Proof of Theorem CSS

(⇒) Suppose $\vect{u}\in\spn{\set{\vectorlist{u}{k}}}$. Then we know there are scalars, $\scalarlist{a}{k}$, such that \begin{equation*} \vect{u}=\lincombo{a}{u}{k} \end{equation*}

Then, by Theorem LTLC, \begin{align*} \vectrep{B}{\vect{u}}&=\vectrep{B}{\lincombo{a}{u}{k}}\\ &=a_1\vectrep{B}{\vect{u}_1}+a_2\vectrep{B}{\vect{u}_2}+a_3\vectrep{B}{\vect{u}_3}+\cdots+a_k\vectrep{B}{\vect{u}_k} \end{align*} which says that $\vectrep{B}{\vect{u}}\in\spn{\set{\vectrep{B}{\vect{u}_1},\,\vectrep{B}{\vect{u}_2},\,\vectrep{B}{\vect{u}_3},\,\ldots,\,\vectrep{B}{\vect{u}_k}}}$.

(⇐) Suppose that $\vectrep{B}{\vect{u}}\in\spn{\set{\vectrep{B}{\vect{u}_1},\,\vectrep{B}{\vect{u}_2},\,\vectrep{B}{\vect{u}_3},\,\ldots,\,\vectrep{B}{\vect{u}_k}}}$. Then there are scalars $\scalarlist{b}{k}$ such that \begin{equation*} \vectrep{B}{\vect{u}}=b_1\vectrep{B}{\vect{u}_1}+b_2\vectrep{B}{\vect{u}_2}+b_3\vectrep{B}{\vect{u}_3}+\cdots+b_k\vectrep{B}{\vect{u}_k} \end{equation*}

Recall that $\vectrepname{B}$ is invertible (Theorem VRILT), so \begin{align*} \vect{u}&=\lt{I_U}{\vect{u}}&&\knowl{./knowls/definition.IDLT.knowl}{\text{Definition IDLT}}\\ &=\lt{\left(\compose{\ltinverse{\vectrepname{B}}}{\vectrepname{B}}\right)}{\vect{u}}&&\knowl{./knowls/definition.IVLT.knowl}{\text{Definition IVLT}}\\ &=\lt{\ltinverse{\vectrepname{B}}}{\lt{\vectrepname{B}}{\vect{u}}}&&\knowl{./knowls/definition.LTC.knowl}{\text{Definition LTC}}\\ &=\lt{\ltinverse{\vectrepname{B}}}{b_1\vectrep{B}{\vect{u}_1}+b_2\vectrep{B}{\vect{u}_2}+\cdots+b_k\vectrep{B}{\vect{u}_k}}\\ &=b_1\lt{\ltinverse{\vectrepname{B}}}{\vectrep{B}{\vect{u}_1}}+b_2\lt{\ltinverse{\vectrepname{B}}}{\vectrep{B}{\vect{u}_2}} +\cdots+b_k\lt{\ltinverse{\vectrepname{B}}}{\vectrep{B}{\vect{u}_k}}&&\knowl{./knowls/theorem.LTLC.knowl}{\text{Theorem LTLC}}\\ &=b_1\lt{I_U}{\vect{u}_1}+b_2\lt{I_U}{\vect{u}_2}+\cdots+b_k\lt{I_U}{\vect{u}_k}&&\knowl{./knowls/definition.IVLT.knowl}{\text{Definition IVLT}}\\ &=\lincombo{b}{u}{k}&&\knowl{./knowls/definition.IDLT.knowl}{\text{Definition IDLT}} \end{align*} which says that $\vect{u}\in\spn{\set{\vectorlist{u}{k}}}$.