Proof of Theorem KILT

(⇒) We assume $T$ is injective and we need to establish that two sets are equal (Definition SE). Since the kernel is a subspace (Theorem KLTS), $\set{\zerovector}\subseteq\krn{T}$. To establish the opposite inclusion, suppose $\vect{x}\in\krn{T}$. \begin{align*} \lt{T}{\vect{x}} &=\zerovector&&\knowl{./knowls/definition.KLT.knowl}{\text{Definition KLT}}\\ &=\lt{T}{\zerovector} &&\knowl{./knowls/theorem.LTTZZ.knowl}{\text{Theorem LTTZZ}} \end{align*}

We can apply Definition ILT to conclude that $\vect{x}=\zerovector$. Therefore $\krn{T}\subseteq\set{\zerovector}$ and by Definition SE, $\krn{T}=\set{\zerovector}$.

(⇐) To establish that $T$ is injective, appeal to Definition ILT and begin with the assumption that $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$. Then \begin{align*} \lt{T}{\vect{x}-\vect{y}} &=\lt{T}{\vect{x}}-\lt{T}{\vect{y}} &&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\zerovector &&\text{Hypothesis} \end{align*}

So $\vect{x}-\vect{y}\in\krn{T}$ by Definition KLT and with the hypothesis that the kernel is trivial we conclude that $\vect{x}-\vect{y}=\zerovector$. Then \begin{align*} \vect{y} =\vect{y}+\zerovector =\vect{y}+\left(\vect{x}-\vect{y}\right) =\vect{x} \end{align*} thus establishing that $T$ is injective by Definition ILT.