Proof of Theorem VRI

We will appeal to Theorem KILT. 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 $\vect{u}\in\krn{\vectrepname{B}}$. We write $\vect{u}$ as a linear combination of the vectors in the basis $B$ where the scalars are the components of the vector representation, $\lt{\vectrepname{B}}{\vect{u}}$. \begin{align*} \vect{u} &= \vectorentry{\lt{\vectrepname{B}}{\vect{u}}}{1}\vect{u}_1+ \vectorentry{\lt{\vectrepname{B}}{\vect{u}}}{2}\vect{u}_2+ \cdots+ \vectorentry{\lt{\vectrepname{B}}{\vect{u}}}{n}\vect{u}_n &&\knowl{./knowls/definition.VR.knowl}{\text{Definition VR}}\\ &= \vectorentry{\zerovector}{1}\vect{u}_1+ \vectorentry{\zerovector}{2}\vect{u}_2+ \cdots+ \vectorentry{\zerovector}{n}\vect{u}_n &&\knowl{./knowls/definition.KLT.knowl}{\text{Definition KLT}}\\ &= 0\vect{u}_1+ 0\vect{u}_2+ \cdots+ 0\vect{u}_n &&\knowl{./knowls/definition.ZCV.knowl}{\text{Definition ZCV}}\\ &=\zerovector+\zerovector+\cdots+\zerovector &&\knowl{./knowls/theorem.ZSSM.knowl}{\text{Theorem ZSSM}}\\ &=\zerovector &&\knowl{./knowls/property.Z.knowl}{\text{Property Z}} \end{align*}

Thus an arbitrary vector, $\vect{u}$, from the kernel ,$\krn{\vectrepname{B}}$, must equal the zero vector of $U$. So $\krn{\vectrepname{B}}=\set{\zerovector}$ and by Theorem KILT, $\vectrepname{B}$ is injective.