Proof of Theorem VRLT

We will take a novel approach in this proof. We will construct another function, which we will easily determine is a linear transformation, and then show that this second function is really $\vectrepname{B}$ in disguise. Here we go.

Since $B$ is a basis, we can define $\ltdefn{T}{V}{\complex{n}}$ to be the unique linear transformation such that $\lt{T}{\vect{v}_i}=\vect{e}_i$, $1\leq i\leq n$, as guaranteed by Theorem LTDB, and where the $\vect{e}_i$ are the standard unit vectors (Definition SUV). Then suppose for an arbitrary $\vect{w}\in V$ we have, \begin{align*} \vectorentry{\lt{T}{\vect{w}}}{i} &= \vectorentry{\lt{T}{\sum_{j=1}^{n}\vectorentry{\vectrep{B}{\vect{w}}}{j}\vect{v}_j}}{i} &&\knowl{./knowls/definition.VR.knowl}{\text{Definition VR}}\\ &= \vectorentry{\sum_{j=1}^{n}\vectorentry{\vectrep{B}{\vect{w}}}{j}\lt{T}{\vect{v}_j}}{i} &&\knowl{./knowls/theorem.LTLC.knowl}{\text{Theorem LTLC}}\\ &= \vectorentry{\sum_{j=1}^{n}\vectorentry{\vectrep{B}{\vect{w}}}{j}\vect{e}_j}{i}\\ &= \sum_{j=1}^{n}\vectorentry{\vectorentry{\vectrep{B}{\vect{w}}}{j}\vect{e}_j}{i} &&\knowl{./knowls/definition.CVA.knowl}{\text{Definition CVA}}\\ &= \sum_{j=1}^{n}\vectorentry{\vectrep{B}{\vect{w}}}{j}\vectorentry{\vect{e}_j}{i} &&\knowl{./knowls/definition.CVSM.knowl}{\text{Definition CVSM}}\\ &= \vectorentry{\vectrep{B}{\vect{w}}}{i}\vectorentry{\vect{e}_i}{i} + \sum_{\substack{j=1\\j\neq i}}^{n}\vectorentry{\vectrep{B}{\vect{w}}}{j}\vectorentry{\vect{e}_j}{i} &&\knowl{./knowls/property.CC.knowl}{\text{Property CC}}\\ &= \vectorentry{\vectrep{B}{\vect{w}}}{i}\left(1\right) + \sum_{\substack{j=1\\j\neq i}}^{n}\vectorentry{\vectrep{B}{\vect{w}}}{j}\left(0\right) &&\knowl{./knowls/definition.SUV.knowl}{\text{Definition SUV}}\\ &= \vectorentry{\vectrep{B}{\vect{w}}}{i} \end{align*}

As column vectors, Definition CVE implies that $\lt{T}{\vect{w}}=\vectrep{B}{\vect{w}}$. Since $\vect{w}$ was an arbitrary element of $V$, as functions $T=\vectrepname{B}$. Now, since $T$ is known to be a linear transformation, it must follow that $\vectrepname{B}$ is also a linear transformation.