Proof of Theorem LTDB

To prove the existence of $T$, we construct a function and show that it is a linear transformation (Proof Technique C). Suppose $\vect{w}\in U$ is an arbitrary element of the domain. Then by Theorem VRRB there are unique scalars $\scalarlist{a}{n}$ such that \begin{equation*} \vect{w}=\lincombo{a}{u}{n} \end{equation*}

Then define the function $T$ by \begin{equation*} \lt{T}{\vect{w}}=\lincombo{a}{v}{n} \end{equation*}

It should be clear that $T$ behaves as required for $n$ inputs from $B$. Since the scalars provided by Theorem VRRB are unique, there is no ambiguity in this definition, and $T$ qualifies as a function with domain $U$ and codomain $V$ (i.e. $T$ is well-defined). But is $T$ a linear transformation as well?

Let $\vect{x}\in U$ be a second element of the domain, and suppose the scalars provided by Theorem VRRB (relative to $B$) are $\scalarlist{b}{n}$. Then \begin{align*} \lt{T}{\vect{w}+\vect{x}}&= \lt{T}{ a_1\vect{u}_1+ \cdots+ a_n\vect{u}_n+ b_1\vect{u}_1+ \cdots+ b_n\vect{u}_n }\\ &= \lt{T}{ \left(a_1+b_1\right)\vect{u}_1+ \cdots+ \left(a_n+b_n\right)\vect{u}_n } &&\knowl{./knowls/definition.VS.knowl}{\text{Definition VS}}\\ &= \left(a_1+b_1\right)\vect{v}_1+ \cdots+ \left(a_n+b_n\right)\vect{v}_n &&\text{Definition of $T$}\\ &= a_1\vect{v}_1+ \cdots+ a_n\vect{v}_n+ b_1\vect{v}_1+ \cdots+ b_n\vect{v}_n &&\knowl{./knowls/definition.VS.knowl}{\text{Definition VS}}\\ &=\lt{T}{\vect{w}}+\lt{T}{\vect{x}} \end{align*}

Let $\alpha\in\complexes$ be any scalar. Then \begin{align*} \lt{T}{\alpha\vect{w}}&= \lt{T}{\alpha\left(\lincombo{a}{u}{n}\right)}\\ &= \lt{T}{\lincombo{\alpha a}{u}{n}} &&\knowl{./knowls/definition.VS.knowl}{\text{Definition VS}}\\ &=\lincombo{\alpha a}{v}{n} &&\text{Definition of $T$}\\ &=\alpha\left(\lincombo{a}{v}{n}\right) &&\knowl{./knowls/definition.VS.knowl}{\text{Definition VS}}\\ &=\alpha\lt{T}{\vect{w}} \end{align*}

So by Definition LT, $T$ is a linear transformation.

Is $T$ unique (among all linear transformations that take the $\vect{u}_i$ to the $\vect{v}_i$)? Applying Proof Technique U, we posit the existence of a second linear transformation, $\ltdefn{S}{U}{V}$ such that $\lt{S}{\vect{u}_i}=\vect{v}_i$, $1\leq i\leq n$. Again, let $\vect{w}\in U$ represent an arbitrary element of $U$ and let $\scalarlist{a}{n}$ be the scalars provided by Theorem VRRB (relative to $B$). We have, \begin{align*} \lt{T}{\vect{w}}&= \lt{T}{\lincombo{a}{u}{n}} &&\knowl{./knowls/theorem.VRRB.knowl}{\text{Theorem VRRB}}\\ &= a_1\lt{T}{\vect{u}_1}+ a_2\lt{T}{\vect{u}_2}+ a_3\lt{T}{\vect{u}_3}+ \cdots+ a_n\lt{T}{\vect{u}_n} &&\knowl{./knowls/theorem.LTLC.knowl}{\text{Theorem LTLC}}\\ &= a_1\vect{v}_1+ a_2\vect{v}_2+ a_3\vect{v}_3+ \cdots+ a_n\vect{v}_n &&\text{Definition of $T$}\\ &= a_1\lt{S}{\vect{u}_1}+ a_2\lt{S}{\vect{u}_2}+ a_3\lt{S}{\vect{u}_3}+ \cdots+ a_n\lt{S}{\vect{u}_n} &&\text{Definition of $S$}\\ &= \lt{S}{\lincombo{a}{u}{n}} &&\knowl{./knowls/theorem.LTLC.knowl}{\text{Theorem LTLC}}\\ &= \lt{S}{\vect{w}} &&\knowl{./knowls/theorem.VRRB.knowl}{\text{Theorem VRRB}} \end{align*}

So the output of $T$ and $S$ agree on every input, which means they are equal as functions, $T=S$. So $T$ is unique.