Proof of Theorem ILTB

(⇒) Assume $T$ is injective. Since $B$ is a basis, we know $B$ is linearly independent (Definition B). Then Theorem ILTLI says that $C$ is a linearly independent subset of $V$.

(⇐) Assume that $C$ is linearly independent. To establish that $T$ is injective, we will show that the kernel of $T$ is trivial (Theorem KILT). Suppose that $\vect{u}\in\krn{T}$. As an element of $U$, we can write $\vect{u}$ as a linear combination of the basis vectors in $B$ (uniquely). So there are are scalars, $\scalarlist{a}{m}$, such that \begin{equation*} \vect{u}=\lincombo{a}{u}{m} \end{equation*}

Then, \begin{align*} \zerovector &=\lt{T}{\vect{u}} &&\knowl{./knowls/definition.KLT.knowl}{\text{Definition KLT}}\\ &=\lt{T}{\lincombo{a}{u}{m}} &&\knowl{./knowls/definition.SSVS.knowl}{\text{Definition SSVS}}\\ &=a_1\lt{T}{\vect{u}_1}+a_2\lt{T}{\vect{u}_2}+a_3\lt{T}{\vect{u}_3}+\cdots+a_m\lt{T}{\vect{u}_m} &&\knowl{./knowls/theorem.LTLC.knowl}{\text{Theorem LTLC}} \end{align*}

This is a relation of linear dependence (Definition RLD) on the linearly independent set $C$, so the scalars are all zero: $a_1=a_2=a_3=\cdots=a_m=0$. Then \begin{align*} \vect{u}&=\lincombo{a}{u}{m}\\ &=0\vect{u}_1+0\vect{u}_2+0\vect{u}_3+\cdots+0\vect{u}_m&&\knowl{./knowls/theorem.ZSSM.knowl}{\text{Theorem ZSSM}}\\ &=\zerovector+\zerovector+\zerovector+\cdots+\zerovector&&\knowl{./knowls/theorem.ZSSM.knowl}{\text{Theorem ZSSM}}\\ &=\zerovector&&\knowl{./knowls/property.Z.knowl}{\text{Property Z}} \end{align*}

Since $\vect{u}$ was chosen as an arbitrary vector from $\krn{T}$, we have $\krn{T}=\set{\zerovector}$ and Theorem KILT tells us that $T$ is injective.