Proof of Theorem SLTB

(⇒) Assume $T$ is surjective. Since $B$ is a basis, we know $B$ is a spanning set of $U$ (Definition B). Then Theorem SSRLT says that $C$ spans $\rng{T}$. But the hypothesis that $T$ is surjective means $V=\rng{T}$ (Theorem RSLT), so $C$ spans $V$.

(⇐) Assume that $C$ spans $V$. To establish that $T$ is surjective, we will show that every element of $V$ is an output of $T$ for some input (Definition SLT). Suppose that $\vect{v}\in V$. As an element of $V$, we can write $\vect{v}$ as a linear combination of the spanning set $C$. So there are scalars, $\scalarlist{b}{m}$, such that \begin{equation*} \vect{v}=b_1\lt{T}{\vect{u}_1}+b_2\lt{T}{\vect{u}_2}+b_3\lt{T}{\vect{u}_3}+\cdots+b_m\lt{T}{\vect{u}_m} \end{equation*}

Now define the vector $\vect{u}\in U$ by \begin{equation*} \vect{u}=\lincombo{b}{u}{m} \end{equation*}

Then \begin{align*} \lt{T}{\vect{u}}&=\lt{T}{\lincombo{b}{u}{m}}\\ &=b_1\lt{T}{\vect{u}_1}+b_2\lt{T}{\vect{u}_2}+b_3\lt{T}{\vect{u}_3}+\cdots+b_m\lt{T}{\vect{u}_m}&&\knowl{./knowls/theorem.LTLC.knowl}{\text{Theorem LTLC}}\\ &=\vect{v} \end{align*}

So, given any choice of a vector $\vect{v}\in V$, we can design an input $\vect{u}\in U$ to produce $\vect{v}$ as an output of $T$. Thus, by Definition SLT, $T$ is surjective.