Proof of Theorem VRRB

That $\vect{w}$ can be written as a linear combination of the vectors in $B$ follows from the spanning property of the set (Definition SSVS). This is good, but not the meat of this theorem. We now know that for any choice of the vector $\vect{w}$ there exist some scalars that will create $\vect{w}$ as a linear combination of the basis vectors. The real question is: Is there more than one way to write $\vect{w}$ as a linear combination of $\{\vectorlist{v}{m}\}$? Are the scalars $a_1,\,a_2,\,a_3,\,\ldots,\,a_m$ unique? (Proof Technique U)

Assume there are two different linear combinations of $\{\vectorlist{v}{m}\}$ that equal the vector $\vect{w}$. In other words there exist scalars $a_1,\,a_2,\,a_3,\,\ldots,\,a_m$ and $b_1,\,b_2,\,b_3,\,\ldots,\,b_m$ so that \begin{align*} \vect{w}&=\lincombo{a}{v}{m}\\ \vect{w}&=\lincombo{b}{v}{m}. \end{align*}

Then notice that \begin{align*} \zerovector &=\vect{w}+(\vect{-w})&&\knowl{./knowls/property.AI.knowl}{\text{Property AI}}\\ &=\vect{w}+(-1)\vect{w}&&\knowl{./knowls/theorem.AISM.knowl}{\text{Theorem AISM}}\\ &=(\lincombo{a}{v}{m})+\\ &\quad\quad(-1)(\lincombo{b}{v}{m})\\ &=(\lincombo{a}{v}{m})+\\ &\quad\quad (-b_1\vect{v}_1-b_2\vect{v}_2-b_3\vect{v}_3-\ldots-b_m\vect{v}_m)&&\knowl{./knowls/property.DVA.knowl}{\text{Property DVA}}\\ &=(a_1-b_1)\vect{v_1}+(a_2-b_2)\vect{v_2}+(a_3-b_3)\vect{v_3}+\\ &\quad\quad\cdots+(a_m-b_m)\vect{v_m}&&\knowl{./knowls/property.C.knowl}{\text{Property C}}, \knowl{./knowls/property.DSA.knowl}{\text{Property DSA}} \end{align*}

But this is a relation of linear dependence on a linearly independent set of vectors (Definition RLD)! Now we are using the other assumption about $B$, that $\{\vectorlist{v}{m}\}$ is a linearly independent set. So by Definition LI it must happen that the scalars are all zero. That is, \begin{align*} (a_1-b_1)&=0&(a_2-b_2)&=0&(a_3-b_3)&=0&\ldots&&(a_m-b_m)&=0\\ a_1&=b_1&a_2&=b_2&a_3&=b_3&\ldots&&a_m&=b_m. \end{align*}

And so we find that the scalars are unique.