Proof of Theorem RLTS

We can apply the three-part test of Theorem TSS. First, $\zerovector_U\in U$ and $\lt{T}{\zerovector_U}=\zerovector_V$ by Theorem LTTZZ, so $\zerovector_V\in\rng{T}$ and we know that the range is nonempty.

Suppose we assume that $\vect{x},\,\vect{y}\in\rng{T}$. Is $\vect{x}+\vect{y}\in\rng{T}$? If $\vect{x},\,\vect{y}\in\rng{T}$ then we know there are vectors $\vect{w},\,\vect{z}\in U$ such that $\lt{T}{\vect{w}}=\vect{x}$ and $\lt{T}{\vect{z}}=\vect{y}$. Because $U$ is a vector space, additive closure (Property AC) implies that $\vect{w}+\vect{z}\in U$.

Then \begin{align*} \lt{T}{\vect{w}+\vect{z}}&=\lt{T}{\vect{w}}+\lt{T}{\vect{z}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\vect{x}+\vect{y}&&\text{Definition of $\vect{w}$ and $\vect{z}$} \end{align*}

So we have found an input, $\vect{w}+\vect{z}$, which when fed into $T$ creates $\vect{x}+\vect{y}$ as an output. This qualifies $\vect{x}+\vect{y}$ for membership in $\rng{T}$. So we have additive closure.

Suppose we assume that $\alpha\in\complex{\null}$ and $\vect{x}\in\rng{T}$. Is $\alpha\vect{x}\in\rng{T}$? If $\vect{x}\in\rng{T}$, then there is a vector $\vect{w}\in U$ such that $\lt{T}{\vect{w}}=\vect{x}$. Because $U$ is a vector space, scalar closure implies that $\alpha\vect{w}\in U$. Then \begin{align*} \lt{T}{\alpha\vect{w}}&=\alpha\lt{T}{\vect{w}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\alpha\vect{x}&&\text{Definition of $\vect{w}$} \end{align*}

So we have found an input ($\alpha\vect{w}$) which when fed into $T$ creates $\alpha\vect{x}$ as an output. This qualifies $\alpha\vect{x}$ for membership in $\rng{T}$. So we have scalar closure and Theorem TSS tells us that $\rng{T}$ is a subspace of $V$.