\vectorlist{e}{n}}$ consisting of the standard unit vectors as a basis of $\complex{n}$ (Theorem SUVB) and build a matrix representation of $T$ relative to $B$ and $C$. Then \begin{align*} \vectrep{C}{\lt{T}{\vect{e}_i}} &=\vectrep{C}{A\vect{e}_i}\\ &=\vectrep{C}{\vect{A}_i}\\ &=\vect{A}_i \end{align*}

So then the matrix representation of $T$, relative to $B$ and $C$, is simply $\matrixrep{T}{B}{C}=A$. With this observation, the proof becomes a specialization of Theorem IMR, \begin{align*} T\text{ is invertible} \iff\matrixrep{T}{B}{C}\text{ is invertible} \iff A\text{ is invertible} \end{align*}