Proof of Theorem VRS

We will appeal to Theorem RSLT. Suppose $U$ is a vector space of dimension $n$, so vector representation is $\ltdefn{\vectrepname{B}}{U}{\complex{n}}$. Let $B=\set{\vectorlist{u}{n}}$ be the basis of $U$ used in the definition of $\vectrepname{B}$. Suppose $\vect{v}\in\complex{n}$. Define the vector $\vect{u}$ by \begin{equation*} \vect{u} = \vectorentry{\vect{v}}{1}\vect{u}_1+ \vectorentry{\vect{v}}{2}\vect{u}_2+ \vectorentry{\vect{v}}{3}\vect{u}_3+ \cdots+ \vectorentry{\vect{v}}{n}\vect{u}_n \end{equation*}

Then for $1\leq i\leq n$, by Definition VR, \begin{align*} \vectorentry{\vectrep{B}{\vect{u}}}{i} &=\vectorentry{\vectrep{B}{ \vectorentry{\vect{v}}{1}\vect{u}_1+ \vectorentry{\vect{v}}{2}\vect{u}_2+ \vectorentry{\vect{v}}{3}\vect{u}_3+ \cdots+ \vectorentry{\vect{v}}{n}\vect{u}_n }}{i} =\vectorentry{\vect{v}}{i}&&\text{} \end{align*} so the entries of vectors $\vectrep{B}{\vect{u}}$ and $\vect{v}$ are equal and Definition CVE yields the vector equality $\vectrep{B}{\vect{u}}=\vect{v}$. This demonstrates that $\vect{v}\in\rng{\vectrepname{B}}$, so $\complex{n}\subseteq\rng{\vectrepname{B}}$. Since $\rng{\vectrepname{B}}\subseteq\complex{n}$ by Definition RLT, we have $\rng{\vectrepname{B}}=\complex{n}$ and Theorem RSLT says $\vectrepname{B}$ is surjective.